🧠:Okay, let's tackle this problem. So, we need to figure out how many matches can be made from a wooden cube that's 1 meter on each edge. Each match has specific dimensions: 5 cm long, 2 mm wide, and 2 mm high. And we can't lose any material when cutting, which is ideal but probably simplifies things. Hmm, let's start breaking this down.First, the cube has edges of 1 meter. Since the metric system is involved here, converting everything to the same unit would make sense. The matches are given in centimeters and millimeters, so maybe converting the cube's dimensions to centimeters first? Let me think. 1 meter is 100 centimeters. So, the cube is 100 cm x 100 cm x 100 cm. That's straightforward.But the match dimensions: length is 5 cm, which is fine, but the width and height are in millimeters. Let me convert those to centimeters as well for consistency. 2 mm is 0.2 cm, right? Because 1 cm is 10 mm. So, each match is 5 cm long, 0.2 cm wide, and 0.2 cm high. So, the volume of one match would be length × width × height. Let me calculate that. 5 cm * 0.2 cm * 0.2 cm = 5 * 0.04 = 0.2 cubic centimeters. So each match is 0.2 cm³.Now, the volume of the wooden cube. Since it's 100 cm on each edge, the volume is 100 cm * 100 cm * 100 cm = 1,000,000 cm³. That's 1 million cubic centimeters.If each match is 0.2 cm³, then dividing the cube's volume by the match volume should give the number of matches, right? So 1,000,000 cm³ / 0.2 cm³ per match. Let me compute that. 1,000,000 divided by 0.2 is the same as multiplying by 5, so 5,000,000 matches. But wait, that seems like a lot. Is there something I'm missing here?Wait, maybe I should check if the dimensions align in terms of cutting. Because even though the volume calculation gives a theoretical maximum, sometimes you have to consider how you can actually cut the cube into those shapes. If the cube is 100 cm on each side, and each match is 5 cm long, then along each edge of the cube, how many matches can we get?Let me think. If we're cutting the cube into matches, each match has a length of 5 cm. So along the 100 cm edge, we can divide 100 cm by 5 cm to get the number of matches along that dimension. 100 / 5 = 20. So, along the length, we can get 20 matches.But then the cross-sectional dimensions of the match are 2 mm by 2 mm. The cube's other edges are 100 cm, which is 1000 mm. So, converting the cube's dimensions to millimeters: 1000 mm x 1000 mm x 1000 mm. The match is 5 cm long, which is 50 mm, 2 mm wide, and 2 mm high. Wait, maybe converting everything to millimeters might be better here.Let me re-express the cube's dimensions in millimeters: 1 meter is 1000 mm per edge. So, cube is 1000 mm x 1000 mm x 1000 mm. Each match is 50 mm (length) x 2 mm (width) x 2 mm (height). So, to find how many matches fit along each axis:Along the length (let's say the x-axis), 1000 mm / 50 mm = 20. So 20 matches along the x-axis.Along the width (y-axis), 1000 mm / 2 mm = 500. So 500 along the y-axis.Similarly, along the height (z-axis), 1000 mm / 2 mm = 500. So 500 along the z-axis.Therefore, total number of matches would be 20 * 500 * 500 = 20 * 250,000 = 5,000,000. So that's 5 million matches. Wait, that's the same as the volume calculation. So maybe the volume approach was correct. But why?Because volume-wise, the cube is 1000^3 mm³ = 1,000,000,000 mm³. Each match is 50*2*2=200 mm³. So 1,000,000,000 / 200 = 5,000,000. Yep, same answer. So both methods agree. So perhaps the answer is 5 million matches.But let me just confirm once more. The edge of the cube is 1 meter, which is 100 cm or 1000 mm. If we cut the cube into 5 cm (50 mm) segments along one axis, then each segment is a 50 mm thick slab. Then, each slab can be cut into smaller pieces of 2 mm width and 2 mm height. So each slab is 1000 mm x 1000 mm x 50 mm. Wait, no, if we cut along the 1000 mm length into 50 mm pieces, each of those is 50 mm long. Then the remaining dimensions are 1000 mm each. So each 50 mm long piece can be divided into (1000/2) = 500 pieces along the width and (1000/2)=500 along the height. So per 50 mm segment, you get 500*500=250,000 matches. Then, since there are 20 such segments (1000/50=20), total matches are 20*250,000=5,000,000. Yep, that's the same result.Alternatively, if we think of the cube as divided into small match-sized units. Each match occupies a space of 50 mm x 2 mm x 2 mm. So along each axis, the number of divisions is 1000/50 = 20, 1000/2 = 500, 1000/2=500. Multiply them all together: 20*500*500=5,000,000. So that's consistent.Therefore, both the volume approach and the dimensional analysis approach give the same answer, which is reassuring.So, the conclusion is 5,000,000 matches. That's five million matches.But just to be absolutely thorough, is there a possibility that arranging the matches in a different orientation could yield more? For example, if we rotate the matches to use different dimensions? But since all dimensions are multiples and the cube is uniform, the orientation shouldn't matter. Because the cube is symmetrical in all directions. So whether we take the 50 mm along the x, y, or z-axis, the number of divisions would be the same. So rotating the match wouldn't give a better result. For example, if we tried to make the width or height of the match correspond to a different axis, but since the match's width and height are both 2 mm, and the cube's edges are 1000 mm, we can always divide each axis into 2 mm increments regardless of orientation. So, regardless of how we orient the match, the number should remain the same. So, the answer is indeed 5,000,000.Another way to think about it: The cube can be divided into small rectangular prisms each of size 50mm x 2mm x 2mm. The number of such prisms is (1000/50)*(1000/2)*(1000/2) = 20*500*500=5,000,000. So, that's another confirmation.Therefore, the answer must be 5,000,000 matches.

✅:To determine how many matches can be made from a 1-meter wooden cube, we analyze the dimensions and calculate as follows:1. Convert units for consistency: - Cube edge length: (1,text{m} = 100,text{cm} = 1000,text{mm}). - Match dimensions: (5,text{cm} = 50,text{mm}), (2,text{mm}), and (2,text{mm}).2. Volume-based approach: - Cube volume: (1000,text{mm} times 1000,text{mm} times 1000,text{mm} = 1{,}000{,}000{,}000,text{mm}^3). - Match volume: (50,text{mm} times 2,text{mm} times 2,text{mm} = 200,text{mm}^3). - Total matches: (frac{1{,}000{,}000{,}000,text{mm}^3}{200,text{mm}^3} = 5{,}000{,}000).3. Dimensional analysis: - Along the length ((50,text{mm})): (frac{1000,text{mm}}{50,text{mm}} = 20). - Along the width ((2,text{mm})): (frac{1000,text{mm}}{2,text{mm}} = 500). - Along the height ((2,text{mm})): (frac{1000,text{mm}}{2,text{mm}} = 500). - Total matches: (20 times 500 times 500 = 5{,}000{,}000).Both methods confirm the same result. Answer: (boxed{5000000}) matches can be made.

🔑:1. Convert the volume of the wooden cube from cubic meters to cubic centimeters: Given the edge length of the cube is (1 , mathrm{m}), we convert this to centimeters: [ 1 , mathrm{m} = 100 , mathrm{cm} ] The volume of the cube in cubic centimeters: [ text{Volume of the cube} = (100 , mathrm{cm})^3 = 100^3 , mathrm{cm}^3 = 1,000,000 , mathrm{cm}^3 ]2. Determine the dimensions and volume of one matchstick: The matchstick's dimensions are given as: - Length: (5 , mathrm{cm}) - Width: (2 , mathrm{mm} = 0.2 , mathrm{cm}) - Height: (2 , mathrm{mm} = 0.2 , mathrm{cm}) Calculate the volume of one matchstick: [ text{Volume of one matchstick} = 5 , mathrm{cm} times 0.2 , mathrm{cm} times 0.2 , mathrm{cm} = 5 times 0.2 times 0.2 , mathrm{cm}^3 = 0.2 , mathrm{cm}^3 ]3. Calculate the total number of matchsticks that can be obtained from the wooden cube: We divide the total volume of the cube by the volume of one matchstick: [ text{Number of matchsticks} = frac{1,000,000 , mathrm{cm}^3}{0.2 , mathrm{cm}^3} = frac{1,000,000}{0.2} ]4. Perform the division to find the number of matchsticks: [ frac{1,000,000}{0.2} = 1,000,000 times 5 = 5,000,000 ] Conclusion:[boxed{5,000,000}]

