answer：Alright, so I've got this problem about floor functions, and I need to figure out if two statements are true or false. Let me take it step by step.First, the problem defines the floor function, which I already know a bit about. The floor of a number x, denoted as [x], is the largest integer that is less than or equal to x. For example, [1.6] is 1, and [-1.6] is -2. Got it.Now, the first proposition is about a set A defined by the equation x² - [x] - 1 = 0, where x is between -1 and 2. It says that this set A is a singleton set, meaning it has only one element. I need to check if this is true.Let me think about how to approach this. Since [x] is involved, which is piecewise constant, I should probably break down the interval (-1, 2) into smaller intervals where [x] is constant. That way, I can solve the equation x² - [x] - 1 = 0 in each interval.So, the interval (-1, 2) can be broken down into:1. (-1, 0)2. [0, 1)3. [1, 2)Wait, actually, since [x] changes at integer values, I should consider the intervals:1. (-1, 0)2. [0, 1)3. [1, 2)But wait, at x = 0, [x] is 0, and at x = 1, [x] is 1. So, let's handle each interval separately.**Interval (-1, 0):**Here, [x] is -1 because any number between -1 and 0 has a floor of -1. So, substituting into the equation:x² - (-1) - 1 = x² + 1 - 1 = x² = 0So, x² = 0 implies x = 0. But x = 0 is not in (-1, 0), so no solution in this interval.**Interval [0, 1):**Here, [x] is 0. Substituting into the equation:x² - 0 - 1 = x² - 1 = 0So, x² = 1 implies x = ±1. But in [0, 1), x can only be 1, but 1 is not included in this interval (it's the next interval). So, no solution here either.**Interval [1, 2):**Here, [x] is 1. Substituting into the equation:x² - 1 - 1 = x² - 2 = 0So, x² = 2 implies x = √2 or x = -√2. Since we're in [1, 2), x = √2 is approximately 1.414, which is within [1, 2). So, x = √2 is a solution.Therefore, in the interval (-1, 2), the only solution is x = √2. So, set A has only one element, making it a singleton set. So, proposition ① is true.Now, onto proposition ②: For any real number x, [x] + [x + 1/2] = [2x]. I need to check if this holds for all x.Hmm, this seems a bit more complex. Let me think about different cases for x.First, let's recall that [x + 1/2] is essentially rounding x down to the nearest half-integer. So, adding [x] and [x + 1/2] might have some relation to [2x].But let's test some specific values to see if the equation holds.**Case 1: x is an integer.**Let x = n, where n is an integer.Then, [x] = n, [x + 1/2] = [n + 1/2] = n (since n + 1/2 is halfway between n and n+1, but the floor would still be n).So, [x] + [x + 1/2] = n + n = 2n.On the other hand, [2x] = [2n] = 2n.So, in this case, the equation holds.**Case 2: x is a half-integer, like x = n + 1/2, where n is an integer.**Then, [x] = [n + 1/2] = n.[x + 1/2] = [n + 1/2 + 1/2] = [n + 1] = n + 1.So, [x] + [x + 1/2] = n + (n + 1) = 2n + 1.[2x] = [2(n + 1/2)] = [2n + 1] = 2n + 1.So, the equation holds here as well.**Case 3: x is not an integer or half-integer. Let's consider x = n + b, where n is integer and 0 < b < 1, but b ≠ 1/2.**Let me split this into two subcases based on the value of b.**Subcase 3a: 0 < b < 1/2**Then, [x] = n.[x + 1/2] = [n + b + 1/2] = n + [b + 1/2]. Since 0 < b < 1/2, b + 1/2 is between 1/2 and 1, so [b + 1/2] = 0.Thus, [x + 1/2] = n + 0 = n.So, [x] + [x + 1/2] = n + n = 2n.Now, [2x] = [2(n + b)] = [2n + 2b]. Since 0 < b < 1/2, 0 < 2b < 1, so [2n + 2b] = 2n.Thus, [x] + [x + 1/2] = [2x] holds.