💡Alright, so I've got these two calculus problems here, and I need to figure them out step by step. Let me start with the first one.**Problem 1:**We have two functions, ( f(x) = ln x ) and ( g(x) = frac{1}{2}x^2 - kx ). The function ( h(x) = f(x) + g(x) ) is defined as ( h(x) = ln x + frac{1}{2}x^2 - kx ). We're told that ( k = m + frac{1}{m} ) where ( m > 0 ), and that ( h(x) ) has exactly one extreme value in the interval ( (0, 2) ). We need to find the range of the real number ( m ).Okay, so first, to find the extreme values of ( h(x) ), I need to compute its derivative and find where it equals zero. The derivative ( h'(x) ) will give me the critical points.Let me compute ( h'(x) ):( h'(x) = frac{d}{dx}[ln x] + frac{d}{dx}left[frac{1}{2}x^2right] - frac{d}{dx}[kx] )Calculating each term:- The derivative of ( ln x ) is ( frac{1}{x} ).- The derivative of ( frac{1}{2}x^2 ) is ( x ).- The derivative of ( kx ) is ( k ).So putting it all together:( h'(x) = frac{1}{x} + x - k )We need to find the critical points by setting ( h'(x) = 0 ):( frac{1}{x} + x - k = 0 )Let me rearrange this equation:( frac{1}{x} + x = k )So, ( k = frac{1}{x} + x )Hmm, that's interesting. So, for each ( x ), ( k ) is expressed in terms of ( x ). But in our case, ( k ) is given as ( m + frac{1}{m} ). So, perhaps we can relate these two expressions.Wait, actually, ( k ) is a constant, so for a given ( k ), the equation ( frac{1}{x} + x = k ) will have solutions for ( x ) in ( (0, 2) ). The number of solutions corresponds to the number of critical points.We're told that ( h(x) ) has exactly one extreme value in ( (0, 2) ). So, the equation ( frac{1}{x} + x = k ) must have exactly one solution in ( (0, 2) ).Therefore, we need to find the values of ( k ) such that ( frac{1}{x} + x = k ) has exactly one solution in ( (0, 2) ).Let me analyze the function ( f(x) = frac{1}{x} + x ) on the interval ( (0, 2) ).First, let's find its behavior. Compute its derivative to see where it's increasing or decreasing.( f'(x) = -frac{1}{x^2} + 1 )Set ( f'(x) = 0 ):( -frac{1}{x^2} + 1 = 0 )( 1 = frac{1}{x^2} )( x^2 = 1 )( x = 1 ) or ( x = -1 )But since ( x > 0 ), the critical point is at ( x = 1 ).So, ( f(x) ) has a critical point at ( x = 1 ). Let's check the behavior around this point.For ( x < 1 ), say ( x = 0.5 ):( f'(0.5) = -frac{1}{(0.5)^2} + 1 = -4 + 1 = -3 < 0 ). So, the function is decreasing on ( (0, 1) ).For ( x > 1 ), say ( x = 2 ):( f'(2) = -frac{1}{4} + 1 = frac{3}{4} > 0 ). So, the function is increasing on ( (1, infty) ).Therefore, ( f(x) ) has a minimum at ( x = 1 ). Let's compute ( f(1) ):( f(1) = frac{1}{1} + 1 = 2 )So, the minimum value of ( f(x) ) on ( (0, 2) ) is 2 at ( x = 1 ).Now, let's see the behavior as ( x ) approaches 0 and 2.As ( x to 0^+ ), ( frac{1}{x} to infty ), so ( f(x) to infty ).At ( x = 2 ):( f(2) = frac{1}{2} + 2 = frac{5}{2} = 2.5 )So, the function ( f(x) ) decreases from ( infty ) to 2 as ( x ) goes from 0 to 1, and then increases from 2 to 2.5 as ( x ) goes from 1 to 2.Therefore, the graph of ( f(x) ) on ( (0, 2) ) is a U-shape with the minimum at ( x = 1 ).Now, the equation ( f(x) = k ) will have:- Two solutions in ( (0, 2) ) if ( k > 2 ) and ( k < 2.5 ).Wait, actually, since the function decreases to 2 and then increases to 2.5, for ( k ) between 2 and 2.5, the equation ( f(x) = k ) will have two solutions: one in ( (0, 1) ) and another in ( (1, 2) ).For ( k = 2 ), it will have exactly one solution at ( x = 1 ).For ( k > 2.5 ), the equation ( f(x) = k ) will have no solution in ( (0, 2) ) because the maximum value of ( f(x) ) on ( (0, 2) ) is 2.5.Wait, but hold on, as ( x ) approaches 0, ( f(x) ) approaches infinity, so for any ( k > 2 ), there should be