🧠:Alright, let's tackle this problem step by step. So, the question is: Given set A as (-2, 4) and set B as the set of real numbers x that satisfy the quadratic equation x² + a x + 4 = 0. We need to find the range of the real number a such that the number of non-empty subsets of A ∩ B is 1. First, let me parse the problem. The number of non-empty subsets of a set is given by 2ⁿ - 1, where n is the number of elements in the set. If the number of non-empty subsets is 1, that means the set itself has exactly one element because 2¹ - 1 = 1. Therefore, A ∩ B must be a singleton set, i.e., it contains exactly one element. So, our goal is to find all real numbers a for which the quadratic equation x² + a x + 4 = 0 has exactly one solution that lies in the interval (-2, 4).But wait, let me double-check that. If the intersection A ∩ B has exactly one element, then B must contain precisely one element that's within the interval (-2, 4). Since B is the solution set of the quadratic equation, the quadratic must have exactly one real root in (-2, 4). However, quadratics can have two roots, one root, or no real roots depending on the discriminant. So, we need to consider cases where the quadratic has exactly one root in (-2, 4), and possibly another root outside of this interval, or maybe even a repeated root within the interval. Hmm.But wait, the number of non-empty subsets being 1 implies that A ∩ B has exactly one element. So, regardless of how many roots the quadratic has, only one of them is inside (-2, 4). However, if the quadratic has a repeated root (i.e., discriminant zero), then that repeated root would be the only root. So, we have two possibilities:1. The quadratic has two distinct real roots, one of which is in (-2, 4), and the other is outside.2. The quadratic has exactly one real root (a repeated root) which is in (-2, 4).So, we need to find the values of a for which either case occurs. Let's structure this into steps.First, let's recall that a quadratic equation x² + a x + 4 = 0 has real roots if its discriminant is non-negative. The discriminant D = a² - 16. So, D ≥ 0 implies a² ≥ 16, so a ≤ -4 or a ≥ 4. If a² = 16, then the quadratic has a repeated root at x = -a/2. If a² > 16, then there are two distinct roots.But in our problem, we need the intersection A ∩ B to have exactly one element. So, for the case when the quadratic has a repeated root (a² = 16), we need to check if that root is inside (-2, 4). For the case when the quadratic has two distinct roots (a² > 16), we need exactly one of those roots to be in (-2, 4).So, let's break it down:Case 1: Quadratic has a repeated root (i.e., a² = 16). Then, the root is x = -a/2. We need this root to be in (-2, 4). Let's compute x for a = 4 and a = -4.If a = 4, then x = -4/2 = -2. But the interval is open, (-2, 4), so -2 is excluded. Therefore, x = -2 is not in A. Similarly, if a = -4, then x = -(-4)/2 = 2. 2 is in (-2, 4). So, when a = -4, the repeated root is 2, which is in A. Therefore, in this case, A ∩ B = {2}, which has exactly one element. So, a = -4 is a valid solution.But wait, when a = 4, the root is -2, which is not in A. So, A ∩ B would be empty, which is not what we want. So, a = 4 is invalid. Therefore, only a = -4 is valid in this case.Case 2: Quadratic has two distinct real roots (i.e., a² > 16). We need exactly one root in (-2, 4), and the other root outside. So, how do we ensure that exactly one root lies in the interval?One method is to use the concept that if a continuous function (here, f(x) = x² + a x + 4) changes sign over an interval, then there is at least one root in that interval. However, since we need exactly one root in the interval, we can use the following approach: either one root is inside and the other is outside, or both roots are at the endpoints (but since the interval is open, endpoints are excluded). But since the quadratic can have at most two roots, the only way for exactly one root to be inside is that one root is in (-2, 4) and the other is ≤ -2 or ≥ 4.Alternatively, we can use the following conditions:1. f(-2) and f(4) have opposite signs, which would imply at least one root in (-2, 4). But since we need exactly one root, combined with the discriminant being positive, this might not be sufficient. Wait, perhaps another approach.Alternatively, the quadratic equation x² + a x + 4 = 0 can be considered. Let me denote the roots as α and β. We need exactly one of α or β to be in (-2, 4). To translate this into conditions on a, we can use the following reasoning:Suppose one root is in (-2, 4) and the other is outside. Then, the following must hold:Either:- α ∈ (-2, 4) and β ∉ (-2, 4)Or:- β ∈ (-2, 4) and α ∉ (-2, 4)But since α and β are symmetric in terms of the equation, we can just consider one case and combine the conditions.Alternatively, using the location of roots theorems. For a quadratic equation ax² + bx + c = 0 (here, a=1, b=a, c=4), the conditions for exactly one root in an interval (m, n) can be given by f(m) * f(n) < 0, which would imply an odd number of roots in the interval. But since quadratics have at most two roots, it would imply exactly one root. However, this is only valid if the function actually crosses the x-axis, so combined with discriminant conditions.Wait, but in our case, if f(-2) * f(4) < 0, then there is exactly one root in (-2, 4). But since the quadratic can have two roots, if f(-2) and f(4) have opposite signs, there must be exactly one root in (-2, 4). However, this is only true if there are two real roots. Wait, no. Let me think. If the quadratic has two real roots, and f(-2) and f(4) have opposite signs, then there is one root in (-2, 4) and the other outside. If the quadratic has two real roots and f(-2) and f(4) have the same sign, then either both roots are in (-2, 4) or both are outside. So, to ensure exactly one root in (-2, 4), we need two real roots (so discriminant ≥ 0, i.e., a² ≥ 16) and f(-2) * f(4) < 0.Wait, but if the quadratic has a repeated root (discriminant = 0), then f(-2) * f(4) < 0 would not hold because the function touches the x-axis at one point. However, in the case of a repeated root inside the interval, we need to check if that root is inside. But we already considered that in Case 1.So, putting this together:For exactly one root in (-2, 4), we have two possibilities:1. The quadratic has a repeated root in (-2, 4). As found earlier, this happens only when a = -4, giving root x = 2.2. The quadratic has two distinct real roots, with exactly one in (-2, 4). For this, we need the discriminant D > 0 (i.e., |a| > 4) and f(-2) * f(4) < 0.Therefore, our total solution will be the union of the solutions from Case 1 and Case 2.So, let's compute f(-2) and f(4):f(x) = x² + a x + 4f(-2) = (-2)² + a*(-2) + 4 = 4 - 2a + 4 = 8 - 2af(4) = (4)² + a*4 + 4 = 16 + 4a + 4 = 20 + 4aThen, f(-2)*f(4) < 0 implies:(8 - 2a)(20 + 4a) < 0Let's solve this inequality.First, factor out constants:(8 - 2a)(20 + 4a) = 2*(4 - a) * 4*(5 + a) = 8*(4 - a)(5 + a)But since we can just solve (8 - 2a)(20 + 4a) < 0 without factoring out:Multiply the two terms:(8 - 2a)(20 + 4a) = 8*20 + 8*4a - 2a*20 - 2a*4a = 160 + 32a - 40a - 8a² = 160 - 8a - 8a² = -8a² -8a +160But perhaps it's easier to find the critical points.Set each factor to zero:8 - 2a = 0 ⇒ a = 420 + 4a = 0 ⇒ a = -5So, the critical points are a = -5 and a = 4. We can test intervals around these points to determine where the product is negative.The number line is divided into intervals by a = -5 and a = 4:1. a < -52. -5 < a < 43. a > 4Test each interval:1. a < -5: Let's pick a = -6.(8 - 2*(-6)) = 8 +12 = 20(20 + 4*(-6)) = 20 -24 = -4Product: 20*(-4) = -80 < 02. -5 < a < 4: Let's pick a = 0.(8 - 0) = 8(20 + 0) = 20Product: 8*20 = 160 > 03. a > 4: Let's pick a = 5.(8 - 2*5) = 8 -10 = -2(20 + 4*5) = 20 +20 = 40Product: (-2)*40 = -80 < 0Therefore, the inequality (8 - 2a)(20 + 4a) < 0 holds when a < -5 or a > 4.But wait, but we have to consider that in Case 2, we need the quadratic to have two distinct real roots, which requires a² > 16, i.e., |a| > 4. So, a < -4 or a > 4. Therefore, the solutions to the inequality (8 - 2a)(20 + 4a) < 0 are a < -5 or a > 4, but we need to intersect this with the discriminant condition a² > 16 (i.e., a < -4 or a > 4).Therefore, combining these:For a < -5: a < -5 intersects with a < -4, so a < -5.For a > 4: a > 4 is already part of a > 4, so a > 4.Therefore, the solutions for Case 2 are a < -5 or a > 4.But wait, let's confirm this. If a > 4, then discriminant is positive (a² > 16), and f(-2)*f(4) < 0 (from the inequality above), so there is exactly one root in (-2, 4). Similarly, if a < -5, discriminant is positive (since a < -5 implies a² > 25 > 16), and f(-2)*f(4) < 0, so exactly one root in (-2, 4).But we need to verify if these are correct. Let's take an example for a > 4, say a = 5.Then the quadratic is x² +5x +4 = 0. Factoring: (x +1)(x +4) = 0, roots at x = -1 and x = -4. Wait, but wait, x² +5x +4 = (x +1)(x +4). So, roots are -1 and -4. So, A = (-2,4). Then, -1 is in (-2,4), and -4 is not. So, intersection is {-1}, which has one element. So, that's correct. So, for a =5, which is >4, we have one root in the interval.But wait, wait, but when a =5, the quadratic is x² +5x +4=0, which factors to (x+1)(x+4). So, roots are x=-1 and x=-4. So, x=-1 is in (-2,4), and x=-4 is not. So, intersection is {-1}, which is correct.Similarly, take a = -6 (a < -5). Quadratic is x² -6x +4=0. Using quadratic formula: x = [6 ± sqrt(36 -16)]/2 = [6 ± sqrt(20)]/2 = [6 ± 2*sqrt(5)]/2 = 3 ± sqrt(5). sqrt(5) is approx 2.236. So, roots are 3 + 2.236 ≈5.236 and 3 - 2.236 ≈0.764. So, 0.764 is in (-2,4), and 5.236 is not. Therefore, intersection is {0.764}, which is one element. So, that works.Wait, but in this case, a = -6, so the quadratic is x² + (-6)x +4 = x² -6x +4. So, roots at 3 ± sqrt(5). So, 3 - sqrt(5) ≈0.764 is in (-2,4), and 3 + sqrt(5)≈5.236 is outside. So, exactly one root in the interval. Correct.But now, let's check a value between -5 and -4, say a = -4.5. Since a is between -5 and -4, which is part of the inequality solution a < -5 or a >4, but since a is between -5 and -4, which is not in the inequality solution. Let's see what happens.If a = -4.5, discriminant D = (-4.5)^2 -16 =20.25 -16=4.25>0, so two real roots. Compute f(-2) and f(4):f(-2)=8 -2*(-4.5)=8+9=17f(4)=20 +4*(-4.5)=20 -18=2f(-2)*f(4)=17*2=34>0. Therefore, the product is positive, which means f(-2) and f(4) have the same sign. So, either both roots are in (-2,4) or both are outside. Let's compute the roots.Quadratic equation: x² -4.5x +4=0.Solutions: [4.5 ± sqrt(20.25 -16)]/2 = [4.5 ± sqrt(4.25)]/2 ≈ [4.5 ±2.0616]/2.So, (4.5 +2.0616)/2 ≈6.5616/2≈3.2808(4.5 -2.0616)/2≈2.4384/2≈1.2192Both roots are approximately 3.28 and 1.22, which are both in (-2,4). Therefore, A ∩ B would have two elements, which would mean the number of non-empty subsets is 2² -1=3, which is not 1. Therefore, a=-4.5 is invalid, which aligns with our previous conclusion that between -5 and -4, the product f(-2)*f(4) is positive, so no roots or two roots. In this case, two roots inside, which is invalid. Therefore, our condition of a < -5 or a >4 is correct.But wait, let's check another value between -5 and -4. Let's take a = -4.5, which we did, and saw both roots inside. So, that interval a ∈ (-5, -4) would lead to both roots inside the interval. Therefore, invalid for our purpose.Similarly, take a value a < -5, say a=-6. As we saw, one root inside, one outside. So, works.Similarly, take a= -5, which is a boundary. Let's check a=-5.Compute f(-2)=8 -2*(-5)=8 +10=18f(4)=20 +4*(-5)=20 -20=0So, f(4)=0. But the interval is open at 4, so x=4 is not included. So, even though f(4)=0, the root x=4 is excluded. But for a=-5, let's solve the quadratic:x² -5x +4=0. Factors: (x-1)(x-4)=0. Roots x=1 and x=4. But x=4 is excluded, so A ∩ B = {1}, which has one element. But wait, x=4 is not in A, so only x=1 is in A ∩ B. So, even though f(4)=0, since the interval is open, we exclude 4. Therefore, for a=-5, A ∩ B has one element. So, a=-5 should be included?Wait, but according to our earlier inequality, (8 -2a)(20 +4a) <0 when a < -5 or a >4. But when a=-5, the product (8 -2*(-5))(20 +4*(-5))= (8 +10)(20 -20)=18*0=0, which is not less than zero. Therefore, a=-5 is not included in the inequality solution. However, when a=-5, the quadratic has a root at x=4, which is not in A, and x=1, which is in A. Therefore, A ∩ B={1}, which is a singleton. Therefore, the number of non-empty subsets is 1. So, a=-5 should be included.But according to our previous analysis, the Case 2 requires a² >16, i.e., a < -4 or a >4. But a=-5 satisfies a < -4. However, when a=-5, f(4)=0, but x=4 is excluded. So, the intersection still has one element. So, why was a=-5 excluded from the inequality solution?Because the inequality (8 -2a)(20 +4a) <0 is strict. At a=-5, the product is zero, so the inequality does not hold. However, even though f(4)=0, the root x=4 is not in A. So, in this case, even though a=-5 is on the boundary, it still results in exactly one element in A ∩ B. Therefore, we need to consider this case separately.Similarly, when a=4, as before, the root is x=-2, which is excluded. So, A ∩ B is empty, which is invalid. So, a=4 is excluded.So, in addition to the previous cases, we need to check the boundary points a=-5 and a=4.For a=-5:Quadratic equation: x² -5x +4=0. Roots x=1 and x=4. Since x=4 is excluded, A ∩ B={1}, which is valid. So, a=-5 should be included.For a=4:Quadratic equation: x² +4x +4=0. Root x=-2 (double root). But x=-2 is excluded, so A ∩ B is empty. Therefore, a=4 is excluded.Therefore, even though a=-5 is not in the inequality solution (since the product is zero), it still gives a valid case where A ∩ B has one element. Therefore, we need to include a=-5 in our solution.Similarly, let's check a=-5:The quadratic x² -5x +4=0 has roots at 1 and 4. So, only 1 is in A. Therefore, {1} is the intersection. So, non-empty subsets: 1, which is correct.Therefore, our Case 2 should include a ≤ -5 or a >4? Wait, but discriminant for a=-5 is D=25 -16=9>0, so two distinct roots. However, one of them is at x=4, which is excluded. So, the intersection still has one element. Therefore, we need to include a=-5 in the range.Similarly, for a >4, we need to check if when a approaches 4 from above, whether the root enters the interval.Wait, for a approaching 4 from above, say a=5, we saw the roots are -1 and -4. So, -1 is in the interval. Similarly, as a increases beyond 4, the roots move. Wait, maybe we need to analyze how the roots behave as a varies.Alternatively, let's formalize this.We have two cases where A ∩ B has exactly one element:1. The quadratic has a repeated root in (-2,4). This happens when a=-4, giving root x=2.2. The quadratic has two distinct roots, with exactly one in (-2,4). This happens when: a. a < -5 or a >4 (from the inequality (8 -2a)(20 +4a) <0 and discriminant >0) b. Also, when a=-5, which gives one root at x=1 (inside) and x=4 (excluded), hence valid.But wait, when a=-5, the quadratic has roots at x=1 and x=4. Since x=4 is not in A, only x=1 is in A ∩ B. So, it's a valid case. However, according to our earlier analysis, a=-5 is not included in the inequality solution because the product was zero, but it still results in a valid case. Therefore, we need to include a=-5 in the solution.Similarly, when a approaches -5 from below (a=-5 - ε), the roots are both approaching 1 and 4 from the inside and outside? Wait, let me check.If a is approaching -5 from below (i.e., a=-5 - ε where ε is small positive), then the quadratic equation becomes x² + (-5 - ε)x +4=0. Let's compute the roots:x = [5 + ε ± sqrt((5 + ε)^2 -16)]/2= [5 + ε ± sqrt(25 +10ε + ε² -16)]/2= [5 + ε ± sqrt(9 +10ε + ε²)]/2For small ε, sqrt(9 +10ε) ≈3 + (10ε)/(2*3) =3 + (5ε)/3. So, the roots are approximately:[5 + ε ± (3 +5ε/3)]/2So, the larger root:[5 + ε +3 +5ε/3]/2 = [8 + (8ε)/3]/2 =4 + (4ε)/3The smaller root:[5 + ε -3 -5ε/3]/2 = [2 - (2ε)/3]/2 =1 - (ε)/3As ε approaches 0 from the positive side, the larger root approaches 4 from above, and the smaller root approaches 1 from above. However, since a is approaching -5 from below (a=-5 - ε), the roots are 4 + (4ε)/3 and 1 - ε/3. As ε approaches 0, the larger root approaches 4 from above (outside the interval), and the smaller root approaches 1 from above (still inside). Therefore, for a approaching -5 from below, one root approaches 1 (inside) and the other approaches 4 from outside. Therefore, when a=-5, one root is exactly at 4 (excluded), and the other at 1 (included). Hence, a=-5 is a valid case.Therefore, combining all this, the values of a for which A ∩ B has exactly one element are:- a=-4 (giving a repeated root at 2 inside the interval)- a ≤ -5 or a >4 (giving two distinct roots with exactly one inside the interval)But wait, when a=-5, we have a= -5. So, in this case, the quadratic has roots at 1 and 4. Since 4 is excluded, only 1 is in A ∩ B. So, a=-5 is valid. However, according to the discriminant condition for Case 2, we need a² >16, but a=-5 gives a²=25>16, so it's included. However, the inequality (8 -2a)(20 +4a) <0 was for a < -5 or a >4. But when a=-5, the product is zero. So, perhaps we need to adjust our conditions.Alternatively, let's think that in addition to the cases where f(-2)*f(4) <0, we also need to consider when one root is exactly at -2 or 4, but since the interval is open, those roots are excluded. However, if one root is at 4 and the other is inside, then A ∩ B would have one element. Similarly, if one root is at -2 and the other is inside, then A ∩ B would have one element.But since the interval is open, roots at -2 or 4 are excluded. Therefore, if one root is at -2 or 4, it doesn't contribute to A ∩ B. So, in such cases, even though the quadratic has a root at the endpoint, since it's excluded, we have to check if the other root is inside.For example, take a=-5:Quadratic has roots at 1 and 4. 4 is excluded, so intersection is {1}. So, valid.Similarly, take a value of a where one root is at -2. Let's solve for a such that -2 is a root.Plug x=-2 into x² +a x +4=0:(-2)^2 +a*(-2)+4=0 ⇒4 -2a +4=0 ⇒8 -2a=0 ⇒a=4.So, when a=4, the quadratic has a root at x=-2 (double root). But since -2 is excluded, A ∩ B is empty. Therefore, a=4 is invalid.Therefore, the only boundary case where one root is at an endpoint and the other is inside is a=-5. So, a=-5 is valid. Similarly, if we had a value of a where one root is at 4, but that's when a=-5 (since x=4 is a root when a=-5). Wait, let's check:If x=4 is a root, then substituting into x² +a x +4=0:16 +4a +4=0 ⇒4a +20=0 ⇒a=-5. Yes, exactly. So, x=4 is a root only when a=-5. But since 4 is excluded, the other root is x=1, which is inside. So, a=-5 is valid.Therefore, in summary, the valid values of a are:1. a=-4 (repeated root at 2 inside the interval)2. a ≤ -5 or a >4 (two distinct roots, exactly one inside the interval)But wait, when a=-5, even though it's the boundary case where one root is at 4 (excluded), the other root is inside. So, a=-5 is included in the a ≤ -5 category. Therefore, combining everything:The range of a is a=-4, a ≤ -5, or a >4. But need to write this as intervals.So, in terms of real numbers:a ∈ (-∞, -5] ∪ {-4} ∪ (4, ∞)Wait, but is that correct?Wait, when a=-4, it's a single point. So, the range is all real numbers a where a ≤ -5, a=-4, or a >4.But let's verify with examples.Take a=-6 (which is ≤ -5): As before, roots at 3 ± sqrt(5), one inside, one outside. Correct.Take a=-5: Roots at 1 and 4. Only 1 is inside. Correct.Take a=-4: Repeated root at 2. Correct.Take a=5: Roots at -1 and -4. Only -1 is inside. Correct.Take a=4: Roots at -2 (double root). Excluded, so invalid. Correct.Take a= -4.5: Roots both inside, which we saw earlier, which is invalid. Correctly excluded.Take a=3: Discriminant 9 -16 <0, so no real roots. Intersection is empty. Invalid.Therefore, the correct range is a ≤ -5, a=-4, or a >4. But wait, is a=-4 included in a ≤ -5? No, because -4 > -5. So, they are separate.Therefore, the range is:a ∈ (-∞, -5] ∪ {-4} ∪ (4, ∞)But we need to express this in interval notation, including the single point -4.In interval notation, this would be:(-∞, -5] ∪ { -4 } ∪ (4, ∞)However, typically, such ranges are written as the union of intervals and individual points. In this case, since -4 is a single point, we include it as a separate union.But let's confirm whether a=-4 is the only value in Case 1. Yes, because discriminant zero occurs at a=±4. For a=4, the root is at -2, which is excluded. For a=-4, the root is at 2, which is included. Therefore, only a=-4 is valid for Case 1.Therefore, combining Case 1 and Case 2:Case 1: a=-4Case 2: a ≤ -5 or a >4Therefore, the total solution is a=-4 or a ≤ -5 or a >4. But since a=-4 is not included in a ≤ -5 (because -4 > -5), we have to write it separately.Hence, the range is:a ∈ (-∞, -5] ∪ { -4 } ∪ (4, ∞)But in some contexts, this might be written as a ≤ -5 or a=-4 or a >4.However, in the problem statement, it asks for the range of the real number a, so we need to present it appropriately. In Chinese mathematical notation, sometimes they use union symbols and intervals. Since the user might expect interval notation combined with a separate point.Alternatively, since a=-4 is a single point between -5 and 4, it's separate.Therefore, the final answer is the union of three parts: all real numbers less than or equal to -5, the single number -4, and all real numbers greater than 4.But let me check once more for any possible overlaps or mistakes.When a=-4, it's a separate case with a repeated root inside the interval.When a ≤ -5, we have two distinct roots, one inside and one outside.When a >4, we have two distinct roots, one inside and one outside.Therefore, the answer should indeed be a ∈ (-∞, -5] ∪ { -4 } ∪ (4, ∞).But let me check if there's another way to express this. For example, combining (-∞, -5] and {-4} as (-∞, -4] ∪ { -4 } which is redundant, but no, because -4 is greater than -5. So, (-∞, -5] and {-4} are separate.Thus, the correct range is a ≤ -5, a=-4, or a >4.In boxed notation, since the user might expect intervals and individual points, we can write:boxed{(-infty, -5] cup {-4} cup (4, infty)}This notation accurately represents all values of a where A ∩ B has exactly one element, covering both the repeated root case and the two distinct roots cases with one inside the interval.

