answer：First, identify the given set ( S = {1, 2} ).A proper subset of a set is any subset that is not equal to the set itself.To find the number of proper subsets, calculate the total number of subsets using the formula ( 2^n ), where ( n ) is the number of elements in the set.For ( S ), ( n = 2 ), so the total number of subsets is ( 2^2 = 4 ).Subtract 1 to exclude the set itself, resulting in ( 4 - 1 = 3 ) proper subsets.Therefore, the number of proper subsets ( m ) is 3.

answer：Okay, so I have two problems to solve here. The first one is about proving an inequality involving square roots, and the second one is about proving that at least one of three given expressions isn't less than 1. Let me tackle them one by one.Starting with the first problem: Prove that when ( a > 2 ), ( sqrt{a+2} + sqrt{a-2} < 2sqrt{a} ).Hmm, okay. So, I need to show that the sum of these two square roots is less than twice the square root of ( a ). Since ( a > 2 ), both ( a+2 ) and ( a-2 ) are positive, so the square roots are defined. That's good.Maybe I can square both sides to eliminate the square roots. If I square both sides, I have to be careful because squaring inequalities can sometimes introduce issues, but since both sides are positive, it should be okay.Let me write that down:( (sqrt{a+2} + sqrt{a-2})^2 < (2sqrt{a})^2 )Expanding the left side:( (sqrt{a+2})^2 + 2sqrt{a+2}sqrt{a-2} + (sqrt{a-2})^2 < 4a )Simplify the squares:( (a + 2) + 2sqrt{(a+2)(a-2)} + (a - 2) < 4a )Combine like terms:( a + 2 + a - 2 + 2sqrt{a^2 - 4} < 4a )Simplify further:( 2a + 2sqrt{a^2 - 4} < 4a )Subtract ( 2a ) from both sides:( 2sqrt{a^2 - 4} < 2a )Divide both sides by 2:( sqrt{a^2 - 4} < a )Now, since ( a > 2 ), ( a ) is positive, so squaring both sides should preserve the inequality:( a^2 - 4 < a^2 )Subtract ( a^2 ) from both sides:( -4 < 0 )Which is always true. So, that means our original inequality holds. Okay, that wasn't too bad. I just had to square both sides and simplify step by step.Now, moving on to the second problem: Given ( x in mathbb{R} ), ( a = x^2 + frac{1}{2} ), ( b = 2 - x ), ( c = x^2 - x + 1 ), prove that at least one of ( a ), ( b ), ( c ) is not less than 1.Hmm, so I need to show that it's impossible for all three of ( a ), ( b ), and ( c ) to be less than 1 simultaneously. In other words, if I assume that all three are less than 1, I should reach a contradiction.Let me try that approach. Assume that ( a < 1 ), ( b < 1 ), and ( c < 1 ).So,1. ( a = x^2 + frac{1}{2} < 1 )2. ( b = 2 - x < 1 )3. ( c = x^2 - x + 1 < 1 )Let me write down these inequalities:1. ( x^2 + frac{1}{2} < 1 ) implies ( x^2 < frac{1}{2} )2. ( 2 - x < 1 ) implies ( -x < -1 ) which implies ( x > 1 )3. ( x^2 - x + 1 < 1 ) implies ( x^2 - x < 0 ) which factors to ( x(x - 1) < 0 )So, from inequality 1: ( x^2 < frac{1}{2} ) implies ( |x| < frac{sqrt{2}}{2} approx 0.707 ). So, ( x ) is between approximately -0.707 and 0.707.From inequality 2: ( x > 1 ). So, ( x ) is greater than 1.But wait, from inequality 1, ( x ) is less than approximately 0.707, and from inequality 2, ( x ) is greater than 1. These two can't be true at the same time because 1 is greater than 0.707. So, there's already a contradiction here.Therefore, our assumption that all three ( a ), ( b ), and ( c ) are less than 1 is false. Hence, at least one of them must be not less than 1.Alternatively, maybe I can approach this by adding the expressions or something. Let me see.Given ( a = x^2 + frac{1}{2} ), ( b = 2 - x ), ( c = x^2 - x + 1 ).If I add them up: ( a + b + c = x^2 + frac{1}{2} + 2 - x + x^2 - x + 1 ).Simplify:( 2x^2 - 2x + frac{1}{2} + 2 + 1 )Which is ( 2x^2 - 2x + 3.5 ).Hmm, maybe I can write this as ( 2(x^2 - x) + 3.5 ). Completing the square for ( x^2 - x ):( x^2 - x = (x - frac{1}{2})^2 - frac{1}{4} )So, substituting back:( 2[(x - frac{1}{2})^2 - frac{1}{4}] + 3.5 = 2(x - frac{1}{2})^2 - frac{1}{2} + 3.5 = 2(x - frac{1}{2})^2 + 3 )So, ( a + b + c = 2(x - frac{1}{2})^2 + 3 ). Since squares are always non-negative, the minimum value of this expression is 3, achieved when ( x = frac{1}{2} ).So, ( a + b + c geq 3 ).But if all three ( a ), ( b ), ( c ) were less than 1, their sum would be less than 3. But we just saw that their sum is at least 3. Therefore, it's impossible for all three to be less than 1. Hence, at least one of them must be at least 1.That's another way to see it. So, both approaches lead to the same conclusion.Wait, in my first approach, I noticed a contradiction between ( x > 1 ) and ( x < sqrt{1/2} ). So, that was a quicker way to see it. But the second approach, adding them up, also shows that the sum is at least 3, so they can't all be less than 1.Either way, the conclusion is the same. So, the second problem is also proven.I think that's it. Both problems are solved by either manipulating the inequalities directly or by considering the sum of the expressions.