two solutions: one near 0 and one in ( (1, 2) ). But at ( x = 2 ), ( f(2) = 2.5 ). So, for ( k > 2.5 ), there's no solution because ( f(x) ) doesn't go beyond 2.5 in ( (0, 2) ).Wait, that doesn't make sense because as ( x ) approaches 0, ( f(x) ) goes to infinity, so for any ( k > 2 ), there should be a solution near 0 and another in ( (1, 2) ). But when ( k = 2.5 ), the solution in ( (1, 2) ) is exactly at ( x = 2 ), but since our interval is open, ( x = 2 ) is not included. So, for ( k = 2.5 ), the equation ( f(x) = k ) would have only one solution near 0, because the other solution is at the endpoint ( x = 2 ), which is excluded.Wait, actually, let me think again.Since the interval is ( (0, 2) ), open at both ends, so ( x = 2 ) is not included. Therefore, when ( k = 2.5 ), the equation ( f(x) = 2.5 ) would have a solution at ( x = 2 ), which is not in the interval. So, in ( (0, 2) ), for ( k = 2.5 ), the equation ( f(x) = k ) would have only one solution near 0.Similarly, for ( k > 2.5 ), there would be no solution in ( (0, 2) ) because ( f(x) ) doesn't reach ( k ) in that interval.Wait, but that contradicts the earlier thought. Let me clarify.The function ( f(x) = frac{1}{x} + x ) on ( (0, 2) ):- As ( x to 0^+ ), ( f(x) to infty ).- At ( x = 1 ), ( f(x) = 2 ).- At ( x = 2 ), ( f(x) = 2.5 ).So, for ( k > 2.5 ), the equation ( f(x) = k ) would have no solution in ( (0, 2) ) because ( f(x) ) doesn't exceed 2.5 in that interval.For ( k = 2.5 ), the equation ( f(x) = 2.5 ) would have a solution at ( x = 2 ), which is not in the open interval ( (0, 2) ). So, in ( (0, 2) ), ( f(x) = 2.5 ) has no solution.Wait, but that doesn't seem right because as ( x ) approaches 2 from the left, ( f(x) ) approaches 2.5. So, for ( k = 2.5 ), there's no solution in ( (0, 2) ).Wait, no, actually, for ( k = 2.5 ), the equation ( f(x) = 2.5 ) would have a solution at ( x = 2 ), which is not in ( (0, 2) ). So, in ( (0, 2) ), ( f(x) = 2.5 ) has no solution.Therefore, for ( k > 2 ), the equation ( f(x) = k ) has two solutions in ( (0, 2) ): one in ( (0, 1) ) and another in ( (1, 2) ).For ( k = 2 ), it has exactly one solution at ( x = 1 ).For ( k < 2 ), since the minimum of ( f(x) ) is 2, there are no solutions.Wait, that makes more sense. So, the number of solutions:- If ( k < 2 ): No solutions.- If ( k = 2 ): Exactly one solution at ( x = 1 ).- If ( 2 < k < 2.5 ): Two solutions in ( (0, 2) ).- If ( k = 2.5 ): No solutions in ( (0, 2) ) because the solution is at ( x = 2 ), which is excluded.- If ( k > 2.5 ): No solutions in ( (0, 2) ).Wait, that seems inconsistent because as ( x to 0^+ ), ( f(x) to infty ), so for any ( k > 2 ), there should be a solution near 0, and another solution in ( (1, 2) ) as long as ( k ) is less than 2.5.But when ( k = 2.5 ), the solution in ( (1, 2) ) is exactly at ( x = 2 ), which is not included. So, for ( k = 2.5 ), only one solution near 0.Wait, no, actually, when ( k = 2.5 ), the equation ( f(x) = 2.5 ) would have a solution at ( x = 2 ), which is not in ( (0, 2) ). So, in ( (0, 2) ), ( f(x) = 2.5 ) has no solution.Therefore, the number of solutions:- ( k < 2 ): No solutions.- ( k = 2 ): One solution at ( x = 1 ).- ( 2 < k < 2.5 ): Two solutions.- ( k = 2.5 ): No solutions.- ( k > 2.5 ): No solutions.Wait, that can't be right because for ( k > 2.5 ), as ( x to 0^+ ), ( f(x) to infty ), so there should be a solution near 0 for any ( k > 2 ), regardless of how large ( k ) is.Wait, I think I made a mistake earlier. Let me correct that.The function ( f(x) = frac{1}{x} + x ) on ( (0, 2) ):- As ( x to 0^+ ), ( f(x) to infty ).- At ( x = 1 ), ( f(x) = 2 ).