✅:To determine the range of the real number ( a ) such that the number of non-empty subsets of ( A cap B ) is 1, we need to ensure that ( A cap B ) contains exactly one element. This occurs when the quadratic equation ( x^2 + ax + 4 = 0 ) has exactly one root in the interval ((-2, 4)). Key Steps:1. Discriminant Analysis: - The quadratic equation ( x^2 + ax + 4 = 0 ) has real roots if the discriminant ( D = a^2 - 16 geq 0 ), i.e., ( a leq -4 ) or ( a geq 4 ).2. Repeated Root Case: - When ( D = 0 ), the quadratic has a repeated root at ( x = -frac{a}{2} ). - For ( a = -4 ), the root is ( x = 2 ), which is in ((-2, 4)). Thus, ( A cap B = {2} ).3. Distinct Roots Case: - For ( D > 0 ) (i.e., ( |a| > 4 )), we need exactly one root in ((-2, 4)). This requires ( f(-2) cdot f(4) < 0 ): - ( f(-2) = 8 - 2a ) - ( f(4) = 20 + 4a ) - Solving ( (8 - 2a)(20 + 4a) < 0 ) gives ( a < -5 ) or ( a > 4 ).4. Boundary Case: - When ( a = -5 ), the roots are ( x = 1 ) and ( x = 4 ). Since ( x = 4 ) is excluded, ( A cap B = {1} ). Final Range:Combining all cases, the range of ( a ) is:- ( a = -4 ) (repeated root),- ( a leq -5 ) (one root in interval, one outside),- ( a > 4 ) (one root in interval, one outside). Answer:[boxed{(-infty, -5] cup {-4} cup (4, infty)}]

🔑: Problem:设 A=(-2,4), B=left{x in mathbf{R} mid x^{2}+a x+4=0right}. 若 A cap B 的非空子集个数为 1 ，则实数 a 的取值范围是 quad .首先，由已知，A cap B 含有一个元素。设 f(x)=x^{2}+a x+4。我们需要讨论二次方程 x^2 + a x + 4 = 0 在区间 (-2, 4) 的根的情况。1. 计算判别式 Delta 二次方程 x^2 + a x + 4=0 的判别式 Delta 为： [ Delta = a^2 - 4 cdot 1 cdot 4 = a^2 - 16 ]2. 判别存在唯一解的情况（Delta = 0） 当 Delta = 0 时，有且仅有一个实根： [ a^2 - 16 = 0 implies a = pm 4 ] - 若 a = 4，则 f(x) = x^2 + 4x + 4 的唯一解为： [ x = -frac{4}{2} = -2 ] 但 -2 notin (-2,4)，因此这个情况不符合要求，应舍去。 - 若 a = -4，则 f(x) = x^2 - 4x + 4 的唯一解为： [ x = frac{4}{2} = 2 ] 显然，2 in (-2,4)，符合条件，因此 a = -4 满足唯一解的要求。3. 判别存在两个解的情况（Delta > 0） 当 Delta > 0 时，有两个不同的实根。根据二次函数图像性质，确保在区间 (-2, 4) 内有且仅一个解时，需要： [ f(-2) f(4) leq 0 ] 计算 f(-2) 和 f(4)： [ f(-2) = (-2)^2 + (-2)a + 4 = 4 - 2a + 4 = 8 - 2a ] [ f(4) = (4)^2 + 4a + 4 = 16 + 4a + 4 = 20 + 4a ] 所以： [ (8 - 2a)(20 + 4a) leq 0 ] 解不等式： [ (8 - 2a)(20 + 4a) leq 0 implies 160 + 8a - 40a - 8a^2 leq 0 implies -8a^2 - 32a + 160 leq 0 implies a^2 + 4a - 20 geq 0 ] 利用一元二次不等式的求解，我们需找到该二次函数的零点： [ a^2 + 4a - 20 = 0 implies a = frac{-4 pm sqrt{16 + 80}}{2} = frac{-4 pm sqrt{96}}{2} = frac{-4 pm 4sqrt{6}}{2} = -2 pm 2sqrt{6} ] 因此，解得： [ a geq 2sqrt{6} - 2 approx 4 quad text{或}quad a leq -2 - 2sqrt{6} approx -5 ] 最后： [ boxed{(-infty,-5] cup{-4} cup(4,+infty)} ]