answer：Okay, so I have this problem here about sets A and B. Let me read it again: A is defined as all x where x is greater than 0, and B is all x where x is less than 4. I need to find the intersection of A and B, which is A ∩ B. The options are given as A: empty set, B: {x | x > 0}, C: {x | x < 4}, and D: {x | 0 < x < 4}.Hmm, intersection means the elements that are common to both sets, right? So, if A is all positive numbers and B is all numbers less than 4, then the intersection should be the numbers that are both positive and less than 4. That would be numbers between 0 and 4, not including 0 and 4 themselves because in A, x has to be greater than 0, and in B, x has to be less than 4.Let me visualize this on a number line. A starts just above 0 and goes to infinity, while B starts from negative infinity up to just below 4. The overlapping part of these two sets would be from just above 0 to just below 4. So, that's the interval (0, 4).Wait, let me make sure I'm not making a mistake here. If I take any number between 0 and 4, it's definitely in both A and B. For example, 2 is in A because 2 > 0 and in B because 2 < 4. Similarly, 1.5 is in both. What about 0? Well, 0 is not in A because A requires x > 0, so 0 is excluded. Similarly, 4 is not in B because B requires x < 4, so 4 is excluded. So, the intersection can't include 0 or 4.Is there any number that's in both A and B outside of (0, 4)? Let's see. If I pick a number less than 0, like -1, it's not in A because A only includes positive numbers. If I pick a number greater than or equal to 4, like 5, it's not in B because B only goes up to 4. So, there are no numbers outside of (0, 4) that are in both A and B.Therefore, the intersection A ∩ B must be the set of all x such that 0 < x < 4, which is option D.I think that makes sense. I don't see any numbers that would be in both sets outside of that interval. So, yeah, D is the correct answer.

answer：Alright, so I've got this problem here with four propositions, and I need to figure out which ones are true. Let me take them one by one and think through each carefully.Starting with proposition ①: "The negation of the proposition 'Triangles of equal area are congruent'." Okay, so first, I need to understand what the original proposition is saying. It claims that if two triangles have the same area, then they must be congruent. Hmm, I remember that congruent triangles have the same shape and size, meaning all corresponding sides and angles are equal. But just because two triangles have the same area doesn't necessarily mean they are congruent. For example, imagine two triangles with different base lengths and heights but the same area. They wouldn't be congruent because their sides aren't equal. So, the original statement is false. Now, the negation of a false statement should be true, right? So, the negation of "Triangles of equal area are congruent" would be "There exist triangles of equal area that are not congruent," which is definitely true. So, proposition ① is true.Moving on to proposition ②: "The inverse of the proposition 'If ( xy = 1 ), then ( x ) and ( y ) are reciprocals of each other'." Okay, the original proposition is a conditional statement: If ( xy = 1 ), then ( x ) and ( y ) are reciprocals. The inverse of a conditional statement is formed by negating both the hypothesis and the conclusion. So, the inverse would be: If ( xy neq 1 ), then ( x ) and ( y ) are not reciprocals of each other. Wait, is that necessarily true? Let me think. If ( x ) and ( y ) are reciprocals, then by definition ( xy = 1 ). So, if ( xy neq 1 ), then they can't be reciprocals. That seems to make sense. So, the inverse statement is true. Therefore, proposition ② is true.Next, proposition ③: "The contrapositive of the proposition 'If ( A cap B = B ), then ( A subseteq B )'." Okay, the original statement is: If the intersection of sets ( A ) and ( B ) is equal to ( B ), then ( A ) is a subset of ( B ). I need to find the contrapositive of this. The contrapositive of "If P, then Q" is "If not Q, then not P." So, the contrapositive would be: If ( A nsubseteq B ), then ( A cap B neq B ). Is this true? Let's see. If ( A ) is not a subset of ( B ), that means there is at least one element in ( A ) that is not in ( B ). Therefore, the