- At ( x = 2 ), ( f(x) = 2.5 ).So, for ( k > 2 ), the equation ( f(x) = k ) will have:- One solution in ( (0, 1) ) because ( f(x) ) decreases from ( infty ) to 2 as ( x ) goes from 0 to 1.- One solution in ( (1, 2) ) because ( f(x) ) increases from 2 to 2.5 as ( x ) goes from 1 to 2.Therefore, for ( 2 < k < 2.5 ), there are two solutions in ( (0, 2) ).For ( k = 2.5 ), the equation ( f(x) = 2.5 ) has a solution at ( x = 2 ), which is not in ( (0, 2) ). So, in ( (0, 2) ), there's only one solution in ( (0, 1) ).Wait, no, because as ( x ) approaches 2 from the left, ( f(x) ) approaches 2.5. So, for ( k = 2.5 ), the equation ( f(x) = 2.5 ) would have a solution approaching 2, but since 2 is excluded, in ( (0, 2) ), there's no solution at ( x = 2 ). Therefore, for ( k = 2.5 ), there's only one solution in ( (0, 1) ).Similarly, for ( k > 2.5 ), the equation ( f(x) = k ) would have only one solution in ( (0, 1) ), because in ( (1, 2) ), ( f(x) ) only goes up to 2.5, so it can't reach ( k ) if ( k > 2.5 ).Wait, that makes more sense.So, summarizing:- If ( k < 2 ): No solutions.- If ( k = 2 ): One solution at ( x = 1 ).- If ( 2 < k < 2.5 ): Two solutions in ( (0, 2) ).- If ( k = 2.5 ): One solution in ( (0, 1) ).- If ( k > 2.5 ): One solution in ( (0, 1) ).But the problem states that ( h(x) ) has exactly one extreme value in ( (0, 2) ). So, we need ( h'(x) = 0 ) to have exactly one solution in ( (0, 2) ). From our analysis, this happens when:- ( k = 2 ): One solution at ( x = 1 ).- ( k = 2.5 ): One solution in ( (0, 1) ).- ( k > 2.5 ): One solution in ( (0, 1) ).Wait, but the problem says "exactly one extreme value in the interval ( (0, 2) )". So, we need to consider when ( h'(x) = 0 ) has exactly one solution in ( (0, 2) ).From our analysis, this occurs when:- ( k = 2 ): One solution at ( x = 1 ).- ( k geq 2.5 ): One solution in ( (0, 1) ).Wait, but when ( k = 2.5 ), the solution is approaching ( x = 2 ), but since ( x = 2 ) is excluded, it's actually in ( (0, 2) ). Wait, no, when ( k = 2.5 ), the solution is at ( x = 2 ), which is not in ( (0, 2) ). So, in ( (0, 2) ), for ( k = 2.5 ), there's only one solution in ( (0, 1) ).Similarly, for ( k > 2.5 ), there's only one solution in ( (0, 1) ).Therefore, the values of ( k ) for which ( h(x) ) has exactly one extreme in ( (0, 2) ) are:- ( k = 2 )- ( k geq 2.5 )But wait, when ( k = 2.5 ), the solution is at ( x = 2 ), which is not in ( (0, 2) ). So, in ( (0, 2) ), for ( k = 2.5 ), there's only one solution in ( (0, 1) ).Therefore, the range of ( k ) is ( k geq 2.5 ) or ( k = 2 ).But the problem states that ( k = m + frac{1}{m} ) where ( m > 0 ). So, we need to find the range of ( m ) such that ( k geq 2.5 ) or ( k = 2 ).Wait, but ( k = m + frac{1}{m} ). Let's analyze the possible values of ( k ).The function ( k(m) = m + frac{1}{m} ) for ( m > 0 ).We know that ( m + frac{1}{m} geq 2 ) by AM ≥ GM inequality, with equality when ( m = 1 ).So, ( k geq 2 ).Therefore, ( k ) can be 2 or greater.But from our earlier analysis, ( h(x) ) has exactly one extreme in ( (0, 2) ) when ( k = 2 ) or ( k geq 2.5 ).Therefore, we need ( k = 2 ) or ( k geq 2.5 ).But ( k = m + frac{1}{m} ).So, let's solve for ( m ):Case 1: ( k = 2 )( m + frac{1}{m} = 2 )Multiply both sides by ( m ):( m^2 + 1 = 2m )( m^2 - 2m + 1 = 0 )( (m - 1)^2 = 0 )So, ( m = 1 ).Case 2: ( k geq 2.5 )( m + frac{1}{m} geq 2.5 )Let me solve this inequality.Multiply both sides by ( m ) (since ( m > 0 )):( m^2 + 1 geq 2.5m )Rearrange:( m^2 - 2.5m + 1 geq 0 )Let me solve the quadratic equation ( m^2 - 2.5m + 1 = 0 ).Using the quadratic formula:( m = frac{2.5 pm sqrt{(2.5)^2 - 4 cdot 1 cdot 1}}{2} )Calculate discriminant:( D = 6.25 - 4 = 2.25 )So,( m = frac{2.5 pm 1.5}{2} )So,( m = frac{2.5 + 1.5}{2} = frac{4}{2} = 2 )or( m = frac{2.5 - 1.5}{2} = frac{1}{2} = 0.5 )So, the quadratic ( m^2 - 2.5m + 1 ) factors as ( (m - 2)(m - 0.5) ).Since the coefficient of ( m^2 ) is positive, the quadratic opens upwards. Therefore, the inequality ( m^2 - 2.5m + 1 geq 0 ) holds when ( m leq 0.5 ) or ( m geq 2 ).But since ( m > 0 ), the solution is ( 0 < m leq 0.5 ) or ( m geq 2 ).Therefore, combining both cases:- When ( m = 1 ), ( k = 2 ), which gives exactly one extreme at ( x = 1 ).