🧠:Okay, so I need to find the limit of (6 - 5/cosx) raised to the power of cotangent squared x as x approaches 0. Hmm, let's start by understanding the components of this function.First, let's recall that cotangent x is cosx/sinx, so cotangent squared x would be (cos²x)/(sin²x). As x approaches 0, sinx behaves like x, so sinx ≈ x, and thus sin²x ≈ x². Similarly, cosx approaches 1 as x approaches 0, so cos²x approaches 1. Therefore, cot²x would be approximately 1/x² as x approaches 0. So the exponent, cot²x, is going to infinity as x approaches 0. Now, looking at the base of the exponent, which is (6 - 5/cosx). Let's evaluate this as x approaches 0. Since cosx approaches 1, 5/cosx approaches 5. Therefore, the base becomes 6 - 5 = 1. So we have a situation where the base is approaching 1 and the exponent is approaching infinity. This is an indeterminate form of type 1^∞, which means we need to use a standard technique to evaluate such limits, typically involving taking the natural logarithm and then applying L'Hospital's Rule after transforming it into a 0/0 or ∞/∞ form.Let me outline the steps:1. Let L be the original limit: L = lim_{x→0} (6 - 5/cosx)^{cot²x}2. Take the natural logarithm of both sides: ln L = lim_{x→0} [cot²x * ln(6 - 5/cosx)]3. Evaluate the limit of cot²x * ln(6 - 5/cosx) as x approaches 0.So, first, let's compute ln(6 - 5/cosx) as x approaches 0. Let's denote the base as f(x) = 6 - 5/cosx. As x→0, cosx ≈ 1 - x²/2, so 1/cosx ≈ 1 + x²/2. Therefore, 5/cosx ≈ 5 + (5/2)x². Then f(x) ≈ 6 - 5 - (5/2)x² = 1 - (5/2)x². Therefore, ln(f(x)) ≈ ln(1 - (5/2)x²) ≈ - (5/2)x² - ( (5/2 x²)^2 )/2 + ... using the Taylor series expansion ln(1 + y) ≈ y - y²/2 + y³/3 - ... for small y. Since x is approaching 0, the higher order terms can be neglected, so approximately ln(f(x)) ≈ -5/2 x².But wait, let me check that again. If f(x) ≈ 1 - (5/2)x², then ln(f(x)) ≈ ln(1 - 5/2 x²) ≈ -5/2 x² - ( (-5/2 x²)^2 )/2 + ... which would be -5/2 x² - (25/8 x^4)/2 + ... So actually, the leading term is -5/2 x², and the next term is of order x^4, which can be neglected as x approaches 0. So ln(f(x)) ≈ -5/2 x².Meanwhile, cot²x is (cos²x)/(sin²x) ≈ (1 - x²)/x² ≈ 1/x² - 1 as x→0. But more precisely, since sinx ≈ x - x³/6 + ..., so sin²x ≈ x² - x^4/3 + ..., and cos²x ≈ 1 - x² + ..., so cot²x ≈ (1 - x²)/(x² - x^4/3) ≈ (1/x²)(1 - x²)/(1 - x²/3) ≈ 1/x² * [1 - x² - (-x²/3)] using the approximation 1/(1 - ε) ≈ 1 + ε for small ε. So that would be 1/x² * (1 - x² + x²/3) = 1/x² * (1 - 2x²/3). Therefore, cot²x ≈ 1/x² - 2/3. But maybe this is overcomplicating. Since sinx ~ x, cosx ~1, so cotx ~ 1/x, so cot²x ~1/x². So perhaps, for the leading term, we can take cot²x ≈ 1/x². Therefore, the product cot²x * ln(f(x)) ≈ (1/x²)( -5/2 x² ) = -5/2. So then, the limit of the logarithm would be -5/2, so L = e^{-5/2}.But wait, that seems too straightforward. Let me verify this step by step more carefully.First, let's compute f(x) = 6 - 5/cosx. Let's expand cosx as a Taylor series around x=0: cosx = 1 - x²/2 + x^4/24 - ... So 1/cosx = 1 + x²/2 + (5x^4)/24 + ... (using the expansion 1/(1 - y) ≈ 1 + y + y² + ... for small y, so here y = x²/2 - x^4/24 + ..., so 1/cosx ≈ 1 + (x²/2) + (x²/2)^2 + ... ≈ 1 + x²/2 + x^4/8 + ...). Wait, actually, maybe a better way is to use the known expansion of secx, which is 1 + x²/2 + 5x^4/24 + ... So 1/cosx = secx ≈ 1 + x²/2 + 5x^4/24 + ... Therefore, 5/cosx ≈ 5 + (5/2)x² + (25/24)x^4 + ... So f(x) = 6 - 5/cosx ≈ 6 - 5 - (5/2)x² - (25/24)x^4 + ... = 1 - (5/2)x² - (25/24)x^4 + ... Therefore, ln(f(x)) = ln(1 - (5/2)x² - (25/24)x^4 + ...) ≈ [ - (5/2)x² - (25/24)x^4 ] - [ ( - (5/2)x² )^2 ] / 2 + ... ≈ - (5/2)x² - (25/24)x^4 - (25/8)x^4 / 2 + ... Wait, perhaps this is getting too complicated. Alternatively, using the expansion ln(1 + y) ≈ y - y²/2 + y³/3 - ..., where y = - (5/2)x² - (25/24)x^4 + ... So up to the x² term, ln(f(x)) ≈ - (5/2)x² - (25/24)x^4 - [ ( -5/2 x² )^2 ] / 2 = - (5/2)x² - (25/24)x^4 - (25/8)x^4 + ... Combining like terms: - (5/2)x² - [25/24 + 25/8]x^4. Let's compute 25/24 + 25/8 = 25/24 + 75/24 = 100/24 = 25/6. So ln(f(x)) ≈ - (5/2)x² - (25/6)x^4 - ... However, since we are multiplying by cot²x, which is approximately 1/x² - 2/3 + ... (from the earlier expansion), perhaps we need to consider higher-order terms. But maybe the leading term is sufficient.But let's see. Let's compute the product cot²x * ln(f(x)). If cot²x ≈ 1/x² - 2/3 + ... and ln(f(x)) ≈ -5/2 x² -25/6 x^4 + ..., then multiplying these together:(1/x² - 2/3 + ...) * (-5/2 x² -25/6 x^4 + ...) = (1/x²)(-5/2 x²) + (1/x²)(-25/6 x^4) + (-2/3)(-5/2 x²) + ... = (-5/2) + (-25/6 x²) + (5/3 x²) + ... Combining the x² terms: (-25/6 + 5/3) x² = (-25/6 + 10/6) x² = (-15/6)x² = -5/2 x². So the expansion would be -5/2 -5/2 x² + ... Therefore, as x approaches 0, the limit of cot²x * ln(f(x)) would approach -5/2. Therefore, the original limit is e^{-5/2}.But let me check this by another approach, maybe using substitution. Let's set t = x, so as t approaches 0, we can rewrite the expression. Alternatively, since we have 1^∞ form, another method is to write the expression as [1 + (f(x) -1)]^{g(x)}, where f(x) approaches 1 and g(x) approaches infinity. Then, the limit is e^{lim_{x→0} (f(x)-1)*g(x)} if (f(x)-1)*g(x) approaches a finite limit.In this case, f(x) is (6 -5/cosx) which approaches 1, and g(x) is cot²x. Therefore, the limit would be e^{lim_{x→0} (6 -5/cosx -1) * cot²x} = e^{lim_{x→0} (5 -5/cosx) * cot²x}. Wait, but 6 -5/cosx -1 = 5 -5/cosx? Wait, no. 6 -5/cosx -1 = 5 -5/cosx? Wait, that's 5(1 -1/cosx). But 1 -1/cosx = -(secx -1). So, (5 -5/cosx) = -5(secx -1). Therefore, the exponent becomes lim_{x→0} -5(secx -1) * cot²x. Let's compute this.First, secx -1 = 1/cosx -1 ≈ (1 + x²/2) -1 = x²/2. Therefore, secx -1 ≈ x²/2. Then, cot²x ≈ 1/x². Therefore, (secx -1) * cot²x ≈ (x²/2)*(1/x²) = 1/2. Therefore, multiplying by -5 gives -5*(1/2) = -5/2. Therefore, the limit is e^{-5/2}, same as before.Therefore, both methods lead to the same result. Let's confirm this with a direct computation using L’Hospital’s Rule. Let's take the logarithm of the original function:ln L = lim_{x→0} cot²x * ln(6 -5/cosx). Let's rewrite cot²x as (cos²x)/(sin²x). Therefore, the expression becomes [cos²x / sin²x] * ln(6 -5/cosx). Let's make the substitution x→0. Let me denote t = x. As t→0, sin t ≈ t, cos t ≈ 1 - t²/2. Let me express everything in terms of t. Let's consider t approaching 0 from the right (since cotx is defined for x approaching 0 from both sides, but since the function is even in x, the limit from both sides should be the same).Let me substitute variables to make it easier. Let’s set y = x. So as y approaches 0, we can write:ln L = lim_{y→0} [ (cos²y)/(sin²y) ] * ln(6 -5/cosy).Let’s consider the expression [ (cos²y)/(sin²y) ] * ln(6 -5/cosy). As y→0, siny ≈ y, cosy ≈ 1 - y²/2. Let's compute the leading terms. First, cos²y ≈ 1 - y², sin²y ≈ y². Therefore, (cos²y)/(sin²y) ≈ (1 - y²)/y² ≈ 1/y² - 1. But let's keep it as [1 - y² + ...]/y² = 1/y² - 1 + ... So this term is approximately 1/y². Then, ln(6 -5/cosy) ≈ ln(6 -5(1 + y²/2 + ... )) = ln(6 -5 - (5/2)y² + ...) = ln(1 - (5/2)y² + ...) ≈ - (5/2)y² - [ (5/2 y²)^2 ] / 2 + ... So approximately -5/2 y². Therefore, multiplying the two parts gives (1/y²)( -5/2 y² ) = -5/2. Therefore, ln L = -5/2, so L = e^{-5/2}.But to apply L’Hospital’s Rule, perhaps we need to write the limit in the form 0/0 or ∞/∞. Let me see. Let's consider the exponent:lim_{x→0} cot²x * ln(6 -5/cosx) = lim_{x→0} [ln(6 -5/cosx)] / tan²x.Wait, since cot²x = 1/tan²x, so [ln(6 -5/cosx)] / tan²x. As x→0, tanx ≈ x, so tan²x ≈ x². The numerator ln(6 -5/cosx) ≈ ln(1 -5/2 x²) ≈ -5/2 x². So the expression becomes (-5/2 x²)/x² = -5/2. Therefore, the limit is indeed -5/2, so the original limit is e^{-5/2}.But let me check using L’Hospital’s Rule step by step. Let’s write the exponent as:lim_{x→0} [ln(6 -5/cosx)] / tan²x.Let’s set this up as a fraction where both numerator and denominator approach 0. Let's verify:As x→0, ln(6 -5/cosx) approaches ln(1) = 0, and tan²x approaches 0. So it's 0/0 form. Therefore, we can apply L’Hospital’s Rule. Take derivatives of numerator and denominator.Let’s compute the derivative of the numerator: d/dx [ln(6 -5/cosx)] = [1/(6 -5/cosx)] * d/dx [6 -5/cosx] = [1/(6 -5/cosx)] * [5 sinx / cos²x].The derivative of the denominator: d/dx [tan²x] = 2 tanx * sec²x.Therefore, applying L’Hospital’s Rule once, the limit becomes:lim_{x→0} [5 sinx / ( (6 -5/cosx) cos²x ) ] / [2 tanx sec²x ].Simplify this expression. Let's rewrite it:= lim_{x→0} [5 sinx / ( (6 -5/cosx) cos²x ) ] * [1/(2 tanx sec²x ) ]Simplify step by step. First, note that tanx = sinx/cosx, and secx = 1/cosx, so sec²x = 1/cos²x.Therefore, the denominator part after L’Hospital becomes 2*(sinx/cosx)*(1/cos²x) = 2 sinx / cos³x.Therefore, the entire expression is:[5 sinx / ( (6 -5/cosx) cos²x ) ] / [2 sinx / cos³x ] = [5 sinx / ( (6 -5/cosx) cos²x ) ] * [cos³x / (2 sinx ) ].The sinx terms cancel out, and cos²x in the denominator cancels with cos³x in the numerator, leaving cosx. So we have:5 * cosx / [2(6 -5/cosx)].Now, let's simplify the denominator 6 -5/cosx. Let's factor out 1/cosx:6 -5/cosx = (6 cosx -5)/cosx.Therefore, the expression becomes:5 cosx / [2*(6 cosx -5)/cosx] = 5 cosx * cosx / [2(6 cosx -5)] = 5 cos²x / [2(6 cosx -5)].Now, take the limit as x→0. cosx approaches 1, so cos²x approaches 1, and 6 cosx -5 approaches 6*1 -5 =1. Therefore, the limit becomes 5*1 / [2*1] = 5/2.Wait, but that contradicts our previous conclusion of -5/2. Hmm, where did I go wrong here?Wait, no, because in the exponent, we have [ln(6 -5/cosx)] / tan²x, which after applying L’Hospital once gives us 5 cos²x / [2(6 cosx -5)]. But when x approaches 0, this tends to 5*1 / [2(6*1 -5)] = 5 / [2(1)] = 5/2. But this is supposed to be the limit of the exponent, which was supposed to be -5/2. There's a discrepancy here. This suggests that there's a mistake in the application of L’Hospital’s Rule.Wait, let's check the differentiation steps again.Original numerator: ln(6 -5/cosx). Its derivative is:[1/(6 -5/cosx)] * derivative of (6 -5/cosx). The derivative of 6 is 0, the derivative of -5/cosx is -5*(-sinx)/cos²x = 5 sinx / cos²x. So the derivative of the numerator is [5 sinx / cos²x] / (6 -5/cosx).Denominator: tan²x. Its derivative is 2 tanx * sec²x. So derivative is 2 (sinx/cosx) * (1/cos²x) = 2 sinx / cos³x.Therefore, the ratio after applying L’Hospital’s Rule is:[5 sinx / (cos²x (6 -5/cosx))] / [2 sinx / cos³x] = [5 sinx / (cos²x (6 -5/cosx))] * [cos³x / (2 sinx)] = [5 / (6 -5/cosx)] * [cosx / 2] = (5 cosx) / [2(6 cosx -5)].Ah, here's the mistake. Earlier, I incorrectly wrote the denominator as 6 cosx -5, but 6 -5/cosx multiplied by cosx is 6 cosx -5. Therefore, the expression simplifies to (5 cosx)/(2(6 cosx -5)).Then, taking the limit as x→0, cosx→1, so the expression becomes (5*1)/(2*(6*1 -5)) = 5/(2*(1)) = 5/2. Wait, but this suggests the limit is 5/2, which contradicts our previous conclusion of -5/2. So where is the error here?Wait a second. The original exponent is [ln(6 -5/cosx)] / tan²x. But as x approaches 0, 6 -5/cosx approaches 1, so ln(6 -5/cosx) approaches ln(1) = 0, and tan²x approaches 0. So applying L’Hospital’s Rule is valid. However, after differentiation, we get the limit of (5 cosx)/(2(6 cosx -5)) as x approaches 0. Plugging x=0, we get 5/(2*(6 -5)) =5/(2*1)=5/2. Therefore, the exponent limit is 5/2? But earlier calculations suggested -5/2.This inconsistency indicates a mistake in one of the approaches. Let's track back.Original function: (6 -5/cosx)^{cot²x}. As x approaches 0, 6 -5/cosx approaches 6 -5*1=1, so the base approaches 1. The exponent cot²x approaches infinity. So it's 1^∞ indeterminate form. When taking the natural logarithm, we have cot²x * ln(6 -5/cosx). Let me compute this expression numerically for a very small x, say x=0.001 radians.Compute cos(0.001) ≈ 0.9999998333, so 1/cosx ≈1.0000001667. Then 5/cosx ≈5.0000008335. Therefore, 6 -5/cosx≈6 -5.0000008335≈0.9999991665. Then ln(6 -5/cosx)≈ln(0.9999991665)≈-0.0000008335 (since ln(1 - y)≈-y for small y). Then cot²x = (cosx/sinx)^2. For x=0.001, sinx≈0.0009999998333, so cotx≈cosx/sinx≈0.9999998333/0.0009999998333≈1000.0001667, so cot²x≈1000000.333. Then the product cot²x * ln(6 -5/cosx)≈1000000.333 * (-0.0000008335)≈-0.8335, which is approximately -5/6? Wait, no. Wait, 0.8335 is approximately 5/6 ≈0.8333. But -0.8335 is approximately -5/6. However, according to previous calculations, we expected -5/2. But this numerical example gives approximately -0.8335, which is about -5/6. This suggests that either my analytical approach was wrong or there is a miscalculation here.Wait, but wait, let's check the numerical calculation again.Take x = 0.001 radians.Compute cos(x) = cos(0.001) ≈ 1 - 0.001²/2 + 0.001^4/24 ≈ 1 - 0.0000005 + 0.000000000004 ≈0.9999995.So 1/cosx ≈1.0000005.Then 5/cosx≈5*1.0000005≈5.0000025.Therefore, 6 -5/cosx≈6 -5.0000025≈0.9999975.Then ln(0.9999975)≈-0.0000025 (since ln(1 - y)≈-y for small y, which here y=0.0000025).Then cotx = cosx/sinx. sin(0.001)≈0.0009999998333, so cotx≈0.9999995 / 0.0009999998333≈1000.0001667. Therefore, cot²x≈(1000.0001667)^2≈1,000,000.333.Then multiply by ln(6 -5/cosx): 1,000,000.333 * (-0.0000025)≈-2.5. Which is approximately -2.5 = -5/2. Ah, that's correct! So the numerical example gives approximately -5/2, which matches our analytical result. So where did I go wrong in the L’Hospital’s Rule application?Wait, let's go back to the L’Hospital step.After applying L’Hospital’s Rule once, we had:lim_{x→0} (5 cosx) / [2(6 cosx -5)].But wait, when x=0, cosx=1, so this becomes 5*1 / [2(6*1 -5)] =5/[2(1)] =5/2. But according to the numerical calculation, the limit should be -5/2. So clearly, there's a mistake here. What's wrong?Wait, the problem is that when we take the derivative of the numerator and denominator, we might have to apply L’Hospital’s Rule more than once. Wait, no. Let's check the direction of the limit. Wait, in the exponent expression:Original exponent: cot²x * ln(6 -5/cosx). When we rewrite this as [ln(6 -5/cosx)] / tan²x, and both numerator and denominator approach 0, then applying L’Hospital’s Rule gives a limit of 5/2. But according to the numerical example, the actual limit is -5/2. This contradiction suggests that somewhere, there is a sign error.Wait, let's check the derivative of the numerator again. The numerator is ln(6 -5/cosx). Its derivative is [1/(6 -5/cosx)] * derivative of (6 -5/cosx). The derivative of -5/cosx is -5*(-sinx)/cos²x = 5 sinx / cos²x. So the derivative is positive, 5 sinx / [cos²x (6 -5/cosx)]. The denominator after L’Hospital’s Rule is derivative of tan²x, which is 2 tanx sec²x. All terms here are positive as x approaches 0. Therefore, the ratio after L’Hospital’s Rule is positive. However, in reality, ln(6 -5/cosx) is negative because 6 -5/cosx <1 when x approaches 0, since 5/cosx >5 when cosx <1 (which it is for x≠0). Therefore, ln(6 -5/cosx) is negative, and tan²x is positive, so the entire expression is negative. However, after applying L’Hospital’s Rule, we got a positive limit, which contradicts. Therefore, there must be a mistake in the sign.Ah! Wait a second. When I take the derivative of the numerator, let's double-check:d/dx [ln(6 -5/cosx)] = [1/(6 -5/cosx)] * d/dx(6 -5/cosx).But d/dx(6 -5/cosx) = 0 -5 * d/dx(1/cosx) = -5 * (sinx / cos²x). Therefore, the derivative is -5 sinx / cos²x. Then divided by (6 -5/cosx), so the derivative is [-5 sinx / cos²x] / (6 -5/cosx).Earlier, I thought the derivative was positive, but actually it's negative. That was the mistake! So the correct derivative of the numerator is [-5 sinx / cos²x] / (6 -5/cosx).Then, the derivative of the denominator tan²x is 2 tanx sec²x = 2 (sinx / cosx)(1 / cos²x) = 2 sinx / cos³x.Therefore, after applying L’Hospital’s Rule, the limit becomes:[ (-5 sinx / cos²x ) / (6 -5/cosx) ] / [2 sinx / cos³x ] =Multiply numerator and denominator:[ (-5 sinx / cos²x ) / (6 -5/cosx) ] * [ cos³x / (2 sinx ) ] =The sinx terms cancel, cos²x cancels with cos³x leaving cosx:= [ -5 / (6 -5/cosx) ] * [ cosx / 2 ] == (-5 cosx ) / [2 (6 cosx -5) ]Now, taking the limit as x→0:cosx approaches 1, so numerator is -5*1 = -5.Denominator: 2*(6*1 -5) =2*(1)=2.Thus, the limit is -5/2, which matches our previous results. Therefore, my earlier mistake was missing the negative sign in the derivative of the numerator. Therefore, the correct limit of the exponent is -5/2, so the original limit is e^{-5/2}.Therefore, the final answer is e^{-5/2}, which can be written as 1/e^{5/2} or 1/√(e^5), but typically left as e^{-5/2}.Final AnswerThe limit is boxed{e^{-frac{5}{2}}}.

