answer：First, I need to calculate the square of the complex number (1 + 2i). To do this, I'll use the formula for squaring a binomial: ((a + b)^2 = a^2 + 2ab + b^2).Applying this to (1 + 2i), I get:[(1 + 2i)^2 = 1^2 + 2 times 1 times 2i + (2i)^2 = 1 + 4i + 4i^2]Since (i^2 = -1), this simplifies to:[1 + 4i + 4(-1) = 1 + 4i - 4 = -3 + 4i]Now that I have the squared complex number (-3 + 4i), the next step is to find its conjugate. The conjugate of a complex number (a + bi) is (a - bi).Applying this to (-3 + 4i), the conjugate is:[-3 - 4i]Therefore, the conjugate of the complex number ((1 + 2i)^2) is (-3 - 4i).

answer：Alright, so I have this function ( f(x) = -x^2 + ax + 2(x^2 - x)ln x ). There are two parts to the problem. Let me tackle them one by one.**Part I: When ( a = 2 ), determine the intervals on which ( f(x) ) is monotonic.**Okay, so first, I need to find where the function is increasing or decreasing. That means I need to find its derivative and analyze the sign of the derivative.Given ( f(x) = -x^2 + 2x + 2(x^2 - x)ln x ).Let me compute the derivative ( f'(x) ).The derivative of ( -x^2 ) is ( -2x ).The derivative of ( 2x ) is ( 2 ).Now, for the term ( 2(x^2 - x)ln x ), I need to use the product rule. Let me denote ( u = 2(x^2 - x) ) and ( v = ln x ).Then, ( u' = 2(2x - 1) ) and ( v' = frac{1}{x} ).So, the derivative of ( uv ) is ( u'v + uv' ).Putting it all together:( f'(x) = -2x + 2 + [2(2x - 1)ln x + 2(x^2 - x)cdot frac{1}{x}] ).Simplify the last term:( 2(x^2 - x)cdot frac{1}{x} = 2(x - 1) ).So, now the derivative becomes:( f'(x) = -2x + 2 + 2(2x - 1)ln x + 2(x - 1) ).Combine like terms:First, combine the constants and linear terms:-2x + 2 + 2(x - 1) = -2x + 2 + 2x - 2 = 0.Wait, that's interesting. So, the linear and constant terms cancel out. So, we're left with:( f'(x) = 2(2x - 1)ln x ).So, ( f'(x) = 2(2x - 1)ln x ).Now, to find where ( f(x) ) is increasing or decreasing, we need to analyze the sign of ( f'(x) ).So, ( f'(x) = 2(2x - 1)ln x ).The critical points occur where ( f'(x) = 0 ) or where ( f'(x) ) is undefined. Since ( ln x ) is defined for ( x > 0 ), the domain is ( (0, infty) ).Setting ( f'(x) = 0 ):( 2(2x - 1)ln x = 0 ).So, either ( 2x - 1 = 0 ) or ( ln x = 0 ).Solving ( 2x - 1 = 0 ) gives ( x = frac{1}{2} ).Solving ( ln x = 0 ) gives ( x = 1 ).So, critical points at ( x = frac{1}{2} ) and ( x = 1 ).Now, let's analyze the sign of ( f'(x) ) in the intervals determined by these critical points: ( (0, frac{1}{2}) ), ( (frac{1}{2}, 1) ), and ( (1, infty) ).1. **Interval ( (0, frac{1}{2}) ):**Choose a test point, say ( x = frac{1}{4} ).Compute ( 2x - 1 = 2(frac{1}{4}) - 1 = frac{1}{2} - 1 = -frac{1}{2} ).Compute ( ln frac{1}{4} = ln 0.25 approx -1.386 ).So, ( (2x - 1)ln x = (-frac{1}{2})(-1.386) = positive ).Multiply by 2: ( f'(x) = positive ).So, ( f(x) ) is increasing on ( (0, frac{1}{2}) ).2. **Interval ( (frac{1}{2}, 1) ):**Choose a test point, say ( x = frac{3}{4} ).Compute ( 2x - 1 = 2(frac{3}{4}) - 1 = frac{3}{2} - 1 = frac{1}{2} ).Compute ( ln