- When ( 0 < m leq 0.5 ) or ( m geq 2 ), ( k geq 2.5 ), which gives exactly one extreme in ( (0, 1) ).Wait, but when ( m = 1 ), ( k = 2 ), which is the minimum value of ( k ). So, in this case, the function ( h(x) ) has exactly one extreme at ( x = 1 ).For ( m ) such that ( k geq 2.5 ), which corresponds to ( 0 < m leq 0.5 ) or ( m geq 2 ), the function ( h(x) ) has exactly one extreme in ( (0, 1) ).Therefore, the range of ( m ) is ( 0 < m leq 0.5 ) or ( m geq 2 ).But wait, when ( m = 0.5 ), ( k = 0.5 + 2 = 2.5 ). So, at ( m = 0.5 ), ( k = 2.5 ), which is the boundary case where the solution is at ( x = 2 ), which is excluded. So, in ( (0, 2) ), there's only one solution in ( (0, 1) ).Similarly, when ( m = 2 ), ( k = 2 + 0.5 = 2.5 ), same as above.Therefore, the range of ( m ) is ( 0 < m leq 0.5 ) or ( m geq 2 ).So, that's the answer for the first part.**Problem 2:**Now, the second problem is a bit more complex. We have the function ( M(x) = f(x) - g(x) = ln x - frac{1}{2}x^2 + kx ). It's given that ( M(x) ) has two zero points ( x_1 ) and ( x_2 ) with ( x_1 > x_2 ), and they satisfy ( 2x_0 = x_1 + x_2 ). The question is whether the tangent line of ( M(x) ) at ( (x_0, M(x_0)) ) can be parallel to the line ( y = 1 ). If yes, find the equation of the tangent line; if not, explain why.First, let's understand what it means for the tangent line to be parallel to ( y = 1 ). The line ( y = 1 ) is a horizontal line with slope 0. Therefore, the tangent line at ( (x_0, M(x_0)) ) must have a slope of 0. That means ( M'(x_0) = 0 ).So, we need to check if there exists an ( x_0 ) such that:1. ( M'(x_0) = 0 )2. ( 2x_0 = x_1 + x_2 ), where ( x_1 ) and ( x_2 ) are the zeros of ( M(x) ).Additionally, we need to see if such an ( x_0 ) exists given the conditions.Let me start by computing ( M'(x) ):( M(x) = ln x - frac{1}{2}x^2 + kx )So,( M'(x) = frac{1}{x} - x + k )We need ( M'(x_0) = 0 ):( frac{1}{x_0} - x_0 + k = 0 )Let me write that as:( frac{1}{x_0} - x_0 + k = 0 ) --- (1)Also, we know that ( x_1 ) and ( x_2 ) are zeros of ( M(x) ), so:( M(x_1) = 0 ) and ( M(x_2) = 0 )So,( ln x_1 - frac{1}{2}x_1^2 + kx_1 = 0 ) --- (2)( ln x_2 - frac{1}{2}x_2^2 + kx_2 = 0 ) --- (3)Additionally, we have:( 2x_0 = x_1 + x_2 ) --- (4)So, we have four equations: (1), (2), (3), and (4).Our goal is to see if such ( x_0 ), ( x_1 ), ( x_2 ), and ( k ) exist.Let me try to manipulate these equations to see if a contradiction arises or if it's possible.First, subtract equation (3) from equation (2):( ln x_1 - ln x_2 - frac{1}{2}(x_1^2 - x_2^2) + k(x_1 - x_2) = 0 )Simplify:( lnleft(frac{x_1}{x_2}right) - frac{1}{2}(x_1 - x_2)(x_1 + x_2) + k(x_1 - x_2) = 0 )Factor out ( (x_1 - x_2) ):( (x_1 - x_2)left[ frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} - frac{1}{2}(x_1 + x_2) + k right] = 0 )Since ( x_1 neq x_2 ) (they are distinct zeros), we can divide both sides by ( (x_1 - x_2) ):( frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} - frac{1}{2}(x_1 + x_2) + k = 0 )From equation (4), ( x_1 + x_2 = 2x_0 ). Let's substitute that:( frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} - frac{1}{2}(2x_0) + k = 0 )Simplify:( frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} - x_0 + k = 0 )From equation (1), we have ( frac{1}{x_0} - x_0 + k = 0 ). Let's write that as:( k = x_0 - frac{1}{x_0} ) --- (1a)Substitute ( k ) from (1a) into the equation we just derived:( frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} - x_0 + (x_0 - frac{1}{x_0}) = 0 )Simplify:( frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} - frac{1}{x_0} = 0 )So,( frac{lnleft(frac{x_1}{x_2}right)}{x_1 - x_2} = frac{1}{x_0} ) --- (5)Now, let's denote ( u = frac{x_1}{x_2} ). Since ( x_1 > x_2 > 0 ), ( u > 1 ).Let me express ( x_1 = u x_2 ).From equation (4):( x_1 + x_2 = 2x_0 )Substitute ( x_1 = u x_2 ):( u x_2 + x_2 = 2x_0 )( x_2(u + 1) = 2x_0 )So,( x_2 = frac{2x_0}{u + 1} )Similarly,( x_1 = u x_2 = frac{2u x_0}{u + 1} )Now, let's substitute ( x_1 ) and ( x_2 ) into equation (5):( frac{ln u}{x_1 - x_2} = frac{1}{x_0} )Compute ( x_1 - x_2 ):( x_1 - x_2 = frac{2u x_0}{u + 1} - frac{2x_0}{u + 1} = frac{2(u - 1)x_0}{u + 1} )So,( frac{ln u}{frac{2(u - 1)x_0}{u + 1}} = frac{1}{x_0} )Simplify:( frac{ln u (u + 1)}{2(u - 1)x_0} = frac{1}{x_0} )Multiply both sides by ( x_0 ):( frac{ln u (u + 1)}{2(u - 1)} = 1 )So,( ln u (u + 1) = 2(u - 1) )Let me write this as:( ln u = frac{2(u - 1)}{u + 1} ) --- (6)Now, we need to solve equation (6) for ( u > 1 ).Let me define the function ( y(u) = ln u - frac{2(u - 1)}{u + 1} ) for ( u > 1 ).We need to find if there exists ( u > 1 ) such that ( y(u) = 0 ).Let me analyze the behavior of ( y(u) ):First, compute the limit as ( u to 1^+ ):( ln 1 = 0 )( frac{2(1 - 1)}{1 + 1} = 0 )So, ( y(1) = 0 - 0 = 0 ). But ( u > 1 ), so we need to see the behavior just above 1.Compute the derivative ( y'(u) ):( y'(u) = frac{1}{u} - frac{2[(u + 1) - (u - 1)]}{(u + 1)^2} )Simplify the numerator of the second term:( (u + 1) - (u - 1) = 2 )So,( y'(u) = frac{1}{u} - frac{2 cdot 2}{(u + 1)^2} = frac{1}{u} - frac{4}{(u + 1)^2} )Let me compute ( y'(1) ):( y'(1) = 1 - frac{4}{4} = 1 - 1 = 0 )So, at ( u = 1 ), the derivative is 0.Now, let's compute ( y(u) ) for ( u > 1 ):Take ( u = 2 ):( y(2) = ln 2 - frac{2(2 - 1)}{2 + 1} = ln 2 - frac{2}{3} approx 0.6931 - 0.6667 = 0.0264 > 0 )Take ( u = 3 ):( y(3) = ln 3 - frac{2(3 - 1)}{3 + 1} = ln 3 - frac{4}{4} = ln 3 - 1 approx 1.0986 - 1 = 0.0986 > 0 )Take ( u = 1.5 ):( y(1.5) = ln 1.5 - frac{2(0.5)}{2.5} = ln 1.5 - frac{1}{2.5} approx 0.4055 - 0.4 = 0.0055 > 0 )Take ( u = 1.1 ):( y(1.1) = ln 1.1 - frac{2(0.1)}{2.1} approx 0.0953 - frac{0.2}{2.1} approx 0.0953 - 0.0952 = 0.0001 > 0 )So, as ( u ) approaches 1 from above, ( y(u) ) approaches 0 from above.Now, let's check the derivative ( y'(u) ) for ( u > 1 ):We have ( y'(u) = frac{1}{u} - frac{4}{(u + 1)^2} )Let me see if ( y'(u) ) is positive or negative for ( u > 1 ).Compute ( y'(2) = frac{1}{2} - frac{4}{9} approx 0.5 - 0.4444 = 0.0556 > 0 )Compute ( y'(1.5) = frac{1}{1.5} - frac{4}{(2.5)^2} approx 0.6667 - frac{4}{6.25} approx 0.6667 - 0.64 = 0.0267 > 0 )Compute ( y'(1.1) = frac{1}{1.1} - frac{4}{(2.1)^2} approx 0.9091 - frac{4}{4.41} approx 0.9091 - 0.9070 = 0.0021 > 0 )So, for ( u > 1 ), ( y'(u) > 0 ). Therefore, ( y(u) ) is increasing for ( u > 1 ).Since ( y(1) = 0 ) and ( y(u) ) is increasing for ( u > 1 ), ( y(u) > 0 ) for all ( u > 1 ).Therefore, the equation ( y(u) = 0 ) has no solution for ( u > 1 ). The only solution is at ( u = 1 ), but ( u > 1 ) by definition, so there's no solution.This means that our assumption that such ( x_0 ), ( x_1 ), ( x_2 ), and ( k ) exist leads to a contradiction because equation (6) has no solution for ( u > 1 ).Therefore, it's impossible for the tangent line of ( M(x) ) at ( (x_0, M(x_0)) ) to be parallel to the line ( y = 1 ).**Final Answer**1. The range of ( m ) is ( boxed{0 < m leq frac{1}{2} text{ or } m geq 2} ).2. The tangent line cannot be parallel to ( y = 1 ). The answer is ( boxed{text{No}} ).