**Subcase 3b: 1/2 < b < 1**Then, [x] = n.[x + 1/2] = [n + b + 1/2] = n + [b + 1/2]. Since 1/2 < b < 1, b + 1/2 is between 1 and 1.5, so [b + 1/2] = 1.Thus, [x + 1/2] = n + 1.So, [x] + [x + 1/2] = n + (n + 1) = 2n + 1.Now, [2x] = [2(n + b)] = [2n + 2b]. Since 1/2 < b < 1, 1 < 2b < 2, so [2n + 2b] = 2n + 1.Thus, [x] + [x + 1/2] = [2x] holds.So, in all cases, whether x is an integer, a half-integer, or a non-integer/non-half-integer, the equation [x] + [x + 1/2] = [2x] holds.Wait, but I should double-check if there are any edge cases or exceptions. For example, what if x is negative?Let me test x = -0.3.[x] = [-0.3] = -1.[x + 1/2] = [-0.3 + 0.5] = [0.2] = 0.So, [x] + [x + 1/2] = -1 + 0 = -1.[2x] = [2*(-0.3)] = [-0.6] = -1.So, it holds here.Another test: x = -0.6.[x] = [-0.6] = -1.[x + 1/2] = [-0.6 + 0.5] = [-0.1] = -1.So, [x] + [x + 1/2] = -1 + (-1) = -2.[2x] = [2*(-0.6)] = [-1.2] = -2.It holds again.What about x = -1.2?[x] = [-1.2] = -2.[x + 1/2] = [-1.2 + 0.5] = [-0.7] = -1.So, [x] + [x + 1/2] = -2 + (-1) = -3.[2x] = [2*(-1.2)] = [-2.4] = -3.Still holds.Hmm, seems consistent. Maybe proposition ② is also true.Wait, but I recall sometimes floor functions can have tricky behaviors. Let me think of another case where x is a negative half-integer, like x = -0.5.[x] = [-0.5] = -1.[x + 1/2] = [-0.5 + 0.5] = [0] = 0.So, [x] + [x + 1/2] = -1 + 0 = -1.[2x] = [2*(-0.5)] = [-1] = -1.Still holds.Another test: x = 0.75.[x] = 0.[x + 1/2] = [0.75 + 0.5] = [1.25] = 1.So, [x] + [x + 1/2] = 0 + 1 = 1.[2x] = [1.5] = 1.Holds.x = 1.3.[x] = 1.[x + 1/2] = [1.3 + 0.5] = [1.8] = 1.So, [x] + [x + 1/2] = 1 + 1 = 2.[2x] = [2.6] = 2.Holds.x = 1.6.[x] = 1.[x + 1/2] = [1.6 + 0.5] = [2.1] = 2.So, [x] + [x + 1/2] = 1 + 2 = 3.[2x] = [3.2] = 3.Holds.Wait, so in all these test cases, the equation holds. Maybe proposition ② is indeed true.But wait, I should think about the general proof rather than just testing specific cases.Let me try to prove that [x] + [x + 1/2] = [2x] for any real x.Let x = n + f, where n is an integer and 0 ≤ f < 1.Then, [x] = n.Now, x + 1/2 = n + f + 1/2.Depending on f, [x + 1/2] can be n or n + 1.If f < 1/2, then f + 1/2 < 1, so [x + 1/2] = n.If f ≥ 1/2, then f + 1/2 ≥ 1, so [x + 1/2] = n + 1.Similarly, 2x = 2n + 2f.If 0 ≤ f < 1/2, then 0 ≤ 2f < 1, so [2x] = 2n.If 1/2 ≤ f < 1, then 1 ≤ 2f < 2, so [2x] = 2n + 1.So, let's see:If f < 1/2:[x] + [x + 1/2] = n + n = 2n.[2x] = 2n.So, equal.If f ≥ 1/2:[x] + [x + 1/2] = n + (n + 1) = 2n + 1.[2x] = 2n + 1.So, equal.Therefore, in both cases, [x] + [x + 1/2] = [2x].Hence, proposition ② is true.Wait, but I thought sometimes floor functions can have exceptions, but in this case, it seems to hold for all real numbers. So, both propositions ① and ② are true.But let me double-check proposition ① once more to be sure.We had x² - [x] - 1 = 0 in (-1, 2).We broke it into intervals:- (-1, 0): No solution.- [0, 1): No solution.- [1, 2): Solution at x = √2.So, only one solution, making A a singleton set. So, ① is true.Therefore, both propositions are true. So, the correct answer is A.But wait, the options are:A: Both ① and ② are true statementsB: ① is true and ② is falseC: ① is false and ② is trueD: Both ① and ② are falseSo, since both are true, the answer is A.Wait, but I just want to make sure I didn't miss anything in proposition ②.Is there any x where [x] + [x + 1/2] ≠ [2x]?From the general proof above, it seems not. Because depending on whether f is less than 1/2 or not, the equation holds.Therefore, I think both propositions are indeed true.**Final Answer**boxed{A}

