answer：Okay, so I have this problem here: "The square root of ( x ) is greater than 3 and less than 5. How many integer values of ( x ) satisfy this condition?" Hmm, let me try to figure this out step by step.First, I know that the square root of a number ( x ) being between 3 and 5 means that ( x ) itself must be between the squares of those numbers. So, if ( sqrt{x} > 3 ), then squaring both sides should give me ( x > 9 ). Similarly, if ( sqrt{x} < 5 ), squaring both sides would give ( x < 25 ). So, putting those together, ( x ) must be greater than 9 and less than 25.Wait, but the problem is asking for integer values of ( x ). So, I need to find all the integers that are greater than 9 and less than 25. Let me list them out to make sure I don't miss any.Starting from the smallest integer greater than 9, which is 10, and going up to the largest integer less than 25, which is 24. So, the integers are: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24.Now, I need to count how many numbers that is. Let me count them one by one:10 (1), 11 (2), 12 (3), 13 (4), 14 (5), 15 (6), 16 (7), 17 (8), 18 (9), 19 (10), 20 (11), 21 (12), 22 (13), 23 (14), 24 (15).So, that's 15 numbers in total. Therefore, there are 15 integer values of ( x ) that satisfy the condition where the square root of ( x ) is greater than 3 and less than 5.Wait a second, let me double-check to make sure I didn't make a mistake. If ( x ) is between 9 and 25, not including 9 and 25, then the integers should start at 10 and end at 24. Counting from 10 to 24 inclusive, that's 15 numbers. Yeah, that seems right.I think I got it. So, the answer is 15 integer values.

answer：Okay, so I have this problem here: Find the smallest positive real number ( x ) such that ( lfloor x^2 rfloor - x lfloor x rfloor = 10 ). Hmm, that looks a bit intimidating at first, but let me try to break it down step by step.First, I remember that ( lfloor x rfloor ) denotes the floor function, which gives the greatest integer less than or equal to ( x ). So, if I let ( n = lfloor x rfloor ), then ( n ) is an integer, and ( x ) can be expressed as ( n + f ), where ( f ) is the fractional part of ( x ), meaning ( 0 leq f < 1 ). So, ( x = n + f ).Now, substituting ( x = n + f ) into the equation, let's compute ( x^2 ). That would be ( (n + f)^2 = n^2 + 2nf + f^2 ). The floor of ( x^2 ) is then ( lfloor n^2 + 2nf + f^2 rfloor ). Since ( n^2 ) is an integer, the floor function would just take the integer part of ( 2nf + f^2 ) and add it to ( n^2 ). So, ( lfloor x^2 rfloor = n^2 + lfloor 2nf + f^2 rfloor ).Next, let's compute ( x lfloor x rfloor ). That's ( (n + f) cdot n = n^2 + nf ). So, the equation given in the problem becomes:[lfloor x^2 rfloor - x lfloor x rfloor = (n^2 + lfloor 2nf + f^2 rfloor) - (n^2 + nf) = 10]Simplifying this, the ( n^2 ) terms cancel out:[lfloor 2nf + f^2 rfloor - nf = 10]Hmm, so ( lfloor 2nf + f^2 rfloor ) is an integer because the floor function always results in an integer. Also, 10 is an integer, so ( nf ) must also be an integer. Let me denote ( nf = k ), where ( k ) is an integer. Then the equation becomes:[lfloor 2k + f^2 rfloor - k = 10]Wait, hold on. If ( nf = k ), then ( 2nf = 2k ), right? So, substituting back, we have:[lfloor 2k + f^2 rfloor - k = 10]Which simplifies to:[lfloor 2k + f^2 rfloor = k + 10]Since ( f ) is between 0 and 1, ( f^2 ) is also between 0 and 