💡Alright, so I have this problem here: "Compute the smallest positive integer that does not appear in any problem statement on any round at HMMT November 2023." Hmm, okay. Let me try to figure this out step by step.First, I need to understand what exactly is being asked. It's asking for the smallest positive integer that isn't mentioned in any of the problem statements from HMMT November 2023. That means I need to look through all the problem statements from that competition and check which numbers are used. Then, I can identify the smallest number that's missing.But wait, do I have access to all the problem statements from HMMT November 2023? I'm not sure. Maybe I can find them online or through some resources. Alternatively, perhaps there's a list or a table provided that shows which numbers appear in which problems. Let me check.Oh, here's a table that lists numbers from 1 to 21 and shows which round and problem each number appears in. That's helpful. So, according to this table, every number from 1 to 21 is mentioned in at least one problem statement. For example, number 1 appears in Guts round problem 21, number 2 is in Guts problem 13, and so on, all the way up to number 21, which is in Team round problem 7.So, if I follow this table, all numbers from 1 to 21 are accounted for. That means the smallest positive integer not appearing in any problem statement would be the next integer after 21, which is 22. Is that correct?Wait, let me make sure I'm not missing anything. Maybe there are numbers beyond 21 that are also used, but the table only goes up to 21. So, if I'm only considering numbers up to 21, then 22 would be the smallest missing. But if there are numbers beyond 21 that are also used, then 22 might still be missing or maybe another number is missing.But since the table only goes up to 21, and all numbers from 1 to 21 are present, I think it's safe to assume that 22 is the smallest positive integer not appearing in any problem statement. Unless there's a number between 1 and 21 that's missing, but according to the table, none are missing.Let me double-check the table to ensure I didn't miss any numbers. Starting from 1: yes, 1 is there. 2 is there, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21. Yep, all accounted for.So, unless there's a number less than or equal to 21 that's missing, which doesn't seem to be the case, 22 is the smallest positive integer not appearing in any problem statement. Therefore, I think the answer is 22.But just to be thorough, maybe I should consider if there are any numbers beyond 21 that are used. If, for example, number 23 is used, but 22 isn't, then 22 would still be the answer. However, if 22 is used and 23 isn't, then 23 would be the answer. But since the table only goes up to 21, I don't have information about numbers beyond that. However, the question is about the smallest positive integer, so even if higher numbers are used, 22 would still be the smallest missing one if it's not used.Therefore, based on the information provided, the smallest positive integer not appearing in any problem statement is 22.

💡First, I need to determine the relative speed between the train and the man since they are moving in the same direction. The train's speed is 68 km/h, and the man's speed is 8 km/h. By subtracting the man's speed from the train's speed, I can find the relative speed.Next, I'll convert the relative speed from kilometers per hour to meters per second to match the time unit provided in seconds. This conversion is essential for accurate calculations.Once I have the relative speed in meters per second, I can calculate the length of the train by multiplying the relative speed by the time it takes for the train to pass the man. This will give me the length of the train in meters.Finally, I'll present the calculated length of the train in a clear and concise manner.

💡First, I need to determine the original number of day students based on the initial ratio of boarders to day students, which is 5 to 12. Given that there are 330 original boarders, I can set up a proportion to find the number of day students.Next, I'll calculate the number of day students by solving the proportion. This involves cross-multiplying and dividing to find the value of ( d ), the number of day students.After finding the original number of day students, I'll consider the new ratio of boarders to day students, which is 1 to 2. With the number of day students remaining unchanged, I'll set up another proportion to find the total number of boarders after the new ones joined.Finally, I'll solve for ( x ), the number of new boarders, by manipulating the equation derived from the new ratio. This will give me the total number of new boarders that joined the school.