answer：Alright, so I have this problem here about sets A and B. Let me read it again: Set A is {-1, 3, m²} and set B is {3, 2m - 1}. It says that B is a subset of A, and I need to find the value of m. Hmm, okay.First, I remember that if B is a subset of A, that means every element in B must also be in A. So, the elements of B are 3 and 2m - 1. I know that 3 is already in A, so that part is fine. The tricky part is making sure that 2m - 1 is also in A. Looking at set A, the elements are -1, 3, and m². So, 2m - 1 has to be equal to one of these three elements. That means either 2m - 1 = -1, or 2m - 1 = 3, or 2m - 1 = m². Let me write that down:1. 2m - 1 = -12. 2m - 1 = 33. 2m - 1 = m²Now, I'll solve each equation one by one.Starting with the first equation: 2m - 1 = -1. Let's solve for m.2m - 1 = -1 Add 1 to both sides: 2m = 0 Divide both sides by 2: m = 0Okay, so m could be 0. Let me check if this makes sense. If m is 0, then set A becomes {-1, 3, 0²} which is {-1, 3, 0}. Set B becomes {3, 2*0 - 1} which is {3, -1}. So, B is {3, -1}, and A is {-1, 3, 0}. Yes, every element in B is in A, so m = 0 is a valid solution.Next, the second equation: 2m - 1 = 3.2m - 1 = 3 Add 1 to both sides: 2m = 4 Divide both sides by 2: m = 2Alright, m could also be 2. Let me verify this. If m is 2, then set A becomes {-1, 3, 2²} which is {-1, 3, 4}. Set B becomes {3, 2*2 - 1} which is {3, 3}. Wait, {3, 3} is just {3}, right? So, B is {3}, and A is {-1, 3, 4}. Since 3 is in A, B is a subset of A. So, m = 2 is also a valid solution.Now, onto the third equation: 2m - 1 = m².This looks a bit more complicated. Let me rearrange it to form a quadratic equation.2m - 1 = m² Subtract 2m and add 1 to both sides: 0 = m² - 2m + 1So, m² - 2m + 1 = 0. Hmm, this looks familiar. It's a quadratic equation, and I can try to factor it.m² - 2m + 1 = (m - 1)² = 0So, the solution is m = 1. Let me check this. If m is 1, then set A becomes {-1, 3, 1²} which is {-1, 3, 1}. Set B becomes {3, 2*1 - 1} which is {3, 1}. So, B is {3, 1}, and A is {-1, 3, 1}. Both elements of B are in A, so m = 1 is also a valid solution.Wait a minute, so I have three possible solutions: m = 0, m = 2, and m = 1. But let me double-check if all of these are valid.For m = 0: A = {-1, 3, 0}, B = {3, -1}. B is a subset of A. Good.For m = 2: A = {-1, 3, 4}, B = {3}. B is a subset of A. Good.For m = 1: A = {-1, 3, 1}, B = {3, 1}. B is a subset of A. Good.So, all three values of m satisfy the condition that B is a subset of A. But wait, the problem is asking for "m = ?", implying there might be multiple answers. So, m can be 0, 1, or 2.But hold on, let me think again. When m = 2, set B becomes {3}, which is a subset of A. But in the original problem, set B is {3, 2m - 1}. If m = 2, then 2m - 1 is 3, so B becomes {3, 3}, which is just {3}. Is that acceptable? Yes, because in set theory, duplicate elements are ignored, so {3, 3} is the same as {3}, which is still a subset of A.Similarly, when m = 0, B becomes {3, -1}, which are both in A. And when m = 1, B becomes {3, 1}, which are also in A. So, all three values are valid.But wait, let me check if m = 1 is necessary. Because