✅:To calculate the limit of the function:lim _{x rightarrow 0}left(6-frac{5}{cos x}right)^{operatorname{ctg}^{2} x}we start by recognizing that as (x rightarrow 0), the base (6 - frac{5}{cos x}) approaches 1 and the exponent (operatorname{ctg}^2 x) (which is (cot^2 x)) approaches infinity. This is an indeterminate form of type (1^infty), so we use the natural logarithm transformation and L'Hospital's Rule.1. Let (L) be the original limit: [ L = lim_{x rightarrow 0} left(6 - frac{5}{cos x}right)^{cot^2 x} ]2. Take the natural logarithm of both sides: [ ln L = lim_{x rightarrow 0} left[cot^2 x cdot lnleft(6 - frac{5}{cos x}right)right] ]3. Evaluate the limit of the exponent: [ lim_{x rightarrow 0} left[cot^2 x cdot lnleft(6 - frac{5}{cos x}right)right] ]4. Approximate (ln(6 - frac{5}{cos x})) using the Taylor series expansion around (x = 0): [ 6 - frac{5}{cos x} approx 1 - frac{5}{2}x^2 implies lnleft(1 - frac{5}{2}x^2right) approx -frac{5}{2}x^2 ]5. Approximate (cot^2 x) as (x rightarrow 0): [ cot^2 x approx frac{1}{x^2} ]6. Multiply the approximations: [ cot^2 x cdot lnleft(6 - frac{5}{cos x}right) approx frac{1}{x^2} cdot left(-frac{5}{2}x^2right) = -frac{5}{2} ]7. Confirm using L'Hospital's Rule on the transformed limit: [ lim_{x rightarrow 0} frac{lnleft(6 - frac{5}{cos x}right)}{tan^2 x} ] Applying L'Hospital's Rule twice, we find the limit of the exponent is (-frac{5}{2}).8. Exponentiate the result to find the original limit: [ L = e^{-frac{5}{2}} ]Thus, the limit is:[boxed{e^{-frac{5}{2}}}]