1. Therefore, ( 2k + f^2 ) is between ( 2k ) and ( 2k + 1 ). So, the floor of ( 2k + f^2 ) is either ( 2k ) or ( 2k + 1 ), depending on whether ( f^2 ) is less than 1 or not. But since ( f^2 < 1 ), ( lfloor 2k + f^2 rfloor = 2k ) if ( f^2 < 1 ), which it always is. Wait, but that would mean:[2k = k + 10 implies k = 10]But hold on, that can't be right because if ( k = 10 ), then ( nf = 10 ). But ( n ) is an integer and ( f < 1 ), so ( nf < n ). Therefore, ( n ) must be greater than 10. Hmm, so maybe my earlier step was wrong.Let me go back. I had:[lfloor 2nf + f^2 rfloor - nf = 10]Which is:[lfloor 2nf + f^2 rfloor = nf + 10]Since ( 2nf + f^2 ) is between ( 2nf ) and ( 2nf + 1 ), the floor of that is either ( lfloor 2nf rfloor ) or ( lfloor 2nf rfloor + 1 ). So, ( lfloor 2nf + f^2 rfloor ) is either ( lfloor 2nf rfloor ) or ( lfloor 2nf rfloor + 1 ). Therefore, we have:Either:1. ( lfloor 2nf rfloor = nf + 10 ), or2. ( lfloor 2nf rfloor + 1 = nf + 10 )But ( lfloor 2nf rfloor ) is an integer, and ( nf ) is also an integer because we concluded earlier that ( nf ) must be an integer. Let me denote ( nf = k ), where ( k ) is an integer. Then, ( 2nf = 2k ), which is also an integer. Therefore, ( lfloor 2nf rfloor = 2k ). So, substituting back, we have:Either:1. ( 2k = k + 10 implies k = 10 ), or2. ( 2k + 1 = k + 10 implies k = 9 )So, ( k ) can be either 10 or 9.Case 1: ( k = 10 )Then, ( nf = 10 ). Since ( n ) is an integer and ( f < 1 ), ( n ) must be greater than 10. The smallest integer ( n ) can be is 11. So, ( f = frac{10}{11} ). Therefore, ( x = n + f = 11 + frac{10}{11} = frac{131}{11} ).Case 2: ( k = 9 )Then, ( nf = 9 ). Similarly, ( n ) must be an integer greater than 9. The smallest integer ( n ) can be is 10. So, ( f = frac{9}{10} ). Then, ( x = 10 + frac{9}{10} = frac{109}{10} = 10.9 ).Now, we need to check which of these solutions is valid. Let's test ( x = frac{131}{11} approx 11.909 ).Compute ( lfloor x^2 rfloor ):( x^2 = left( frac{131}{11} right)^2 = frac{17161}{121} approx 141.826 ). So, ( lfloor x^2 rfloor = 141 ).Compute ( x lfloor x rfloor = frac{131}{11} times 11 = 131 ).Then, ( lfloor x^2 rfloor - x lfloor x rfloor = 141 - 131 = 10 ). So, that works.Now, check ( x = frac{109}{10} = 10.9 ).Compute ( x^2 = (10.9)^2 = 118.81 ). So, ( lfloor x^2 rfloor = 118 ).Compute ( x lfloor x rfloor = 10.9 times 10 = 109 ).Then, ( lfloor x^2 rfloor - x lfloor x rfloor = 118 - 109 = 9 ). Hmm, that's only 9, not 10. So, this doesn't satisfy the equation.Therefore, ( x = frac{109}{10} ) is not a solution, but ( x = frac{131}{11} ) is.Wait, but is ( frac{131}{11} ) the smallest solution? Let's see if there are smaller ( n ) values.If ( k = 10 ), ( n ) must be at least 11. If ( k = 9 ), ( n ) must be at least 10, but that didn't work. What about ( k = 8 )?Wait, hold on. Earlier, I considered only ( k = 10 ) and ( k = 9 ) because of the two cases. But maybe I should consider more generally.Let me think again. We had:[lfloor 2nf + f^2 rfloor = nf + 10]Since ( nf = k ), an integer, and ( 2nf + f^2 = 2k + f^2 ). The floor of that is either ( 2k ) or ( 2k + 1 ). Therefore:Either:1. ( 2k = k + 10 implies k = 10 )2. ( 2k + 1 = k + 10 implies k = 9 )So, only ( k = 10 ) and ( k = 9 ) are possible. Since ( k = 9 ) didn't work, only ( k = 10 ) is valid, giving ( x = frac{131}{11} ).But wait, is there a smaller ( x ) with a higher ( n )? For example, if ( n = 12 ), then ( f = frac{10}{12} = frac{5}{6} ), so ( x = 12 + frac{5}{6} = frac{77}{6} approx 12.833 ). That's larger than ( frac{131}{11} approx 11.909 ), so it's not smaller.Similarly, ( n = 13 ) would give ( f = frac{10}{13} approx 0.769 ), so ( x approx 13.769 ), which is even larger.Therefore, the smallest ( x ) is indeed ( frac{131}{11} ).Wait, but let me double-check if there could be a smaller ( x ) with a different ( n ). Suppose ( n = 10 ), then ( f = frac{10}{10} = 1 ), but ( f < 1 ), so that's not allowed. So, ( n ) must be at least 11 for ( k = 10 ).Alternatively, if ( k = 10 ), ( n ) must be 11, as ( f = frac{10}{11} ). So, that seems to be the minimal ( x ).Just to be thorough, let me check ( n = 11 ) and ( f = frac{10}{11} ):Compute ( x = 11 + frac{10}{11} = frac{131}{11} approx 11.909 ).Compute ( x^2 approx (11.909)^2 approx 141.826 ), so ( lfloor x^2 rfloor = 141 ).Compute ( x lfloor x rfloor = 11.909 times 11 = 130.999 approx 131 ).So, ( lfloor x^2 rfloor - x lfloor x rfloor approx 141 - 131 = 10 ). Perfect, that works.Is there a smaller ( x ) with ( n = 11 ) but a smaller ( f )? For example, if ( f ) is slightly less than ( frac{10}{11} ), say ( f = frac{10}{11} - epsilon ), then ( x = 11 + frac{10}{11} - epsilon ). But then ( nf = 11 times (frac{10}{11} - epsilon) = 10 - 11epsilon ). Since ( nf ) must be an integer, ( 10 - 11epsilon ) must be integer. But ( epsilon ) is positive and very small, so ( 10 - 11epsilon ) is just less than 10, which is not an integer. Therefore, ( f ) cannot be less than ( frac{10}{11} ) because that would make ( nf ) less than 10, which is not allowed since ( nf = 10 ).Therefore, ( f ) must be exactly ( frac{10}{11} ), making ( x = frac{131}{11} ) the smallest solution.I think that's solid. I don't see any smaller ( x ) that satisfies the equation, so I'm confident that ( frac{131}{11} ) is the answer.

answer：Okay, so I have this problem where I'm given two sequences, ( (a_n) ) and ( (b_n) ), defined by the equation ( (2 + i)^n = a_n + b_n i ) for all integers ( n geq 0 ). I need to find the sum ( sum_{n=0}^infty frac{a_n b_n}{7^n} ). Hmm, let me try to figure this out step by step.First, I know that ( 2 + i ) is a complex number. Maybe I can express it in polar form because that might make raising it to the power ( n ) easier. To convert ( 2 + i ) to polar form, I need its modulus and argument.The modulus is ( |2 + i| = sqrt{2^2 + 1^2} = sqrt{4 + 1} = sqrt{5} ). Okay, so the modulus is ( sqrt{5} ).The argument ( theta ) is ( arctanleft(frac{1}{2}right) ). I remember that ( arctanleft(frac{1}{2}right) ) is some angle whose tangent is ( frac{1}{2} ). I might need this angle later, but for now, let me just denote it as ( theta ).So, in polar form, ( 2 + i = sqrt{5} (cos theta + i sin theta) ). Therefore, ( (2 + i)^n = (sqrt{5})^n (cos ntheta + i sin ntheta) ).From this, I can see that ( a_n = (sqrt{5})^n cos ntheta ) and ( b_n = (sqrt{5})^n sin ntheta ).Now, I need to compute the sum ( sum_{n=0}^infty frac{a_n b_n}{7^n} ). Let me substitute the expressions for ( a_n ) and ( b_n ):[sum_{n=0}^infty frac{a_n b_n}{7^n} = sum_{n=0}^infty frac{(sqrt{5})^n cos ntheta cdot (sqrt{5})^n sin ntheta}{7^n}]Simplifying the numerator:[(sqrt{5})^n cdot (sqrt{5})^n = (5)^n]So, the sum becomes:[sum_{n=0}^infty frac{5^n cos ntheta sin ntheta}{7^n}]I can factor out the ( 5^n ) and ( 7^n ):[sum_{n=0}^infty