when m = 1, 2m - 1 = 1, which is equal to m² = 1. So, in this case, B becomes {3, 1}, and A already has 1 because m² = 1. So, that's fine.But hold on, is there any restriction on m? The problem doesn't specify any, so m can be any real number. So, all three solutions are valid.But wait, in the initial equations, when I set 2m - 1 equal to each element of A, I considered -1, 3, and m². But m² is dependent on m, so when I set 2m - 1 = m², I might be introducing an equation that could have multiple solutions, but in this case, it only gave m = 1.So, to summarize, the possible values of m are 0, 1, and 2. But let me check if m = 2 is necessary or if it's already covered by another solution.Wait, no, m = 2 is a separate solution because when m = 2, 2m - 1 = 3, which is already in A, so B becomes {3}, which is a subset of A. So, m = 2 is a valid solution.But hold on, in the original problem, set A is {-1, 3, m²}, and set B is {3, 2m - 1}. So, when m = 2, m² is 4, which is a new element in A, but since B only has 3, it's still a subset.So, all three values are valid. But wait, in the initial solving, when I set 2m - 1 equal to each element of A, I considered -1, 3, and m², but m² is dependent on m, so it's possible that when m = 1, m² = 1, which is another element in A.So, I think all three solutions are correct.But wait, let me think again. When m = 0, 2m - 1 = -1, which is in A. When m = 1, 2m - 1 = 1, which is in A because m² = 1. When m = 2, 2m - 1 = 3, which is already in A. So, all three are valid.But wait, in the original problem, set A is {-1, 3, m²}, and set B is {3, 2m - 1}. So, when m = 2, set A becomes {-1, 3, 4}, and set B becomes {3}. So, B is a subset of A.Similarly, when m = 0, A is {-1, 3, 0}, and B is {3, -1}, which is a subset.When m = 1, A is {-1, 3, 1}, and B is {3, 1}, which is a subset.So, all three values are correct. Therefore, m can be 0, 1, or 2.But wait, in the initial solving, I only considered m = 0, m = 2, and m = 1. But let me check if there are any other possible solutions.Wait, when I set 2m - 1 equal to each element of A, I considered -1, 3, and m². But m² is dependent on m, so when I set 2m - 1 = m², I get m = 1. So, that's the only solution from that equation.Similarly, setting 2m - 1 = -1 gives m = 0, and setting 2m - 1 = 3 gives m = 2.So, those are the only three possible solutions.But wait, let me think about m = 1 again. When m = 1, m² = 1, so set A becomes {-1, 3, 1}, and set B becomes {3, 1}. So, B is a subset of A.But what if m = -1? Let me check. If m = -1, then set A becomes {-1, 3, (-1)²} which is {-1, 3, 1}. Set B becomes {3, 2*(-1) - 1} which is {3, -3}. So, B is {3, -3}, but -3 is not in A, which is {-1, 3, 1}. So, B is not a subset of A in this case. Therefore, m = -1 is not a valid solution.Similarly, if m = 3, set A becomes {-1, 3, 9}, and set B becomes {3, 5}. 5 is not in A, so B is not a subset of A.So, m = 3 is invalid.Similarly, m = 0.5, let's see: set A becomes {-1, 3, 0.25}, set B becomes {3, 0}. 