frac{3}{4} approx -0.2877 ).So, ( (2x - 1)ln x = (frac{1}{2})(-0.2877) = negative ).Multiply by 2: ( f'(x) = negative ).So, ( f(x) ) is decreasing on ( (frac{1}{2}, 1) ).3. **Interval ( (1, infty) ):**Choose a test point, say ( x = 2 ).Compute ( 2x - 1 = 4 - 1 = 3 ).Compute ( ln 2 approx 0.693 ).So, ( (2x - 1)ln x = 3 times 0.693 = positive ).Multiply by 2: ( f'(x) = positive ).So, ( f(x) ) is increasing on ( (1, infty) ).Therefore, summarizing:- Increasing on ( (0, frac{1}{2}) ) and ( (1, infty) ).- Decreasing on ( (frac{1}{2}, 1) ).**Part II: Assume that for ( x in (0, +infty) ), the inequality ( f(x) + x^2 > 0 ) holds true for all ( x ). Find the minimum integer value of ( a ).**So, ( f(x) + x^2 > 0 ) for all ( x > 0 ).Given ( f(x) = -x^2 + ax + 2(x^2 - x)ln x ).So, ( f(x) + x^2 = (-x^2 + ax + 2(x^2 - x)ln x) + x^2 = ax + 2(x^2 - x)ln x ).Therefore, the inequality becomes:( ax + 2(x^2 - x)ln x > 0 ) for all ( x > 0 ).Let me factor out ( x ):( x(a + 2(x - 1)ln x) > 0 ).Since ( x > 0 ), the inequality reduces to:( a + 2(x - 1)ln x > 0 ) for all ( x > 0 ).So, ( a > -2(x - 1)ln x ) for all ( x > 0 ).Therefore, to satisfy this for all ( x > 0 ), ( a ) must be greater than the maximum value of ( -2(x - 1)ln x ).Let me define ( g(x) = -2(x - 1)ln x ). We need to find the maximum of ( g(x) ) over ( x > 0 ).So, ( g(x) = -2(x - 1)ln x ).To find its maximum, compute its derivative and set it to zero.Compute ( g'(x) ):First, expand ( g(x) ):( g(x) = -2x ln x + 2 ln x ).So, ( g'(x) = -2 ln x - 2x cdot frac{1}{x} + frac{2}{x} ).Simplify:( g'(x) = -2 ln x - 2 + frac{2}{x} ).Set ( g'(x) = 0 ):( -2 ln x - 2 + frac{2}{x} = 0 ).Multiply both sides by ( x ) to eliminate the denominator:( -2x ln x - 2x + 2 = 0 ).Simplify:( -2x ln x - 2x + 2 = 0 ).Divide both sides by -2:( x ln x + x - 1 = 0 ).So, ( x ln x + x - 1 = 0 ).This equation might not have an elementary solution, so perhaps we can analyze it numerically or see if we can find a critical point.Alternatively, let's analyze the behavior of ( g(x) ).First, let's consider the limits as ( x ) approaches 0 and infinity.As ( x to 0^+ ):( ln x to -infty ), but ( x to 0 ).So, ( (x - 1) ) approaches -1, and ( ln x ) approaches -infty.So, ( g(x) = -2(x - 1)ln x approx -2(-1)(-infty) = -2 times infty = -infty ).Wait, that seems conflicting. Let me compute more carefully:( x to 0^+ ):( x - 1 approx -1 ).( ln x to -infty ).So, ( g(x) = -2(-1)(-infty) = -2 times infty = -infty ).So, as ( x to 0^+ ), ( g(x) to -infty ).As ( x to infty ):( x - 1 approx x ).( ln x to infty ).So, ( g(x) = -2(x)ln x to -infty ).So, ( g(x) ) tends to -infty at both ends.Therefore, the maximum must occur somewhere in between.We can check ( x = 1 ):( g(1) = -2(1 - 1)ln 1 = 0 ).Also, let's check ( x = e ) (Euler's number, approximately 2.718):( g(e) = -2(e - 1)ln e = -2(e - 1)(1) = -2(e - 1) approx -2(1.718) approx -3.436 ).Negative.What about ( x = frac{1}{2} ):( g(frac{1}{2}) = -2(frac{1}{2} - 1)ln frac{1}{2} = -2(-frac{1}{2})(-ln 2) = -2(-frac{1}{2})(-0.693) ).Compute step by step:First, ( x - 1 = -frac{1}{2} ).