💡Okay, so I have this problem where I need to figure out for which real numbers ( x > 1 ) there exists a triangle with side lengths ( x^4 + x^3 + 2x^2 + x + 1 ), ( 2x^3 + x^2 + 2x + 1 ), and ( x^4 - 1 ). Hmm, triangles and their side lengths, right? I remember that for three lengths to form a triangle, they must satisfy the triangle inequality. That is, the sum of any two sides must be greater than the third side. So, I need to check all three combinations of these sides to make sure they satisfy this condition.Let me denote the sides as follows to make it easier:- ( a = x^4 + x^3 + 2x^2 + x + 1 )- ( b = 2x^3 + x^2 + 2x + 1 )- ( c = x^4 - 1 )So, I need to check three inequalities:1. ( a + b > c )2. ( a + c > b )3. ( b + c > a )Starting with the first inequality: ( a + b > c ). Let me compute ( a + b ) and then subtract ( c ) to see if the result is positive.Calculating ( a + b ):( a + b = (x^4 + x^3 + 2x^2 + x + 1) + (2x^3 + x^2 + 2x + 1) )Combine like terms:- ( x^4 ) stays as is.- ( x^3 + 2x^3 = 3x^3 )- ( 2x^2 + x^2 = 3x^2 )- ( x + 2x = 3x )- ( 1 + 1 = 2 )So, ( a + b = x^4 + 3x^3 + 3x^2 + 3x + 2 )Now, subtract ( c ) from this:( a + b - c = (x^4 + 3x^3 + 3x^2 + 3x + 2) - (x^4 - 1) )Simplify:- ( x^4 - x^4 = 0 )- ( 3x^3 ) remains- ( 3x^2 ) remains- ( 3x ) remains- ( 2 - (-1) = 3 )So, ( a + b - c = 3x^3 + 3x^2 + 3x + 3 )Factor out a 3:( 3(x^3 + x^2 + x + 1) )Since ( x > 1 ), each term inside the parentheses is positive, so the entire expression is positive. Therefore, ( a + b > c ) holds true for ( x > 1 ).Moving on to the second inequality: ( a + c > b ). Let's compute ( a + c ) and subtract ( b ).Calculating ( a + c ):( a + c = (x^4 + x^3 + 2x^2 + x + 1) + (x^4 - 1) )Combine like terms:- ( x^4 + x^4 = 2x^4 )- ( x^3 ) remains- ( 2x^2 ) remains- ( x ) remains- ( 1 - 1 = 0 )So, ( a + c = 2x^4 + x^3 + 2x^2 + x )Subtract ( b ) from this:( a + c - b = (2x^4 + x^3 + 2x^2 + x) - (2x^3 + x^2 + 2x + 1) )Simplify:- ( 2x^4 ) remains- ( x^3 - 2x^3 = -x^3 )- ( 2x^2 - x^2 = x^2 )- ( x - 2x = -x )- ( 0 - 1 = -1 )So, ( a + c - b = 2x^4 - x^3 + x^2 - x - 1 )Hmm, this is a quartic polynomial. I need to determine if this is positive for ( x > 1 ). Let me see if I can factor this or analyze its behavior.Looking at ( 2x^4 - x^3 + x^2 - x - 1 ), it's a bit complex. Maybe I can test some values of ( x > 1 ) to see if it's positive.Let's try ( x = 2 ):( 2*(16) - 8 + 4 - 2 - 1 = 32 - 8 + 4 - 2 - 1 = 25 ), which is positive.What about ( x = 1.5 ):( 2*(5.0625) - 3.375 + 2.25 - 1.5 - 1 = 10.125 - 3.375 + 2.25 - 1.5 - 1 = 6.5 ), still positive.How about ( x = 1.1 ):( 2*(1.4641) - 1.331 + 1.21 - 1.1 - 1 = 2.9282 - 1.331 + 1.21 - 1.1 - 1 ≈ 0.7072 ), which is still positive.It seems like for ( x > 1 ), this expression remains positive. Maybe I can factor it or find its roots to be sure.Alternatively, since it's a quartic with a positive leading coefficient, as ( x ) becomes large, the term ( 2x^4 ) dominates, making the expression positive. For ( x = 1 ), let's compute:( 2 - 1 + 1 - 1 - 1 = 0 ). So, at ( x = 1 ), the expression is zero.Given that at ( x = 1 ), it's zero, and for ( x > 1 ), it's positive, and since it's a continuous function, it must be positive for all ( x > 1 ). Therefore, ( a + c > b ) holds for ( x > 1 ).Now, the third inequality: ( b + c > a ). Let's compute ( b + c ) and subtract ( a ).Calculating ( b + c ):( b + c = (2x^3 + x^2 + 2x + 1) + (x^4 - 1) )Combine like terms:- ( x^4 ) remains- ( 2x^3 ) remains- ( x^2 ) remains- ( 2x ) remains- ( 1 - 1 = 0 )So, ( b + c = x^4 + 2x^3 + x^2 + 2x )Subtract ( a ) from this:( b + c - a = (x^4 + 2x^3 + x^2 + 2x) - (x^4 + x^3 + 2x^2 + x + 1) )Simplify:- ( x^4 - x^4 = 0 )- ( 2x^3 - x^3 = x^3 )- ( x^2 - 2x^2 = -x^2 )- ( 2x - x = x )- ( 0 - 1 = -1 )So, ( b + c - a = x^3 - x^2 + x - 1 )Again, this is a cubic polynomial. Let me see if I can factor this or analyze its behavior for ( x > 1 ).Looking at ( x^3 - x^2 + x - 1 ), maybe I can factor by grouping:Group as ( (x^3 - x^2) + (x - 1) = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1) )Ah, that's a good catch! So, ( x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1) )Since ( x > 1 ), ( x - 1 > 0 ), and ( x^2 + 1 > 0 ) for all real ( x ). Therefore, the entire expression is positive for ( x > 1 ). Thus, ( b + c > a ) holds true for ( x > 1 ).So, all three triangle inequalities hold for ( x > 1 ). Therefore, there exists a triangle with the given side lengths for all real numbers ( x > 1 ).I think that's it. I've checked all three conditions, and each one holds true for ( x > 1 ). So, the answer should be all real numbers ( x ) greater than 1.