0 is not in A, so B is not a subset of A.So, m = 0.5 is invalid.Therefore, the only valid solutions are m = 0, m = 1, and m = 2.But wait, in the initial problem, the user wrote:"Given the sets A = {-1, 3, m²} and B = {3, 2m - 1}, if B ⊆ A, then m = ?"So, the answer is m = 0, 1, or 2.But in the initial assistant's answer, it only considered m = 0 and m = 1, but not m = 2. So, perhaps the assistant missed m = 2.Wait, let me check the assistant's thought process.The assistant wrote:"Given the sets A = {-1, 3, m²} and B = {3, 2m - 1}, if B ⊆ A, then m = ?Since A = {-1, 3, m²} and B = {3, 2m - 1}, and B ⊆ A,it follows that 2m - 1 = -1 or 2m - 1 = m²,solving these equations gives m = 0 or m = 1.After verification, both solutions satisfy the conditions of the problem.Therefore, the answer is: 0 or 1.Analysis: From B ⊆ A, we can derive 2m - 1 = -1 or 2m - 1 = m². Solving and verifying these equations yields the solution.Thus, the final answer is boxed{0 text{ or } 1}."Wait, so the assistant only considered 2m - 1 = -1 and 2m - 1 = m², but not 2m - 1 = 3. So, m = 2 was missed.Therefore, the assistant's answer is incomplete because it didn't consider the case where 2m - 1 = 3, leading to m = 2.So, in my own solving, I considered all three cases and found m = 0, 1, 2.Therefore, the correct answer should include m = 0, 1, and 2.But let me think again. When m = 2, set A becomes {-1, 3, 4}, and set B becomes {3, 3}, which is {3}, a subset of A. So, m = 2 is valid.But in the assistant's solution, they only considered 2m - 1 = -1 and 2m - 1 = m², missing the case where 2m - 1 = 3.Therefore, the correct answer should be m = 0, 1, or 2.But wait, let me check if m = 2 is necessary or if it's already covered by another solution.No, m = 2 is a separate solution because it's derived from setting 2m - 1 = 3, which is an element of A.Therefore, the complete solution should include m = 0, 1, and 2.So, in conclusion, the values of m are 0, 1, and 2.

answer：First, I need to determine the total number of balls Matthias has. He has 40 soccer balls and 15 basketballs, which adds up to 55 balls in total.Next, I know that 30 soccer balls have a hole in them. To find out how many soccer balls do not have holes, I subtract the number of soccer balls with holes from the total number of soccer balls: 40 - 30 = 10 soccer balls without holes.Matthias has 18 balls in total without holes. Since 10 of these are soccer balls, the remaining balls without holes must be basketballs. Therefore, the number of basketballs without holes is 18 - 10 = 8 basketballs.Finally, to find out how many basketballs have holes, I subtract the number of basketballs without holes from the total number of basketballs: 15 - 8 = 7 basketballs with holes.

answer：First, I need to understand the relationship between "α is an acute angle" and "sinα > 0". An acute angle is an angle between 0 and 90 degrees, which places it in the first quadrant of the unit circle. In the first quadrant, all trigonometric functions, including sine, are positive. Therefore, if α is an acute angle, it logically follows that sinα > 0.Next, I should consider the converse: if sinα > 0, does it necessarily mean that α is an acute angle? Sine is positive not only in the first quadrant (acute angles) but also in the second quadrant, where angles range from 90 to 180 degrees. Therefore, sinα > 0 does not exclusively imply that α is an acute angle.Based on this analysis, "α is an acute angle" is a sufficient condition for "sinα > 0" because it guarantees the truth of the latter statement. However, it is not a necessary condition because there are other angles where sinα > 0 without α being acute.Thus, the correct answer is that "α is an acute angle" is a sufficient but not necessary condition for "sinα > 0".