( ln frac{1}{2} = -ln 2 approx -0.693 ).So, ( g(frac{1}{2}) = -2(-frac{1}{2})(-0.693) = -2 times frac{1}{2} times 0.693 = -0.693 ).Negative.Wait, so at ( x = 1 ), ( g(x) = 0 ), which is higher than at ( x = frac{1}{2} ) and ( x = e ).Let me check another point, say ( x = 2 ):( g(2) = -2(2 - 1)ln 2 = -2(1)(0.693) approx -1.386 ).Negative.Wait, so maybe the maximum is at ( x = 1 ), where ( g(x) = 0 ).But let me check another point, say ( x = 0.5 ):Wait, I already did ( x = 0.5 ), it's negative.Wait, maybe I need to check between 0 and 1.Wait, let me try ( x = 0.1 ):( g(0.1) = -2(0.1 - 1)ln 0.1 = -2(-0.9)(-2.3026) = -2 times 0.9 times 2.3026 approx -4.144 ).Negative.Wait, so seems like ( g(x) ) is always negative except at ( x = 1 ), where it's zero.But wait, let me check ( x ) slightly less than 1, say ( x = 0.9 ):( g(0.9) = -2(0.9 - 1)ln 0.9 = -2(-0.1)(-0.10536) approx -2 times 0.1 times 0.10536 approx -0.021 ).Negative.Similarly, ( x = 1.1 ):( g(1.1) = -2(1.1 - 1)ln 1.1 = -2(0.1)(0.09531) approx -0.019 ).Negative.Wait, so it seems that ( g(x) ) reaches its maximum at ( x = 1 ), where it's zero.Therefore, the maximum value of ( g(x) ) is 0.So, for ( a + 2(x - 1)ln x > 0 ) for all ( x > 0 ), we need ( a > -2(x - 1)ln x ) for all ( x > 0 ).Since the maximum of ( -2(x - 1)ln x ) is 0, ( a ) must be greater than 0.Therefore, the minimum integer value of ( a ) is 1.But wait, let me verify this.If ( a = 1 ), then ( a + 2(x - 1)ln x > 0 ) for all ( x > 0 ).At ( x = 1 ), ( 1 + 2(0)ln 1 = 1 > 0 ).At ( x ) approaching 0, ( 1 + 2(x - 1)ln x ).As ( x to 0^+ ), ( (x - 1) approx -1 ), ( ln x to -infty ), so ( 2(x - 1)ln x approx 2(-1)(-infty) = +infty ). So, ( 1 + infty = infty > 0 ).Wait, that seems conflicting with earlier analysis.Wait, no, hold on. ( g(x) = -2(x - 1)ln x ), which we found had a maximum of 0. So, ( a > 0 ).But when ( a = 1 ), ( a + 2(x - 1)ln x = 1 + 2(x - 1)ln x ).Wait, but ( 2(x - 1)ln x = -g(x) ).So, ( 1 - g(x) > 0 ).Since ( g(x) leq 0 ), ( 1 - g(x) geq 1 ), which is always positive.Wait, that seems correct.But earlier, when I considered ( x to 0^+ ), ( g(x) to -infty ), so ( 1 - g(x) to infty ), which is positive.At ( x = 1 ), ( g(x) = 0 ), so ( 1 - 0 = 1 > 0 ).At ( x to infty ), ( g(x) to -infty ), so ( 1 - g(x) to infty ).Therefore, ( a = 1 ) satisfies the inequality.But wait, what if ( a = 0 )?Then, ( 0 + 2(x - 1)ln x > 0 ).But ( 2(x - 1)ln x ) is not always positive.At ( x = 1 ), it's 0.At ( x > 1 ), ( x - 1 > 0 ), ( ln x > 0 ), so positive.At ( x < 1 ), ( x - 1 < 0 ), ( ln x < 0 ), so positive as well.Wait, hold on.Wait, ( 2(x - 1)ln x ).For ( x > 1 ), ( x - 1 > 0 ), ( ln x > 0 ), so positive.For ( x < 1 ), ( x - 1 < 0 ), ( ln x < 0 ), so positive.At ( x = 1 ), it's 0.So, ( 2(x - 1)ln x geq 0 ) for all ( x > 0 ).Therefore, ( a + 2(x - 1)ln x > 0 ) for all ( x > 0 ) if ( a > -2(x - 1)ln x ).But since ( 2(x - 1)ln x geq 0 ), the minimum value of ( -2(x - 1)ln x ) is negative infinity, but the maximum is 0.Wait, so