intersection ( A cap B ) would not contain all elements of ( A ), so it can't be equal to ( B ). Wait, actually, ( A cap B ) is always a subset of ( B ), so if ( A ) is not a subset of ( B ), then ( A cap B ) would be a proper subset of ( B ), meaning it's not equal to ( B ). So, the contrapositive is true. Therefore, proposition ③ is true.Finally, proposition ④: "The contrapositive of the proposition 'If ( m > 1 ), then the equation ( x^2 - 2x + m = 0 ) has real roots'." Okay, let's break this down. The original statement is: If ( m > 1 ), then the quadratic equation ( x^2 - 2x + m = 0 ) has real roots. To find the contrapositive, we need to negate both the hypothesis and conclusion and switch them. So, the contrapositive would be: If the equation ( x^2 - 2x + m = 0 ) does not have real roots, then ( m leq 1 ). Is this true? Let's check the discriminant of the quadratic equation to determine when it has real roots. The discriminant ( D ) is given by ( D = b^2 - 4ac ). For the equation ( x^2 - 2x + m = 0 ), ( a = 1 ), ( b = -2 ), and ( c = m ). So, ( D = (-2)^2 - 4(1)(m) = 4 - 4m ). For the equation to have real roots, the discriminant must be greater than or equal to zero: ( 4 - 4m geq 0 ), which simplifies to ( m leq 1 ). Therefore, if the equation does not have real roots, it means ( D < 0 ), so ( m > 1 ). Wait, that seems contradictory. If ( m > 1 ), then ( D < 0 ), meaning no real roots. So, the original statement is saying that if ( m > 1 ), then there are real roots, but actually, if ( m > 1 ), there are no real roots. Therefore, the original statement is false. Now, the contrapositive of a false statement is also false. So, proposition ④ is false.Wait, let me double-check that. The original statement: If ( m > 1 ), then the equation has real roots. But as we saw, if ( m > 1 ), the discriminant is negative, so there are no real roots. Therefore, the original statement is false. The contrapositive of a statement is logically equivalent to the original statement. So, if the original is false, the contrapositive is also false. Therefore, proposition ④ is false.So, summarizing my thoughts:- Proposition ①: True- Proposition ②: True- Proposition ③: True- Proposition ④: FalseTherefore, the true propositions are ①, ②, and ③.

answer：Okay, so I have this problem where Xiaohong throws two balls vertically upwards, one after the other, with a 1-second interval between them. Both balls reach the same maximum height 1.1 seconds after being thrown. I need to find the time ( t ) after the first ball is thrown when both balls are at the same height.First, I remember that the height of a ball thrown upwards is a quadratic function of time. The general form of this function is ( h(t) = -frac{1}{2}gt^2 + v_0 t + h_0 ), where ( g ) is the acceleration due to gravity, ( v_0 ) is the initial velocity, and ( h_0 ) is the initial height. Since both balls are thrown from the same height, ( h_0 ) is the same for both.It's given that both balls reach the same maximum height 1.1 seconds after being thrown. I know that the time to reach maximum height for a projectile is given by ( t_{text{max}} = frac{v_0}{g} ). So, for both balls, ( 1.1 = frac{v_0}{g} ), which means ( v_0 = 1.1g ).Now, let's denote the time after the first ball is thrown as ( t ). The first ball has been in the air for ( t ) seconds, while the second ball, which was thrown 1 second later, has been in the air for ( t - 1 ) seconds.The height of the first ball at time ( t ) is:[ h_1(t) = -frac{1}{2}g t^2 + 1.1g t + h_0 ]The height of the second ball at time ( t ) (which is ( t - 1 ) seconds after it was thrown) is:[ h_2(t) = -frac{1}{2}g (t - 1)^2 + 1.1g (t - 1) + h_0 ]To find when their heights are equal, set ( h_1(t) = h_2(t) ):[ -frac{1}{2}g t^2 + 1.1g t + h_0 = -frac{1}{2}g (t - 1)^2 + 1.1g (t - 1) + h_0 ]Simplifying this equation by expanding and canceling terms:[ -frac{1}{2}g t^2 + 1.1g t = -frac{1}{2}g (t^2 - 2t + 1) + 1.1g t - 1.1g ][ -frac{1}{2}g t^2 + 1.1g t = -frac{1}{2}g t^2 + g t - frac{1}{2}g + 1.1g t - 1.1g ]Subtracting ( -frac{1}{2}g t^2 + 1.1g t ) from both sides:[ 0 = g t - frac{1}{2}g - 1.1g ][ 0 = g t - 1.6g ][ t = 1.6 text{ seconds} ]So, the time ( t ) when both balls are at the same height is 1.6 seconds after the first ball is thrown.