🔑:We are given the limit:[lim_{x to 0}left(6 - frac{5}{cos x}right)^{operatorname{ctg}^2 x}]1. Use the exponential transformation: Let's remember that any expression of the form a^b can be written as e^{b ln a}. Thus, [ left(6 - frac{5}{cos x}right)^{operatorname{ctg}^2 x} = expleft( operatorname{ctg}^2 x lnleft(6 - frac{5}{cos x}right) right) ]2. Transform the limit inside the exponent: Applying the limit, we get: [ lim_{x to 0}left(6 - frac{5}{cos x}right)^{operatorname{ctg}^2 x} = expleft(lim_{x to 0} operatorname{ctg}^2 x lnleft(6 - frac{5}{cos x}right) right) ]3. Use the equivalence operatorname{ctg} x = frac{1}{tan x} and tan x sim x as x to 0: Given that tan x sim x when x to 0, we have: [ operatorname{ctg}^2 x = left(frac{1}{tan x}right)^2 sim left(frac{1}{x}right)^2 = frac{1}{x^2} ]4. Rewrite using the equivalence: [ lim_{x to 0} operatorname{ctg}^2 x lnleft(6 - frac{5}{cos x}right) = lim_{x to 0} frac{lnleft(6 - frac{5}{cos x}right)}{x^2} ]5. Transform the logarithmic expression: We need to simplify lnleft(6 - frac{5}{cos x}right). Notice that: [ frac{1}{cos x} = 1 + frac{x^2}{2} + O(x^4) text{ as } x to 0 ] Therefore, [ 6 - frac{5}{cos x} = 6 - frac{5}{1 + frac{x^2}{2}} approx 6 - 5 left(1 - frac{x^2}{2}right) = 6 - 5 + frac{5x^2}{2} = 1 + frac{5x^2}{2} ]6. Apply the logarithm approximation ln(1 + u) approx u text{ for small } u: Hence, [ lnleft(6 - frac{5}{cos x}right) approx lnleft(1 + frac{5x^2}{2}right) approx frac{5x^2}{2} ]7. Simplify the ratio inside the limit: [ lim_{x to 0} frac{lnleft(6 - frac{5}{cos x}right)}{x^2} approx lim_{x to 0} frac{frac{5x^2}{2}}{x^2} = lim_{x to 0} frac{5}{2} = frac{5}{2} ]8. Calculate the exponential: [ expleft( frac{5}{2} right) = e^{frac{5}{2}} ]However, recalling our transformation step, there should be a negative in front of our operatorname{ctg}^2 x. Correcting, we have:[expleft( - frac{5}{2} right) = e^{ - frac{5}{2}}] Conclusion:The limit is:[boxed{e^{- frac{5}{2}}}]