if ( a > 0 ), then ( a + 2(x - 1)ln x > 0 ) because ( 2(x - 1)ln x geq 0 ).But if ( a = 0 ), then ( 2(x - 1)ln x geq 0 ), so ( f(x) + x^2 = 0 + 2(x - 1)ln x geq 0 ). But at ( x = 1 ), it's 0, so the inequality is not strict. The problem says ( f(x) + x^2 > 0 ) for all ( x ). So, at ( x = 1 ), it's equal to 0, which violates the strict inequality.Therefore, ( a ) must be greater than 0. Since ( a ) must be an integer, the minimum integer value is 1.Wait, but earlier I thought ( g(x) ) had a maximum of 0, so ( a > 0 ). So, the minimum integer is 1.Yes, that seems consistent.**Final Answer**Ⅰ. The function is increasing on ( left(0, frac{1}{2}right) ) and ( (1, +infty) ), and decreasing on ( left(frac{1}{2}, 1right) ).Ⅱ. The minimum integer value of ( a ) is ( boxed{1} ).

answer：Okay, so I have this quadratic equation: 3x² + 7x + c = 0. I need to find all possible positive integer values of c such that the equation has rational roots. Then, I have to sum those c values. Hmm, let me think about how to approach this.First, I remember that for a quadratic equation ax² + bx + c = 0 to have rational roots, the discriminant must be a perfect square. The discriminant formula is b² - 4ac. So, in this case, the discriminant is 7² - 4*3*c, which simplifies to 49 - 12c. Alright, so 49 - 12c needs to be a perfect square. Let me denote this perfect square as k², where k is a non-negative integer. So, I can write:49 - 12c = k²Now, I can rearrange this equation to solve for c:12c = 49 - k²c = (49 - k²) / 12Since c has to be a positive integer, (49 - k²) must be divisible by 12, and the result must be positive. So, I need to find all integer values of k such that 49 - k² is positive and divisible by 12.Let me list the possible values of k. Since k² must be less than 49, k can be from 0 up to 7 (since 7² = 49, which would make c zero, but c has to be positive, so k must be less than 7). So, k can be 0, 1, 2, 3, 4, 5, or 6.Let me test each k:1. k = 0: c = (49 - 0) / 12 = 49/12 ≈ 4.083 Not an integer, so discard.2. k = 1: c = (49 - 1) / 12 = 48/12 = 4 That's an integer. So, c = 4 is a possible value.3. k = 2: c = (49 - 4) / 12 = 45/12 = 3.75 Not an integer, discard.4. k = 3: c = (49 - 9) / 12 = 40/12 ≈ 3.333 Not an integer, discard.5. k = 4: c = (49 - 16) / 12 = 33/12 = 2.75 Not an integer, discard.6. k = 5: c = (49 - 25) / 12 = 24/12 = 2 That's an integer. So, c = 2 is another possible value.7. k = 6: c = (49 - 36) / 12 = 13/12 ≈ 1.083 Not an integer, discard.So, the only integer values of c that satisfy the condition are 4 and 2. Wait, let me double-check if I missed any k values. I considered k from 0 to 6, which covers all possibilities where k² is less than 49. For k=7, k²=49, which would make c=0, but c has to be positive, so that's not allowed.Therefore, the possible positive integer values of c are 2 and 4. Adding them together: 2 + 4 = 6.So, the sum of all possible positive integer values of c is 6.**Final Answer**The sum of all possible positive integer values of ( c ) is boxed{6}.