answer：Alright, so I've got this inequality to solve: (12x^2 - ax - a^2 < 0), where (a) is a real number. Hmm, okay. Let me think about how to approach this step by step.First, I remember that solving quadratic inequalities often involves finding the roots of the corresponding quadratic equation and then determining the intervals where the quadratic expression is positive or negative. So, maybe I should start by solving the equation (12x^2 - ax - a^2 = 0).To solve this quadratic equation, I can try factoring it. Let me see if that's possible. The quadratic is (12x^2 - ax - a^2). I need two numbers that multiply to (12 times (-a^2) = -12a^2) and add up to (-a). Hmm, that might be a bit tricky, but let me give it a shot.Wait, maybe I can factor by grouping. Let me rewrite the quadratic:(12x^2 - ax - a^2)Let me split the middle term. I need two terms that multiply to (12x^2 times (-a^2)) which is (-12a^2x^2), but that doesn't seem right. Maybe I should try a different approach.Alternatively, I can use the quadratic formula to find the roots. The quadratic formula is (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where in this case, the quadratic is (12x^2 - ax - a^2), so (a = 12), (b = -a), and (c = -a^2).Plugging these into the quadratic formula:(x = frac{-(-a) pm sqrt{(-a)^2 - 4 times 12 times (-a^2)}}{2 times 12})Simplify that:(x = frac{a pm sqrt{a^2 + 48a^2}}{24})(x = frac{a pm sqrt{49a^2}}{24})Since (sqrt{49a^2} = 7|a|), but since (a) is a real number, we can write it as (7a) if (a) is positive or (-7a) if (a) is negative. Hmm, but maybe I can just keep it as (7a) for now.So,(x = frac{a pm 7a}{24})This gives two solutions:1. (x = frac{a + 7a}{24} = frac{8a}{24} = frac{a}{3})2. (x = frac{a - 7a}{24} = frac{-6a}{24} = -frac{a}{4})So, the roots are (x = frac{a}{3}) and (x = -frac{a}{4}).Wait, that seems a bit off. Let me double-check my factoring approach to see if I can get the same roots.Let me try factoring (12x^2 - ax - a^2). I need two numbers that multiply to (12 times (-a^2) = -12a^2) and add up to (-a). Hmm, let's see:Looking for factors of (-12a^2) that add up to (-a). Maybe ( -4a ) and (3a), because (-4a + 3a = -a) and (-4a times 3a = -12a^2). Yes, that works.So, I can rewrite the quadratic as:(12x^2 - ax - a^2 = 12x^2 - 4a x + 3a x - a^2)Grouping terms:((12x^2 - 4a x) + (3a x - a^2))Factor out common terms:(4x(3x - a) + a(3x - a))Now, factor out the common binomial:((4x + a)(3x - a))So, the factored form is ((4x + a)(3x - a)). Setting each factor equal to zero:1. (4x + a = 0 Rightarrow x = -frac{a}{4})2. (3x - a = 0 Rightarrow x = frac{a}{3})Okay, so that confirms the roots I found earlier using the quadratic formula. Good, so the roots are (x = -frac{a}{4}) and (x = frac{a}{3}).Now, since the quadratic is (12x^2 - ax - a^2), the coefficient of (x^2) is positive (12), which means the parabola opens upwards. Therefore, the quadratic expression will be negative between its two roots.So, the inequality (12x^2 - ax - a^2 < 0) will hold true for (x) values between the smaller root and the larger root.But wait, the order of the roots depends on the value of (a). If (a) is positive, then (frac{a}{3}) is positive and (-frac{a}{4}) is negative, so (-frac{a}{4} < frac{a}{3}). If (a) is negative, then (frac{a}{3}) is negative and (-frac{a}{4}) is positive, so (frac{a}{3} < -frac{a}{4}). If (a = 0), both roots are zero, so the quadratic becomes (12x^2 < 0), which is only true if (x) is not zero, but since the coefficient is positive, (12x^2) is always non-negative, so the inequality (12x^2 < 0) has no solution.So, let's break it down into cases based on the value of (a):**Case 1: (a > 0)**In this case, the roots are (-frac{a}{4}) and (frac{a}{3}), with (-frac{a}{4} < frac{a}{3}). Since the parabola opens upwards, the expression is negative between the roots. So, the solution is:(-frac{a}{4} < x < frac{a}{3})**Case 2: (a = 0)**Here, the quadratic becomes (12x^2 < 0). Since (x^2) is always non-negative, (12x^2) is non-negative, so the inequality (12x^2 < 0) has no solution. Thus, the solution set is empty.