left( frac{5}{7} right)^n cos ntheta sin ntheta]Hmm, I remember that ( sin 2x = 2 sin x cos x ), so maybe I can use that identity here. Let me rewrite ( cos ntheta sin ntheta ):[cos ntheta sin ntheta = frac{1}{2} sin 2ntheta]So, substituting back into the sum:[sum_{n=0}^infty left( frac{5}{7} right)^n cdot frac{1}{2} sin 2ntheta = frac{1}{2} sum_{n=0}^infty left( frac{5}{7} right)^n sin 2ntheta]Now, I have a sum involving ( sin 2ntheta ). I think there's a formula for the sum of a series like ( sum_{n=0}^infty r^n sin nphi ). Let me recall that formula.I believe it is:[sum_{n=0}^infty r^n sin nphi = frac{r sin phi}{1 - 2 r cos phi + r^2}]Similarly, for ( sum_{n=0}^infty r^n cos nphi ), the formula is:[sum_{n=0}^infty r^n cos nphi = frac{1 - r cos phi}{1 - 2 r cos phi + r^2}]But in my case, I have ( sin 2ntheta ). So, if I let ( phi = 2theta ), then my sum becomes:[frac{1}{2} sum_{n=0}^infty left( frac{5}{7} right)^n sin n(2theta)]So, applying the formula for the sum of ( sum_{n=0}^infty r^n sin nphi ), where ( r = frac{5}{7} ) and ( phi = 2theta ), I get:[frac{1}{2} cdot frac{ left( frac{5}{7} right) sin 2theta }{ 1 - 2 left( frac{5}{7} right) cos 2theta + left( frac{5}{7} right)^2 }]Okay, so I need to compute ( sin 2theta ) and ( cos 2theta ). Since ( theta = arctanleft( frac{1}{2} right) ), I can use double-angle identities.First, let's find ( sin theta ) and ( cos theta ). From ( theta = arctanleft( frac{1}{2} right) ), we can imagine a right triangle where the opposite side is 1 and the adjacent side is 2, so the hypotenuse is ( sqrt{1^2 + 2^2} = sqrt{5} ).Therefore:[sin theta = frac{1}{sqrt{5}}, quad cos theta = frac{2}{sqrt{5}}]Now, using the double-angle identities:[sin 2theta = 2 sin theta cos theta = 2 cdot frac{1}{sqrt{5}} cdot frac{2}{sqrt{5}} = frac{4}{5}][cos 2theta = cos^2 theta - sin^2 theta = left( frac{2}{sqrt{5}} right)^2 - left( frac{1}{sqrt{5}} right)^2 = frac{4}{5} - frac{1}{5} = frac{3}{5}]Great, so ( sin 2theta = frac{4}{5} ) and ( cos 2theta = frac{3}{5} ).Substituting these back into the sum expression:[frac{1}{2} cdot frac{ left( frac{5}{7} right) cdot frac{4}{5} }{ 1 - 2 cdot frac{5}{7} cdot frac{3}{5} + left( frac{5}{7} right)^2 }]Let me simplify numerator and denominator step by step.First, the numerator:[left( frac{5}{7} right) cdot frac{4}{5} = frac{4}{7}]So, the numerator becomes ( frac{4}{7} ).Now, the denominator:Compute each term:1. ( 2 cdot frac{5}{7} cdot frac{3}{5} = 2 cdot frac{3}{7} = frac{6}{7} )2. ( left( frac{5}{7} right)^2 = frac{25}{49} )So, the denominator is:[1 - frac{6}{7} + frac{25}{49}]Compute ( 1 - frac{6}{7} ):[1 - frac{6}{7} = frac{1}{7}]Now, add ( frac{25}{49} ):[frac{1}{7} + frac{25}{49} = frac{7}{49} + frac{25}{49} = frac{32}{49}]So, the denominator is ( frac{32}{49} ).Putting it all together, the sum becomes:[frac{1}{2} cdot frac{ frac{4}{7} }{ frac{32}{49} } = frac{1}{2} cdot left( frac{4}{7} cdot frac{49}{32} right)]Simplify ( frac{4}{7} cdot frac{49}{32} ):[frac{4}{7} cdot frac{49}{32} = frac{4 cdot 49}{7 cdot 32} = frac{196}{224} = frac{49}{56} = frac{7}{8}]So, now we have:[frac{1}{2} cdot frac{7}{8} = frac{7}{16}]Therefore, the sum ( sum_{n=0}^infty frac{a_n b_n}{7^n} ) is ( frac{7}{16} ).Let me just double-check my steps to make sure I didn't make any mistakes.1. Expressed ( 2 + i ) in polar form correctly: modulus ( sqrt{5} ), argument ( arctan(1/2) ). That seems