answer：Alright, so I need to figure out how many three-digit numbers exist where the second digit is exactly 3 less than the third digit. Hmm, okay, let's break this down step by step.First, let's recall that a three-digit number has the form ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit. Importantly, A can't be zero because otherwise, it wouldn't be a three-digit number. So, A can be any digit from 1 to 9.Now, the problem states that the second digit (B) is 3 less than the third digit (C). So, mathematically, that means B = C - 3. Let me think about what this implies for the possible values of B and C. Since both B and C are single digits (0 through 9), I need to find all pairs where B is exactly 3 less than C. Let's list out the possible values for C and see what B would be in each case:- If C = 3, then B = 0- If C = 4, then B = 1- If C = 5, then B = 2- If C = 6, then B = 3- If C = 7, then B = 4- If C = 8, then B = 5- If C = 9, then B = 6Wait a minute, if C is less than 3, say C = 0, 1, or 2, then B would be negative, which isn't possible because digits can't be negative. So, C can't be less than 3. That means C can only be from 3 to 9, giving us 7 possible values for C, and correspondingly, 7 possible values for B.So, for each valid C (from 3 to 9), there's exactly one B that satisfies B = C - 3. That gives us 7 possible combinations for the last two digits.Now, what about the first digit, A? As I mentioned earlier, A can be any digit from 1 to 9, giving us 9 possible choices for A.Since the choices for A and the pair (B, C) are independent of each other, we can use the multiplication principle to find the total number of three-digit numbers that satisfy the condition. That is, for each of the 9 choices of A, there are 7 possible pairs of (B, C). So, the total number of such three-digit numbers is 9 (choices for A) multiplied by 7 (choices for B and C), which equals 63.Wait, let me double-check that. If C ranges from 3 to 9, that's 7 values. For each of these, B is uniquely determined, so that's 7 pairs. And A has 9 options. So, 9 * 7 = 63. Yeah, that seems right.Is there any chance I'm missing something? Maybe considering leading zeros or something? But no, since A can't be zero, and we've already accounted for that by limiting A to 1-9. And for B and C, we've made sure that B is always a valid digit (0-9) by only considering C from 3-9.So, I think 63 is the correct answer. There are 63 three-digit numbers where the second digit is exactly 3 less than the third digit.