🧠:Okay, let's tackle this problem step by step. The question is asking to simplify the expression √(1 - sin2θ) - √(1 + sin2θ) given that θ is in the interval [5π/4, 3π/2]. The answer choices are all in terms of sinθ or cosθ with coefficients ±2, so we need to manipulate the expression to match one of those forms.First, I remember that expressions involving square roots of 1 ± sin2θ can often be simplified using trigonometric identities. The key identity here might be the double-angle formula for sine: sin2θ = 2sinθcosθ. Let me note that down.So, the original expression is:√(1 - sin2θ) - √(1 + sin2θ)Let me rewrite the sin2θ terms using the identity:√(1 - 2sinθcosθ) - √(1 + 2sinθcosθ)Hmm, but how does that help? Maybe I need to express these square roots in terms of squares. There's another identity that 1 ± sin2θ can be written as (sinθ ± cosθ)². Let me check that:(sinθ + cosθ)² = sin²θ + 2sinθcosθ + cos²θ = 1 + 2sinθcosθ = 1 + sin2θSimilarly, (sinθ - cosθ)² = sin²θ - 2sinθcosθ + cos²θ = 1 - 2sinθcosθ = 1 - sin2θYes! That works. So, we can rewrite each square root as the absolute value of (sinθ ± cosθ). However, since we have square roots, we need to take the principal (non-negative) square root. Therefore:√(1 - sin2θ) = |sinθ - cosθ|√(1 + sin2θ) = |sinθ + cosθ|Therefore, the original expression becomes:|sinθ - cosθ| - |sinθ + cosθ|Now, the critical part is to determine the signs of sinθ - cosθ and sinθ + cosθ in the given interval θ ∈ [5π/4, 3π/2]. Let's analyze the values of sinθ and cosθ in this interval.The interval [5π/4, 3π/2] is in the third and fourth quadrants. Specifically, 5π/4 is in the third quadrant, and 3π/2 is on the negative y-axis. In the third quadrant (π < θ < 3π/2), both sinθ and cosθ are negative. In the fourth quadrant (3π/2 < θ < 2π), sinθ is negative and cosθ is positive. However, our interval ends at 3π/2, so θ ranges from 5π/4 (third quadrant) to 3π/2 (which is the boundary between third and fourth, but actually at 3π/2, sinθ is -1 and cosθ is 0).Let me consider the values of sinθ and cosθ in this interval. Let's pick a sample angle in [5π/4, 3π/2]. For example, θ = 5π/4: sinθ = -√2/2, cosθ = -√2/2. Then sinθ + cosθ = -√2, sinθ - cosθ = 0. Wait, at θ = 5π/4, sinθ - cosθ = (-√2/2) - (-√2/2) = 0? Hmm, that's interesting. Let me check that.Wait, sin(5π/4) = sin(π + π/4) = -sin(π/4) = -√2/2cos(5π/4) = cos(π + π/4) = -cos(π/4) = -√2/2So sinθ - cosθ = (-√2/2) - (-√2/2) = 0And sinθ + cosθ = (-√2/2) + (-√2/2) = -√2Similarly, take θ = 3π/2: sinθ = -1, cosθ = 0sinθ - cosθ = -1 - 0 = -1sinθ + cosθ = -1 + 0 = -1Wait, so between 5π/4 and 3π/2, let's see how sinθ - cosθ and sinθ + cosθ behave.At θ = 5π/4: sinθ - cosθ = 0, sinθ + cosθ = -√2At θ = 3π/2: sinθ - cosθ = -1 - 0 = -1, sinθ + cosθ = -1 + 0 = -1So, let's see for θ between 5π/4 and 3π/2:In the third quadrant (5π/4 to 3π/2), sinθ is negative, cosθ is negative. Let's see when θ increases from 5π/4 to 3π/2:At θ = 5π/4, cosθ = -√2/2, sinθ = -√2/2. As θ increases towards 3π/2, cosθ increases from -√2/2 to 0 (since approaching 3π/2 from the left, cosθ approaches 0 from the negative side). Similarly, sinθ decreases from -√2/2 to -1.Therefore, sinθ - cosθ: At θ = 5π/4, it's 0. As θ increases, sinθ becomes more negative (from -√2/2 to -1) and cosθ becomes less negative (from -√2/2 to 0). So sinθ - cosθ = (more negative) - (less negative) = more negative. Therefore, sinθ - cosθ is negative in (5π/4, 3π/2], and zero at θ=5π/4.Similarly, sinθ + cosθ: At θ=5π/4, it's -√2. As θ increases to 3π/2, sinθ decreases to -1 and cosθ increases to 0. So sinθ + cosθ becomes more negative (since sinθ is decreasing) but cosθ is increasing. Let's compute at θ=7π/4 (but wait 7π/4 is in the fourth quadrant, which is outside our interval). Wait, in our interval, cosθ is negative until θ=3π/2 where it's 0. So in the entire interval [5π/4, 3π/2], cosθ ≤ 0 (negative or zero). So sinθ + cosθ is the sum of two negatives (except at θ=3π/2, where cosθ=0, so sinθ + cosθ = -1 + 0 = -1). Therefore, sinθ + cosθ is always negative in [5π/4, 3π/2].Therefore, the absolute values can be resolved as follows:|sinθ - cosθ|: Since sinθ - cosθ is ≤ 0 in [5π/4, 3π/2], because at θ=5π/4 it's 0, and negative otherwise, the absolute value is -(sinθ - cosθ) = cosθ - sinθ.|sinθ + cosθ|: Since sinθ + cosθ is always negative in this interval (as we saw), the absolute value is -(sinθ + cosθ) = -sinθ - cosθ.Therefore, substituting back into the original expression:|sinθ - cosθ| - |sinθ + cosθ| = (cosθ - sinθ) - (-sinθ - cosθ) = cosθ - sinθ + sinθ + cosθ = 2cosθWait, but that would give 2cosθ, which is option D. But let's check with specific angles to see if this holds.First, at θ = 5π/4:Original expression: √(1 - sin2θ) - √(1 + sin2θ)Compute sin2θ at θ=5π/4: sin(5π/2) = sin(2π + π/2) = sin(π/2) = 1. Wait, wait, wait. Wait 2θ when θ=5π/4 is 2*(5π/4)=5π/2, which is equivalent to π/2 (since 5π/2 - 2π = π/2). So sin(5π/2) = 1. Therefore:√(1 - 1) - √(1 + 1) = √0 - √2 = 0 - √2 = -√2But according to our simplified expression 2cosθ at θ=5π/4: cos(5π/4) = -√2/2, so 2*(-√2/2) = -√2. Which matches. So that works.Now check θ=3π/2:Original expression: √(1 - sin2θ) - √(1 + sin2θ)sin2θ at θ=3π/2: sin(3π) = 0.So expression becomes √(1 - 0) - √(1 + 0) = √1 - √1 = 1 - 1 = 0Simplified expression: 2cosθ at θ=3π/2: cos(3π/2) = 0, so 2*0 = 0. That also matches.Another test at θ=7π/4: Wait, 7π/4 is 315 degrees, which is in the fourth quadrant, but our interval is up to 3π/2 (270 degrees). So let's pick θ=7π/4 is outside the interval. Let's pick θ=3π/2 - π/4 = 5π/4. Wait, that's the same as θ=5π/4, which we already checked.Alternatively, pick θ=7π/4 is not in the interval. Let me pick θ=11π/6 (which is 330 degrees) but that's also outside the interval. Wait, our interval is [5π/4, 3π/2], which is from 225 degrees to 270 degrees. So let's pick another angle in between, say θ= 2π - π/3 = 5π/3, but that's 300 degrees, still outside. Hmm, maybe θ= 7π/4 is 315 degrees, but that's not in the interval.Wait, perhaps take θ=5π/4 + π/6 = 5π/4 + π/6 = 15π/12 + 2π/12 = 17π/12. Which is 17π/12, which is 255 degrees. Let's compute sinθ and cosθ.sin(17π/12): Let's compute. 17π/12 is π + 5π/12. So sin is -sin(5π/12), and cos is -cos(5π/12).sin(5π/12) = sin(π/3 + π/12) = sin(π/3 + π/12) = sin(5π/12). Alternatively, use exact values. sin(5π/12) = sin(75 degrees) = (√6 + √2)/4. Similarly, cos(5π/12) = (√6 - √2)/4. Therefore:sinθ = - (√6 + √2)/4 ≈ -0.9659cosθ = - (√6 - √2)/4 ≈ -0.2588Therefore, sinθ - cosθ = -0.9659 - (-0.2588) = -0.7071sinθ + cosθ = -0.9659 + (-0.2588) = -1.2247So |sinθ - cosθ| = |-0.7071| = 0.7071, which is cosθ - sinθ ≈ -0.2588 - (-0.9659) ≈ 0.7071. Correct.|sinθ + cosθ| = |-1.2247| = 1.2247, which is -sinθ - cosθ ≈ 0.9659 + 0.2588 ≈ 1.2247. Correct.Therefore, the expression becomes 0.7071 - 1.2247 ≈ -0.5176.Now compute 2cosθ: 2*(-0.2588) ≈ -0.5176. Which matches. So that works.Therefore, the simplified expression is 2cosθ, which is option D.Wait, but the answer options are:(A) 2 sinθ(B) -2 sinθ(C) -2 cosθ(D) 2 cosθSo according to our calculation, it's 2cosθ, which is D.But hold on, when I did the absolute values, I concluded:|sinθ - cosθ| = cosθ - sinθ (because sinθ - cosθ ≤ 0)|sinθ + cosθ| = -sinθ - cosθ (because sinθ + cosθ ≤ 0)Therefore, the expression becomes (cosθ - sinθ) - (-sinθ - cosθ) = cosθ - sinθ + sinθ + cosθ = 2cosθ. So yes, that's correct.But wait, when θ is in [5π/4, 3π/2], cosθ is negative except at 3π/2 where it's zero. So 2cosθ would be negative or zero in this interval. Let's check θ=5π/4: 2cosθ = 2*(-√2/2) = -√2, which matches our previous result. θ=3π/2: 2*0 = 0, which also matches.But let me check another angle in the interval. Let's take θ=7π/4 is not in the interval, but θ=3π/2 - π/6 = 4π/3. Wait, 4π/3 is 240 degrees, which is in the interval [5π/4 (225 degrees), 3π/2 (270 degrees)]. Let's compute:θ=4π/3:sinθ = sin(4π/3) = -√3/2cosθ = cos(4π/3) = -1/2Compute original expression:√(1 - sin2θ) - √(1 + sin2θ)First, sin2θ = sin(8π/3) = sin(8π/3 - 2π) = sin(2π/3) = √3/2Thus:√(1 - √3/2) - √(1 + √3/2)Compute numerically:1 - √3/2 ≈ 1 - 0.866 ≈ 0.134√0.134 ≈ 0.3661 + √3/2 ≈ 1 + 0.866 ≈ 1.866√1.866 ≈ 1.366Thus, 0.366 - 1.366 ≈ -1.0Now compute 2cosθ: 2*(-1/2) = -1.0. Which matches. So that's correct.Therefore, the answer is D) 2cosθ.But wait, let me check again. Wait, if θ=4π/3, then 2θ=8π/3, which is equivalent to 2π/3. Sin(2π/3)=√3/2. So that's correct.Original expression: √(1 - √3/2) - √(1 + √3/2) ≈ sqrt(1 - 0.866) - sqrt(1 + 0.866) ≈ sqrt(0.134) - sqrt(1.866) ≈ 0.366 - 1.366 = -1.0Which equals 2cosθ = 2*(-1/2) = -1. Correct.So everything checks out. Therefore, the simplified expression is 2cosθ, which is option D.But wait, the answer choices include -2cosθ as option C. Wait, could there be a mistake?Wait, let's re-examine the absolute value signs.Given that θ is in [5π/4, 3π/2], we determined that sinθ - cosθ is ≤ 0 (from 0 at θ=5π/4 to negative in between and -1 at 3π/2). Therefore, |sinθ - cosθ| = -(sinθ - cosθ) = cosθ - sinθ.Similarly, sinθ + cosθ is always ≤ 0 in this interval. Therefore, |sinθ + cosθ| = -(sinθ + cosθ) = -sinθ - cosθ.Therefore, the expression becomes:(cosθ - sinθ) - (-sinθ - cosθ) = cosθ - sinθ + sinθ + cosθ = 2cosθ.Therefore, the answer is 2cosθ, which is option D.But let's check the answer choices again:(A) 2 sinθ(B) -2 sinθ(C) -2 cosθ(D) 2 cosθYes, D is 2cosθ, which is what we derived. However, when we computed at θ=5π/4, we got -√2, which is 2cosθ = 2*(-√2/2) = -√2. That's correct.At θ=3π/2, 2cosθ = 0, which matches.At θ=4π/3, 2cosθ = -1, which also matches.So seems correct.But wait, I recall sometimes when dealing with square roots and absolute values, depending on the quadrant, signs can flip. Let me verify once more.Alternatively, another approach is to use the formula for √(1 ± sin2θ). Another identity that might be useful is:√(1 ± sin2θ) = |sinθ ± cosθ|But as we did before. Then we need to handle the absolute values based on the quadrant.Another way is to factor the expression.Alternatively, let's consider squaring the entire expression, but that might complicate things because squaring can introduce extraneous solutions. However, given that we have to choose between the options, maybe we can test with specific θ values.Wait, but we already did that. All test cases worked for option D.But let me check if there is a different approach where maybe a sign error creeps in.Alternatively, consider writing 1 ± sin2θ as [sinθ ± cosθ]^2 as before, then taking square roots.But as we have already considered, since θ is in [5π/4, 3π/2], sinθ and cosθ are both negative (in third quadrant) or sinθ is negative and cosθ is zero (at 3π/2).Therefore, let's analyze sinθ - cosθ and sinθ + cosθ:In the interval θ ∈ [5π/4, 3π/2]:From θ=5π/4 to θ=3π/2:At θ=5π/4: sinθ = cosθ = -√2/2, so sinθ - cosθ = 0, sinθ + cosθ = -√2As θ increases past 5π/4 towards 3π/2:- sinθ becomes more negative (from -√2/2 to -1)- cosθ becomes less negative, increasing from -√2/2 to 0Therefore:sinθ - cosθ = (more negative) - (less negative) = more negative - less negative = more negative overall. So from 0 at 5π/4, it becomes negative as θ increases.Similarly, sinθ + cosθ remains negative because both are negative (until θ=3π/2 where cosθ=0, so sinθ + cosθ = -1 + 0 = -1)Therefore:sinθ - cosθ ≤ 0 in [5π/4, 3π/2], so |sinθ - cosθ| = -(sinθ - cosθ) = cosθ - sinθsinθ + cosθ ≤ 0 in [5π/4, 3π/2], so |sinθ + cosθ| = -(sinθ + cosθ) = -sinθ - cosθSubtracting these:|sinθ - cosθ| - |sinθ + cosθ| = (cosθ - sinθ) - (-sinθ - cosθ) = cosθ - sinθ + sinθ + cosθ = 2cosθTherefore, the simplified expression is 2cosθ, which is option D.All steps check out, test cases confirm, so I believe the answer is D) 2 cosθ.Final Answerboxed{D}

✅:Given (theta inleft[frac{5 pi}{4}, frac{3 pi}{2}right]), we need to simplify (sqrt{1 - sin 2theta} - sqrt{1 + sin 2theta}).First, we use the identities:[sqrt{1 - sin 2theta} = |sin theta - cos theta|][sqrt{1 + sin 2theta} = |sin theta + cos theta|]Next, we determine the signs of (sin theta - cos theta) and (sin theta + cos theta) in the interval (theta in left[frac{5 pi}{4}, frac{3 pi}{2}right]).1. For (theta) in the third quadrant (i.e., (frac{5 pi}{4} leq theta < frac{3 pi}{2})): - Both (sin theta) and (cos theta) are negative. - (sin theta - cos theta) is non-positive (negative or zero at (theta = frac{5 pi}{4})). - (sin theta + cos theta) is always negative.2. At (theta = frac{3 pi}{2}): - (sin theta = -1) and (cos theta = 0). - (sin theta - cos theta = -1 - 0 = -1). - (sin theta + cos theta = -1 + 0 = -1).Thus, we resolve the absolute values:[|sin theta - cos theta| = cos theta - sin theta quad text{(since (sin theta - cos theta leq 0))}][|sin theta + cos theta| = -(sin theta + cos theta) quad text{(since (sin theta + cos theta leq 0))}]Substituting these into the original expression:[sqrt{1 - sin 2theta} - sqrt{1 + sin 2theta} = (cos theta - sin theta) - (-(sin theta + cos theta))][= cos theta - sin theta + sin theta + cos theta][= 2 cos theta]Thus, the simplified expression is (boxed{D}).

🔑:: 1. Given (theta in left[frac{5 pi}{4}, frac{3 pi}{2}right]). 2. We need to simplify the expression (sqrt{1-sin 2theta} - sqrt{1+sin 2theta}).3. Let's use the double-angle identity for sine: [sin 2theta = 2 sin theta cos theta.]4. Given (sin 2theta), recognize that: [ 1 - sin 2theta = 1 - 2 sin theta cos theta, ] and [ 1 + sin 2theta = 1 + 2 sin theta cos theta. ]5. By analyzing the interval (theta in left[frac{5pi}{4}, frac{3pi}{2}right]): - In this interval, (theta) is in the third and fourth quadrants of the unit circle. - In the third and fourth quadrants, (cos theta) is non-positive (i.e., (cos theta leq 0)). - (theta) in these quadrants would also imply examining the sign of (sin theta), but we focus on the cosine because of the transformation properties.6. Recall (sqrt{a}) is always non-negative for (a geq 0). We can transform the expression using absolute values associated with cosine and sine: [ sqrt{1 - sin 2theta} = sqrt{1 - 2 sin theta cos theta}, ] and [ sqrt{1 + sin 2theta} = sqrt{1 + 2 sin theta cos theta}. ]7. Factor these equations: [ 1 - 2 sin theta cos theta = (cos theta - sin theta)^2, ] and [ 1 + 2 sin theta cos theta = (cos theta + sin theta)^2. ]8. Taking the square roots: [ sqrt{1 - sin 2theta} = |cos theta - sin theta|, ] and [ sqrt{1 + sin 2theta} = |cos theta + sin theta|. ]9. Simplify the expression (sqrt{1 - sin 2theta} - sqrt{1 + sin 2theta}) using these square roots: [ sqrt{1 - sin 2theta} - sqrt{1 + sin 2theta} = |cos theta - sin theta| - |cos theta + sin theta|. ]10. Because (cos theta) is non-positive in (left[frac{5pi}{4}, frac{3pi}{2}right]), ( cos theta - sin theta ) will yield non-positive results since ( - sin theta ) does not affect its absolute behavior: - (|cos theta - sin theta| = - (cos theta - sin theta) = sin theta - cos theta). - (|cos theta + sin theta| = - (cos theta + sin theta) = - cos theta - sin theta).11. Therefore: [ frac{1}{cos theta}((sin theta - cos theta) - (- cos theta - sin theta)) = 2 cos theta. ] Conclusion:The simplified expression is (2 cos theta).[boxed{D}]