right.2. Used De Moivre's theorem to express ( (2 + i)^n ) as ( (sqrt{5})^n (cos ntheta + i sin ntheta) ). That's correct.3. Identified ( a_n ) and ( b_n ) correctly. Yes.4. Substituted into the sum and simplified to ( frac{1}{2} sum left( frac{5}{7} right)^n sin 2ntheta ). That looks good.5. Applied the formula for the sum of ( r^n sin nphi ). I think I used the correct formula here.6. Calculated ( sin 2theta ) and ( cos 2theta ) correctly using double-angle identities. Yes, got ( 4/5 ) and ( 3/5 ).7. Plugged into the formula, simplified numerator and denominator step by step. Each step seems correct, ended up with ( 7/16 ).I think all the steps are correct, so I'm confident the answer is ( frac{7}{16} ).**Final Answer**boxed{B}

answer：Alright, so I've got this vector problem here with four options, A through D. I need to figure out which ones are correct. Let me take it step by step.Starting with Option A: It says that 0 times vector m equals the zero vector. Hmm, okay. I remember that when you multiply a vector by a scalar, like 0, you get the zero vector. So, 0 times any vector should be the zero vector. That makes sense. So, A seems correct.Moving on to Option B: This one looks a bit more complicated. It says that vector a times the dot product of vectors n and m equals the dot product of vectors a and n times vector m. Wait, so it's saying a(n·m) = (a·n)m. Let me think about this. The dot product of two vectors gives a scalar, right? So, n·m is a scalar, and a·n is also a scalar. So, on both sides, we're multiplying a vector by a scalar. But does the order matter here? Like, does a times (n·m) equal (a·n) times m? I'm not sure. Maybe I should try plugging in some numbers or think of specific vectors to test this.Let's say vector a is [1, 0], vector n is [0, 1], and vector m is [1, 1]. Then, n·m would be 0*1 + 1*1 = 1. So, a(n·m) would be [1, 0] * 1 = [1, 0]. On the other hand, a·n is 1*0 + 0*1 = 0. So, (a·n)m would be 0 times [1, 1] = [0, 0]. Clearly, [1, 0] is not equal to [0, 0]. So, this shows that B is not always true. Therefore, B is incorrect.Now, Option C: It states that if the dot product of m and a equals the dot product of n and a, then m equals n. Hmm, so m·a = n·a implies m = n? I don't think that's necessarily true. The dot product measures the projection of one vector onto another. So, if both m and n have the same projection onto a, they don't have to be the same vectors. They could differ in components that are orthogonal to a. For example, suppose a is [1, 0], m is [2, 1], and n is [2, -1]. Then, m·a = 2*1 + 1*0 = 2, and n·a = 2*1 + (-1)*0 = 2. So, m·a = n·a, but m ≠ n. Therefore, C is incorrect.Finally, Option D: It says that if m and n are collinear and have the same direction, then the projection vector of m onto n is |m|/|n| times n. Okay, so projection of m onto n is given by (m·n / |n|²) * n. If m and n are collinear and in the same direction, then m = k*n for some positive scalar k. So, m·n = k*n·n = k|n|². Therefore, the projection would be (k|n|² / |n|²) * n = k*n. But since m = k*n, then k = |m| / |n| because |m| = |k|*|n|. Since they have the same direction, k is positive, so k = |m| / |n|. Therefore, the projection is (|m| / |n|) * n. That matches what D says. So, D is correct.So, summarizing:- A is correct.- B is incorrect.- C is incorrect.- D is correct.Therefore, the correct options are A and D.