answer：Okay, so I've got this multiple-choice question here about some geometry propositions, and I need to figure out which one is correct. Let me take it step by step because I want to make sure I understand each option properly.First, let's look at option A: "If two lines form equal angles with the same plane, then these two lines are parallel." Hmm, okay. So, imagine I have a plane, like a flat table, and two lines that both make the same angle with this table. Does that mean the lines have to be parallel? I'm not sure. Maybe I can visualize it. If both lines are, say, at 45 degrees to the table, they could be going in different directions but still make the same angle with the table. So, they might not necessarily be parallel. They could intersect or be skew lines if they're in different planes. So, I think A might not be correct.Moving on to option B: "If a line is perpendicular to two lines within a plane, then this line is perpendicular to this plane." Alright, so if I have a line that's perpendicular to two different lines lying on a plane, does that mean it's perpendicular to the entire plane? I remember something about this from my geometry class. I think if a line is perpendicular to two non-parallel lines in a plane, then it is indeed perpendicular to the plane. But wait, does it have to be two non-parallel lines? If the two lines are parallel, then the line might not necessarily be perpendicular to the entire plane. So, the proposition says "two lines within a plane," but it doesn't specify if they are intersecting or not. If they are intersecting, then yes, the line would be perpendicular to the plane. If they are parallel, maybe not. So, I'm a bit confused here. I need to clarify this.Option C: "If a line is parallel to two intersecting planes, then this line is parallel to the intersection line of these two planes." Okay, so imagine two planes that intersect each other, forming a line of intersection. If there's a line that's parallel to both of these planes, does that mean it's parallel to the line where the two planes meet? Let me think. If a line is parallel to both planes, it means it doesn't intersect either plane and maintains a constant distance from each. Since the two planes intersect along a line, the direction of that intersection line must be related to the direction of the original line. I think this might be correct because the line can't be skew to the intersection line if it's parallel to both planes. So, maybe C is correct.Option D: "If two planes are both perpendicular to a third plane, then these two planes are parallel." Hmm, so if I have three planes, and two of them are each perpendicular to the third one, does that mean the two are parallel? Let me visualize. Imagine the third plane is like the floor. If two other planes are both perpendicular to the floor, they could be walls. But walls can intersect each other along a line, like the corner where two walls meet. So, the two walls (planes) are both perpendicular to the floor but are not parallel; they intersect. Therefore, D is not correct.Going back to option B, I think I need to resolve my confusion. The proposition says the line is perpendicular to two lines within the plane. If those two lines are intersecting, then yes, the line is perpendicular to the plane. But if the two lines are parallel, it might not be. So, does the proposition specify that the two lines are intersecting? It just says "two lines within a plane." So, it's possible that the two lines are parallel, which would mean the line isn't necessarily perpendicular to the entire plane. Therefore, B might not always be correct.So, summarizing my thoughts:- A: Not necessarily correct because equal angles don't guarantee parallelism.- B: Only correct if the two lines are intersecting, but since it's not specified, it's not always true.- C: Seems correct because being parallel to both planes implies parallelism to their intersection.- D: Incorrect because two planes perpendicular to a third can intersect.Therefore, the correct answer should be C.

answer：First, I need to calculate the volume of the cylinder. The formula for the volume of a cylinder is ( pi r^2 h ). Given that the radius ( r ) is 10 cm and the height ( h ) is 30 cm, the volume of the cylinder is:[text{Volume of cylinder} = pi (10)^2 (30) = 3000pi text{ cubic cm}.]Next, I'll calculate the volume of one cone. The formula for the volume of a cone is ( frac{1}{3} pi r^2 h ). With a radius of 10 cm and a height of 15 cm, the volume of one cone is:[text{Volume of one cone} = frac{1}{3} pi (10)^2 (15) = 500pi text{ cubic cm}.]Since there are two congruent cones, the total volume occupied by the cones is:[text{Total volume of cones} = 2 times 500pi = 1000pi text{ cubic cm}.]Now, I'll calculate the volume of the sphere. The formula for the volume of a sphere is ( frac{4}{3} pi r^3 ). Given that the diameter of the sphere is 10 cm, the radius ( r ) is 5 cm. Therefore, the volume of the sphere is:[text{Volume of sphere} = frac{4}{3} pi (5)^3 = frac{500}{3}pi text{ cubic cm}.]Finally, to find the volume of space within the cylinder not occupied by the cones or the sphere, I'll subtract the total volumes of the cones and the sphere from the volume of the cylinder:[text{Volume not occupied} = 3000pi - 1000pi - frac{500}{3}pi = left(2000 - frac{500}{3}right)pi = frac{5500}{3}pi text{ cubic cm}.]