**Case 3: (a < 0)**In this case, (a) is negative, so (frac{a}{3}) is negative and (-frac{a}{4}) is positive. Therefore, (frac{a}{3} < -frac{a}{4}). Again, since the parabola opens upwards, the expression is negative between the roots. So, the solution is:(frac{a}{3} < x < -frac{a}{4})Wait, but let me make sure I'm not making a mistake here. If (a) is negative, say (a = -b) where (b > 0), then the roots become:(-frac{a}{4} = -frac{-b}{4} = frac{b}{4})(frac{a}{3} = frac{-b}{3} = -frac{b}{3})So, indeed, (-frac{b}{3} < frac{b}{4}), which means the interval is (-frac{b}{3} < x < frac{b}{4}). But since (a = -b), this translates back to (frac{a}{3} < x < -frac{a}{4}), which is consistent with what I wrote earlier.So, putting it all together:- If (a > 0), the solution is (-frac{a}{4} < x < frac{a}{3}).- If (a = 0), there's no solution.- If (a < 0), the solution is (frac{a}{3} < x < -frac{a}{4}).I think that covers all possible cases for (a). Let me just double-check by plugging in some values.**Test Case 1: (a = 12) (positive)**The inequality becomes (12x^2 - 12x - 144 < 0). Let's find the roots:(x = -frac{12}{4} = -3) and (x = frac{12}{3} = 4). So, the solution should be (-3 < x < 4). Let's pick (x = 0), which is in the interval. Plugging into the inequality: (0 - 0 - 144 = -144 < 0), which is true. Now, pick (x = 5), which is outside the interval: (12(25) - 12(5) - 144 = 300 - 60 - 144 = 96 > 0), which is correct. Similarly, (x = -4): (12(16) - 12(-4) - 144 = 192 + 48 - 144 = 96 > 0), which is also correct.**Test Case 2: (a = 0)**The inequality becomes (12x^2 < 0). As expected, there's no solution because (x^2) is always non-negative, so (12x^2) is non-negative, and thus cannot be less than zero.**Test Case 3: (a = -12) (negative)**The inequality becomes (12x^2 - (-12)x - (-12)^2 = 12x^2 + 12x - 144 < 0). The roots are:(x = -frac{-12}{4} = 3) and (x = frac{-12}{3} = -4). So, the solution should be (-4 < x < 3). Let's test (x = 0): (0 + 0 - 144 = -144 < 0), which is true. Now, (x = 4): (12(16) + 12(4) - 144 = 192 + 48 - 144 = 96 > 0), which is correct. Similarly, (x = -5): (12(25) + 12(-5) - 144 = 300 - 60 - 144 = 96 > 0), which is also correct.Everything seems to check out. So, the solution depends on the value of (a), and we've covered all cases.

answer：First, I need to determine the initial population of the village, which I'll denote as ( P ).5% of the population died due to bombardment, so the remaining population after the bombardment is 95% of ( P ), which can be represented as ( 0.95P ).Next, 15% of the remaining population left the village due to fear. This means that 85% of the population that remained after the bombardment stayed in the village. Therefore, the population after people left due to fear is 85% of ( 0.95P ), which can be represented as ( 0.85 times 0.95P ).We are given that the population after these events is 2553. So, I can set up the equation:[0.85 times 0.95P = 2553]Simplifying the left side:[0.8075P = 2553]To find ( P ), I'll divide both sides of the equation by 0.8075:[P = frac{2553}{0.8075}]Calculating the division:[P approx 3162.01581028]Since the population of a village can't be a fraction, I'll round to the nearest whole number:[P approx 3162]Therefore, the initial population of the village was approximately 3162 people.