🧠:Okay, let's see. I need to find the value of c, which is log(xy), given that log(x/2) = 0.5 and log(y/5) = 0.1. Hmm, logarithms. Let me remember the properties of logarithms. First, log(a/b) is equal to log a - log b. So maybe I can rewrite the given equations using that property. Let me try that. Starting with the first equation: log(x/2) = 0.5. Applying the logarithm property, this becomes log x - log 2 = 0.5. Similarly, the second equation log(y/5) = 0.1 becomes log y - log 5 = 0.1. Okay, so now I have two equations:1. log x - log 2 = 0.52. log y - log 5 = 0.1If I can find log x and log y from these equations, then I can add them together to get log x + log y, which is equal to log(xy) because log a + log b = log(ab). Then c would be that sum. Let me solve the first equation for log x. Adding log 2 to both sides:log x = 0.5 + log 2Similarly, for the second equation, adding log 5 to both sides:log y = 0.1 + log 5Now, let's compute log x + log y:log x + log y = (0.5 + log 2) + (0.1 + log 5) = 0.5 + 0.1 + log 2 + log 5 = 0.6 + log(2*5) [since log a + log b = log(ab)]Wait, 2*5 is 10, and log 10 is 1 if it's base 10. Assuming these are base 10 logarithms, which is common unless specified otherwise. So log 10 = 1. Therefore:log x + log y = 0.6 + 1 = 1.6Therefore, c = log(xy) = 1.6Wait, let me check if I made any mistakes here. Let's verify each step.First, the original equations:log(x/2) = 0.5 → log x - log 2 = 0.5. Correct.log(y/5) = 0.1 → log y - log 5 = 0.1. Correct.Solving for log x and log y:log x = 0.5 + log 2log y = 0.1 + log 5. Correct.Adding them:log x + log y = 0.5 + 0.1 + log 2 + log 5 = 0.6 + log(2*5) = 0.6 + log 10. Since log 10 is 1 (assuming base 10), total is 1.6. So c = 1.6. That seems right.Alternatively, maybe we can solve for x and y first and then compute log(xy). Let me try that approach to double-check.Starting with log(x/2) = 0.5. If we exponentiate both sides with base 10, we get x/2 = 10^0.5. Therefore, x = 2 * 10^0.5.Similarly, log(y/5) = 0.1 → y/5 = 10^0.1 → y = 5 * 10^0.1.Now, compute xy: (2 * 10^0.5) * (5 * 10^0.1) = 2*5 * 10^(0.5 + 0.1) = 10 * 10^0.6 = 10^(1 + 0.6) = 10^1.6Therefore, log(xy) = log(10^1.6) = 1.6. Yep, same result. So c = 1.6. Both methods give the same answer, so that gives me more confidence. Let me check once more.Another way: express log x and log y numerically. If we calculate log 2 and log 5.Wait, log 2 is approximately 0.3010 and log 5 is approximately 0.69897.So log x = 0.5 + 0.3010 ≈ 0.8010log y = 0.1 + 0.69897 ≈ 0.79897Adding them together: 0.8010 + 0.79897 ≈ 1.59997, which is approximately 1.6. So that's consistent.Therefore, c ≈ 1.6. Since the problem didn't specify to approximate, but given the original equations had exact decimal values (0.5 and 0.1), the answer is exact as 1.6, which is 8/5 or 16/10. So 1.6 is exact here.Therefore, c = 1.6. I think that's solid. All methods converge to the same answer, so I can be confident that's correct.

✅:To find ( c ) where ( log xy = c ), given ( log frac{x}{2} = 0.5 ) and ( log frac{y}{5} = 0.1 ):1. Apply logarithm properties to rewrite the equations: - ( log x - log 2 = 0.5 ) ( Rightarrow log x = 0.5 + log 2 ) - ( log y - log 5 = 0.1 ) ( Rightarrow log y = 0.1 + log 5 )2. Sum the logarithms of ( x ) and ( y ): [ log x + log y = (0.5 + log 2) + (0.1 + log 5) ] [ = 0.6 + (log 2 + log 5) ]3. Simplify using logarithmic identities: - ( log 2 + log 5 = log (2 cdot 5) = log 10 = 1 ) (since the logarithm is base 10).4. Combine the results: [ log x + log y = 0.6 + 1 = 1.6 ] Thus, ( log xy = c = 1.6 ).Answer: ( boxed{1.6} )

🔑:1. We start with the given logarithmic equations: [ log frac{x}{2} = 0.5 ] [ log frac{y}{5} = 0.1 ]2. We use the property of logarithms that states the logarithm of a quotient is the difference of the logarithms: [ log frac{x}{2} = log x - log 2 ] [ log frac{y}{5} = log y - log 5 ]3. Substitute these into the given equations: [ log x - log 2 = 0.5 quad text{(1)} ] [ log y - log 5 = 0.1 quad text{(2)} ]4. Solve each equation for (log x) and (log y): From (1): [ log x = 0.5 + log 2 ] From (2): [ log y = 0.1 + log 5 ]5. Next, we need to find (log (xy)): Using the property of logarithms that (log (ab) = log a + log b), we have: [ log (xy) = log x + log y ]6. Substitute the values of (log x) and (log y) from step 4: [ log (xy) = (0.5 + log 2) + (0.1 + log 5) ] Simplify the expression: [ log (xy) = 0.5 + log 2 + 0.1 + log 5 ] [ log (xy) = 0.5 + 0.1 + log 2 + log 5 ] [ log (xy) = 0.6 + (log 2 + log 5) ]7. Using the property of logarithms again that (log (ab) = log a + log b), we find: [ log 2 + log 5 = log (2 times 5) = log 10 ]8. We know that: [ log 10 = 1 ]9. Substitute back into our expression from step 6 and step 7: [ log (xy) = 0.6 + 1 ] [ log (xy) = 1.6 ]Conclusion:Therefore, the value of (c) is:[boxed{1.6}]

🧠:Okay, let's see. I need to find the probability that the random variable X takes a value in the interval (1,2) given its distribution function F(x). Hmm. I remember that the distribution function, also called the cumulative distribution function (CDF), gives the probability that X is less than or equal to a certain value. So F(x) = P(X ≤ x). Right, so if I want the probability that X is between 1 and 2, that's P(1 < X < 2). Wait, but since the interval is open, does it matter if it's open or closed? I think for continuous random variables, the probability at a single point is zero, so whether the interval is open or closed doesn't affect the probability. But let me confirm that. Yes, in continuous distributions, the probability that X is exactly 1 or exactly 2 is zero. So P(1 ≤ X ≤ 2) would be the same as P(1 < X < 2). Therefore, I can calculate it using the CDF. The formula for the probability that X is between a and b is F(b) - F(a). So applying that here, it should be F(2) - F(1). Let me check the given distribution function again. F(x) is defined as 0 when x ≤ 0, x/2 when 0 < x ≤ 2, and 1 when x > 2. So at x = 2, F(2) would be 2/2 = 1. And at x = 1, F(1) would be 1/2. Therefore, subtracting those gives 1 - 1/2 = 1/2. Wait, but let me make sure I'm not missing something here. The distribution function is piecewise defined. Let me verify each part. For x ≤ 0, F(x) = 0. Between 0 and 2, it's linear, x/2. And beyond 2, it's 1. So this looks like a uniform distribution on the interval (0, 2], right? Because the CDF increases linearly from 0 to 1 as x goes from 0 to 2. If that's the case, then the probability density function (PDF) would be the derivative of the CDF. The derivative of x/2 is 1/2, so the PDF f(x) = 1/2 for 0 < x ≤ 2, and 0 otherwise. Then the probability P(1 < X < 2) would be the integral of the PDF from 1 to 2, which is (2 - 1)*(1/2) = 1/2. That matches the earlier result. But just to double-check, maybe there's a point mass somewhere? For example, sometimes distribution functions have jumps, which correspond to discrete probabilities. Let me check if F(x) is continuous. At x = 0, F(0) is 0, and approaching from the right, the limit is 0/2 = 0, so continuous there. At x = 2, F(2) = 1, and approaching from above, it's also 1. So no jumps, meaning it's a continuous distribution. Therefore, the PDF is as I thought, 1/2 on (0,2), and the probability between 1 and 2 is indeed 1/2. Alternatively, thinking in terms of areas, since the PDF is a rectangle from 0 to 2 with height 1/2, the area from 1 to 2 is the width (1) times height (1/2), which is 0.5. Another way to see it: since the total area under the PDF from 0 to 2 is 1 (because 2*1/2 = 1), each half of the interval would contribute half the probability. So from 0 to 1 is 1/2, and 1 to 2 is another 1/2. Wait, but actually, no, that's only if the PDF is uniform over the entire interval. Wait, in this case, the PDF is uniform over (0,2), so the probability from a to b is (b - a)/2. So from 1 to 2, it's (2 - 1)/2 = 1/2. Exactly. So all methods converge to 1/2. Therefore, the probability that X takes a value in (1,2) is 1/2. But let me just make sure that there are no mistakes in interpreting the distribution function. For example, is the CDF defined as x/2 for 0 < x ≤ 2? So at x = 0, it's 0, but immediately to the right of 0, it's x/2. So it's continuous at 0. Similarly, at x = 2, it's 1, and for x > 2, it stays 1. So yes, no jumps. Therefore, all steps seem correct. The answer should be 1/2, or 0.5. Final AnswerThe probability that ( X ) takes a value in the interval ( (1,2) ) is boxed{dfrac{1}{2}}.

✅:To find the probability that the random variable ( X ) takes a value in the interval ( (1, 2) ), we use the given distribution function ( F(x) ).The distribution function ( F(x) ) is defined as:[F(x) = begin{cases}0 & text{if } x leq 0, frac{x}{2} & text{if } 0 < x leq 2, 1 & text{if } x > 2.end{cases}]For a continuous random variable, the probability that ( X ) lies in the interval ( (a, b) ) is given by ( F(b) - F(a) ). Here, we need to find ( P(1 < X < 2) ), which is ( F(2) - F(1) ).Calculating the values of the distribution function at the points 1 and 2:- At ( x = 2 ), ( F(2) = frac{2}{2} = 1 ).- At ( x = 1 ), ( F(1) = frac{1}{2} = 0.5 ).Subtracting these values gives:[P(1 < X < 2) = F(2) - F(1) = 1 - 0.5 = 0.5.]Thus, the probability that ( X ) takes a value in the interval ( (1, 2) ) is (boxed{dfrac{1}{2}}).

🔑:1. Identify the cumulative distribution function (CDF) for the given random variable X: [ F(x)=left{ begin{array}{lc} 0 & text{ if } x leq 0 frac{x}{2} & text{ if } 0 < x leq 2 1 & text{ if } x > 2 end{array} right. ]2. Identify the interval of interest: We need to find the probability that the random variable X takes on a value in the interval (1, 2). Mathematically, this is expressed as P(1 < X < 2).3. Determine the values of the CDF at the endpoints of the interval: According to the given CDF: begin{align*} F(1) &= frac{1}{2} F(2) &= frac{2}{2} = 1 end{align*}4. Use the properties of the CDF to find the probability: The probability that the random variable X lies within the interval (a, b) is given by: [ P(a < X < b) = F(b) - F(a) ]5. Apply the endpoints of the interval to the formula: [ P(1 < X < 2) = F(2) - F(1) ] Substitute the known values: [ P(1 < X < 2) = 1 - frac{1}{2} ]6. Calculate the probability: [ P(1 < X < 2) = frac{1}{2} = 0.5 ] Conclusion: Hence, the probability that the random variable X lies in the interval (1, 2) is: [ boxed{0.5} ]