answer：Alright, so I've got this problem here: I have non-negative real numbers ( x_1, x_2, ldots, x_n ), and I'm given that ( s = sum_{i < j} x_i x_j ). I need to show that at least one of the ( x_i ) has a square that doesn't exceed ( frac{2s}{n^2 - n} ). Hmm, okay, let's break this down.First off, let me understand what ( s ) represents. It's the sum of all products of pairs of these numbers where the first index is less than the second. So, for example, if ( n = 3 ), then ( s = x_1x_2 + x_1x_3 + x_2x_3 ). Got it. So, ( s ) is kind of like the sum of all the pairwise products.Now, the problem is asking me to show that at least one of these ( x_i ) squared is less than or equal to ( frac{2s}{n^2 - n} ). Let me write that down:We need to show that there exists some ( i ) such that ( x_i^2 leq frac{2s}{n^2 - n} ).Hmm. So, maybe I can approach this by contradiction. Let's suppose the opposite: that for every ( i ), ( x_i^2 > frac{2s}{n^2 - n} ). If I can show that this leads to a contradiction, then my original statement must be true.Okay, so assuming that ( x_i^2 > frac{2s}{n^2 - n} ) for all ( i ). Let me see what that implies.Since all ( x_i ) are non-negative, this inequality implies that each ( x_i ) is greater than ( sqrt{frac{2s}{n^2 - n}} ). So, each ( x_i ) is bounded below by this square root.Now, let's think about the sum ( s ). ( s ) is the sum of all pairwise products ( x_i x_j ) for ( i < j ). How many such terms are there? Well, for ( n ) variables, the number of unique pairs is ( binom{n}{2} = frac{n(n-1)}{2} ). So, there are ( frac{n(n-1)}{2} ) terms in the sum ( s ).If each ( x_i ) is greater than ( sqrt{frac{2s}{n^2 - n}} ), then each product ( x_i x_j ) would be greater than ( sqrt{frac{2s}{n^2 - n}} times sqrt{frac{2s}{n^2 - n}} = frac{2s}{n^2 - n} ). So, each term in the sum ( s ) is greater than ( frac{2s}{n^2 - n} ).Therefore, the entire sum ( s ) would be greater than ( frac{n(n-1)}{2} times frac{2s}{n^2 - n} ). Let me compute that:First, ( frac{n(n-1)}{2} times frac{2s}{n^2 - n} ). The 2s in the numerator and the 2 in the denominator will cancel out, leaving ( frac{n(n-1) s}{n^2 - n} ).But ( n^2 - n = n(n - 1) ), so this simplifies to ( frac{n(n-1) s}{n(n - 1)} = s ).So, putting it all together, under our assumption, ( s > s ). Wait, that can't be right. ( s ) can't be greater than itself. That's a contradiction.Therefore, our initial assumption must be wrong. So, it can't be the case that all ( x_i^2 > frac{2s}{n^2 - n} ). Hence, there must be at least one ( x_i ) such that ( x_i^2 leq frac{2s}{n^2 - n} ).Okay, that seems to make sense. I assumed the opposite and arrived at a contradiction, so the original statement holds. But let me double-check my steps to make sure I didn't make any mistakes.First, I defined ( s ) correctly as the sum of all pairwise products. Then, I considered the number of terms in ( s ), which is ( frac{n(n-1)}{2} ). That seems right.I assumed that each ( x_i^2 > frac{2s}{n^2 - n} ), which led me to each ( x_i x_j > frac{2s}{n^2 - n} ). Then, summing over all pairs, I got that ( s ) would be greater than ( s ), which is impossible. So, my reasoning seems solid.Another way to think about it might be using inequalities like the AM-GM inequality or Cauchy-Schwarz, but I think the contradiction approach is straightforward here.Just to recap:1. Assume all ( x_i^2 > frac{2s}{n^2 - n} ).2. Then, each ( x_i x_j > frac{2s}{n^2 - n} ).3. Summing over all pairs, ( s > frac{n(n-1)}{2} times frac{2s}{n^2 - n} = s ).4. Contradiction, so at least one ( x_i^2 leq frac{2s}{n^2 - n} ).Yep, that all checks out. I think I've got it.**Final Answer**boxed{frac{2s}{n^2 - n}}