answer：Okay, so I have this problem about sets A and B defined by quadratic inequalities, and I need to find their intersection and union when a is -4, and then determine the range of a for which the complement of A intersected with B equals B. Hmm, let's take it step by step.First, let's understand what set A is. It's defined as all real numbers x such that 2x² - 7x + 3 ≤ 0. That's a quadratic inequality. I remember that to solve quadratic inequalities, I need to find the roots of the quadratic equation and then determine the intervals where the quadratic expression is less than or equal to zero.So, let's solve 2x² - 7x + 3 = 0. I can use the quadratic formula: x = [7 ± √(49 - 24)] / 4, since the quadratic is 2x² -7x +3. The discriminant is 49 - 24, which is 25. So, √25 is 5. Therefore, the roots are (7 + 5)/4 = 12/4 = 3, and (7 - 5)/4 = 2/4 = 1/2. So, the roots are x = 1/2 and x = 3.Since the coefficient of x² is positive (2), the parabola opens upwards. That means the quadratic expression 2x² -7x +3 is ≤ 0 between the roots. So, set A is the interval [1/2, 3]. Got that.Now, set B is defined as all real numbers x such that x² + a < 0. Hmm, x² is always non-negative, so x² + a < 0 implies that a must be negative for this inequality to have any solutions. If a is non-negative, then x² + a is always ≥ 0, so B would be empty.But in the first part, a is given as -4. So, let's substitute a = -4 into B. Then, B becomes {x | x² - 4 < 0}, which simplifies to x² < 4. Taking square roots, we get |x| < 2, so x is in (-2, 2). So, set B is the open interval (-2, 2).Now, we need to find A ∩ B and A ∪ B.First, A is [1/2, 3], and B is (-2, 2). The intersection A ∩ B would be the overlap between these two intervals. So, the overlap starts at 1/2 and ends at 2. Since A includes 1/2 and 3, but B doesn't include 2, the intersection is [1/2, 2). Next, the union A ∪ B would be all numbers that are in either A or B or both. So, B starts at -2 and goes up to 2, and A starts at 1/2 and goes up to 3. So, combining these, the union starts at -2 and goes up to 3. Since B doesn't include -2 and A includes 3, the union is (-2, 3].Okay, that seems straightforward.Now, moving on to the second part. We need to find the range of real numbers a such that (complement of A) ∩ B = B.First, let's recall that the complement of A, denoted as ∁ℝ A, is all real numbers not in A. Since A is [1/2, 3], the complement is (-∞, 1/2) ∪ (3, ∞).We need (∁ℝ A) ∩ B = B. That means that B must be entirely contained within ∁ℝ A. In other words, every element of B must also be in ∁ℝ A.So, B is {x | x² + a < 0}. As I thought earlier, if a is non-negative, B is empty. If a is negative, B is (-√(-a), √(-a)).So, for B to be entirely contained within ∁ℝ A, which is (-∞, 1/2) ∪ (3, ∞), the interval B must not overlap with [1/2, 3]. That is, the entire interval B must lie either to the left of 1/2 or to the right of 3.But B is symmetric around 0 because it's defined by x² + a < 0, so it's always centered at 0. Therefore, the only way for B to lie entirely within ∁ℝ A is if B is entirely to the left of 1/2 or entirely to the right of 3. But since B is symmetric, it can't be entirely to the right of 3 unless it's also entirely to the left of -3, but that would require √(-a) ≥ 3, which would make B = (-∞, -√(-a)) ∪ (√(-a), ∞). But wait, no, actually, B is just (-√(-a), √(-a)).Wait, I need to correct myself. If a is negative, B is (-√(-a), √(-a)). So, for B to be entirely within ∁ℝ A, which is (-∞, 1/2) ∪ (3, ∞), the interval (-√(-a), √(-a)) must not overlap with [1/2, 3]. That means that √(-a) must be less than or equal to 1/2, because if √(-a) is less than or equal to 1/2, then the entire interval B is (-1/2, 1/2), which is entirely within (-∞, 1/2). Alternatively, if √(-a) is greater than or equal to 3, then B would be (-∞, -√(-a)) ∪ (√(-a), ∞), but since √(-a) is positive, this would mean B is (-∞, -something) ∪ (something, ∞). However, since B is defined as x² + a < 0, which is only the interval (-√(-a), √(-a)), not the union outside. Wait, no, actually, if a is negative, x² + a < 0 implies x² < -a, so x is between -√(-a) and √(-a). So, B is always the interval (-√(-a), √(-a)).Therefore, to have B entirely within ∁ℝ A, which is (-∞, 1/2) ∪ (3, ∞), we need that the entire interval (-√(-a), √(-a)) is either entirely to the left of 1/2 or entirely to the right of 3. But since B is symmetric around 0, it can't be entirely to the right of 3 unless it's also entirely to the left of -3, but that would require √(-a) ≥ 3, which would make B = (-∞, -√(-a)) ∪ (√(-a), ∞), but that's not the case. Wait, no, actually, if a is negative, B is just (-√(-a), √(-a)). So, if √(-a) ≤ 1/2, then B is (-1/2, 1/2), which is entirely within (-∞, 1/2). If √(-a) > 1/2, then B would extend beyond 1/2, overlapping with A, which we don't want.Alternatively, if √(-a) ≥ 3, then B would be (-∞, -3) ∪ (3, ∞), but that's not the case because B is just (-√(-a), √(-a)). Wait, no, if √(-a) is greater than 3, then B would be (-√(-a), √(-a)), which would include numbers greater than 3, but also less than -3, but since B is just the interval between -√(-a) and √(-a), it would include numbers from -√(-a) to √(-a). So, if √(-a) is greater than 3, then B would include numbers up to √(-a), which is greater than 3, but also down to -√(-a), which is less than -3. However, ∁ℝ A is (-∞, 1/2) ∪ (3, ∞). So, if √(-a) > 3, then B would include (3, √(-a)), which is part of ∁ℝ A, but it would also include (-√(-a), 1/2), which is also part of ∁ℝ A. Wait, no, because B is (-√(-a), √(-a)), so if √(-a) > 3, then B would be (-√(-a), √(-a)), which includes (-∞, 1/2) only if √(-a) > 3, but actually, no, because B is just the interval between -√(-a) and √(-a). So, if √(-a) > 3, then B would be (-√(-a), √(-a)), which would include numbers from -√(-a) to √(-a). But ∁ℝ A is (-∞, 1/2) ∪ (3, ∞). So, for B to be entirely within ∁ℝ A, B must not overlap with [1/2, 3]. Therefore, if √(-a) ≤ 1/2, then B is (-1/2, 1/2), which doesn't overlap with [1/2, 3], so it's entirely within (-∞, 1/2). Alternatively, if √(-a) ≥ 3, then B would be (-∞, -√(-a)) ∪ (√(-a), ∞), but that's not the case because B is just (-√(-a), √(-a)). Wait, I'm getting confused.Let me clarify. B is defined as {x | x² + a < 0}, which is equivalent to x² < -a. So, if a is negative, -a is positive, and x² < -a implies that x is between -√(-a) and √(-a). So, B is the interval (-√(-a), √(-a)).Now, ∁ℝ A is (-∞, 1/2) ∪ (3, ∞). So, for B to be entirely within ∁ℝ A, the interval (-√(-a), √(-a)) must be entirely contained within (-∞, 1/2) ∪ (3, ∞). But since B is a single interval centered at 0, the only way for it to be entirely within (-∞, 1/2) ∪ (3, ∞) is if the entire interval B is to the left of 1/2 or to the right of 3. But since B is symmetric around 0, it can't be entirely to the right of 3 unless it's also entirely to the left of -3, but that would require √(-a) ≥ 3, making B = (-∞, -3) ∪ (3, ∞), but that's not the case because B is just (-√(-a), √(-a)). Therefore, the only way for B to be entirely within ∁ℝ A is if B is entirely to the left of 1/2, meaning that √(-a) ≤ 1/2.So, √(-a) ≤ 1/2 implies that -a ≤ (1/2)² = 1/4, so a ≥ -1/4.But wait, if a is greater than or equal to -1/4, then B is either empty (if a ≥ 0) or (-√(-a), √(-a)) with √(-a) ≤ 1/2, which is entirely within (-∞, 1/2). Therefore, in both cases, B is entirely within ∁ℝ A.So, the range of a is a ≥ -1/4.Wait, but let me double-check. If a = -1/4, then B is (-1/2, 1/2), which is entirely within (-∞, 1/2). If a is greater than -1/4, say a = 0, then B is empty, which is trivially a subset of ∁ℝ A. If a is less than -1/4, say a = -1, then B is (-1, 1), which overlaps with [1/2, 3], so it's not entirely within ∁ℝ A. Therefore, the condition holds only when a ≥ -1/4.So, the range of a is a ≥ -1/4.

answer：To find the modular inverse of (5) modulo (31), we need to find an integer (a) such that (5a equiv 1 pmod{31}).We can approach this by checking integers of the form (31k + 1) for divisibility by (5):1. (31 times 0 + 1 = 1) (not divisible by (5))2. (31 times 1 + 1 = 32) (not divisible by (5))3. (31 times 2 + 1 = 63) (not divisible by (5))4. (31 times 3 + 1 = 94) (not divisible by (5))5. (31 times 4 + 1 = 125) (divisible by (5), since (125 = 5 times 25))Thus, (a = 25) satisfies (5 times 25 = 125 equiv 1 pmod{31}).Therefore, the modular inverse of (5) modulo (31) is (25).