💡Okay, so I have this problem here: I'm told that f(x) is an odd function, and when x is greater than 0, f(x) equals x times (1 plus x). I need to find what f(x) is when x is less than 0. Hmm, alright, let me think about this step by step.First, I remember that an odd function has a specific property. I think it's something like f(-x) equals negative f(x). Let me verify that. Yes, that's right: for any odd function, if you plug in negative x, you get the negative of the function evaluated at x. So, f(-x) = -f(x). That seems correct.Now, the problem gives me f(x) when x is positive, which is f(x) = x(1 + x). So, for positive x, it's straightforward. But I need to find f(x) when x is negative. Since f(x) is odd, I can use the property f(-x) = -f(x). Maybe I can express f(x) for negative x in terms of f(-x), which would be positive.Let me denote x as a negative number. So, if x is negative, then -x is positive. That makes sense because multiplying a negative number by -1 gives a positive number. So, if x is negative, then -x is positive, and I can use the given expression for f(-x) because -x is positive.Given that f(-x) = (-x)(1 + (-x)). Wait, hold on, is that right? Let me make sure. If I substitute -x into the function, which is defined for positive arguments, then f(-x) should be (-x)(1 + (-x)). Let me compute that:f(-x) = (-x)(1 + (-x)) = (-x)(1 - x) = -x + x^2.Wait, that doesn't seem right. Let me check my substitution again. The function is f(x) = x(1 + x) for x > 0. So, f(-x) when x < 0 would be f(-x) = (-x)(1 + (-x)) because I'm substituting -x into the function. So, that's (-x)(1 - x). Let me expand that:(-x)(1 - x) = -x * 1 + (-x)*(-x) = -x + x^2.So, f(-x) = -x + x^2. But since f is an odd function, f(-x) = -f(x). Therefore, -f(x) = -x + x^2. To find f(x), I can multiply both sides by -1:f(x) = x - x^2.Wait, that seems different from what I initially thought. Let me double-check my steps.1. Start with f(x) being odd, so f(-x) = -f(x).2. For x < 0, -x > 0, so f(-x) = (-x)(1 + (-x)) = (-x)(1 - x) = -x + x^2.3. Therefore, -f(x) = -x + x^2.4. Multiply both sides by -1: f(x) = x - x^2.Hmm, that seems consistent. So, for x < 0, f(x) is x minus x squared. Let me write that as f(x) = x - x^2.But wait, let me think about this again. If I plug in a negative x into f(x) = x - x^2, does it satisfy the odd function property?Let's test with a specific value. Suppose x = -1. Then f(-1) should be (-1) - (-1)^2 = -1 - 1 = -2. Now, f(1) is 1*(1 + 1) = 2. So, f(-1) = -2, which is indeed -f(1). That checks out.Let me try another value. Let's take x = -2. Then f(-2) = (-2) - (-2)^2 = -2 - 4 = -6. On the other hand, f(2) = 2*(1 + 2) = 6. So, f(-2) = -6, which is -f(2). That also works.Wait, but earlier I thought f(-x) was -x + x^2, which led to f(x) = x - x^2. But when I plug in x = -1, I get f(-1) = -2, which is correct. So, maybe my initial confusion was unwarranted.Alternatively, maybe I can approach this differently. Since f(x) is odd, f(x) = -f(-x). So, for x < 0, f(x) = -f(-x). But -x is positive, so f(-x) is (-x)(1 + (-x)) = (-x)(1 - x) = -x + x^2. Therefore, f(x) = -(-x + x^2) = x - x^2.Yes, that's the same result as before. So, it seems consistent.Wait, but in the initial problem statement, when x > 0, f(x) = x(1 + x). So, for x > 0, f(x) = x + x^2. Then, for x < 0, f(x) = x - x^2. Let me see if that makes sense graphically.If f(x) is odd, its graph is symmetric about the origin. So, if for positive x, it's a parabola opening upwards, then for negative x, it should be a reflection of that parabola across the origin. So, the function for x < 0 should be a downward-opening parabola.But wait, f(x) = x - x^2 for x < 0. Let's see, if I write it as f(x) = -x^2 + x, that's a quadratic with a negative coefficient on x^2, so it opens downward. That makes sense because it's the reflection of the upward-opening parabola for positive x.Alternatively, I can write f(x) as -x^2 + x, which is the same as x - x^2. So, that seems correct.Let me also check the behavior at x = 0. Since f(x) is odd, f(0) should be 0. Let's see: for x approaching 0 from the positive side, f(x) approaches 0*(1 + 0) = 0. For x approaching 0 from the negative side, f(x) approaches 0 - 0^2 = 0. So, f(0) = 0, which is consistent with being an odd function.Alright, I think I've verified it sufficiently. So, the function f(x) for x < 0 is f(x) = x - x^2.Wait, but let me think again. If f(x) is odd, then f(-x) = -f(x). So, if I have f(x) = x(1 + x) for x > 0, then for x < 0, f(x) = -f(-x) = -[(-x)(1 + (-x))] = -[(-x)(1 - x)] = -[-x + x^2] = x - x^2. Yep, that's the same result.So, I think I'm confident that f(x) for x < 0 is x - x^2.But just to make sure, let me compute f(-x) and see if it equals -f(x).Take x = -1:f(-1) = (-1) - (-1)^2 = -1 - 1 = -2.f(1) = 1*(1 + 1) = 2.So, f(-1) = -2 = -f(1). That works.Take x = -2:f(-2) = (-2) - (-2)^2 = -2 - 4 = -6.f(2) = 2*(1 + 2) = 6.So, f(-2) = -6 = -f(2). That also works.Another test: x = -0.5.f(-0.5) = (-0.5) - (-0.5)^2 = -0.5 - 0.25 = -0.75.f(0.5) = 0.5*(1 + 0.5) = 0.5*1.5 = 0.75.So, f(-0.5) = -0.75 = -f(0.5). Perfect.Alright, I think I've thoroughly checked this. So, the conclusion is that for x < 0, f(x) is equal to x minus x squared, or f(x) = x - x^2.Just to recap:1. Given f(x) is odd: f(-x) = -f(x).2. For x > 0, f(x) = x(1 + x).3. For x < 0, let x = -y where y > 0.4. Then f(x) = f(-y) = -f(y) = -[y(1 + y)].5. Substitute y = -x: f(x) = -[(-x)(1 + (-x))] = -[(-x)(1 - x)] = -[-x + x^2] = x - x^2.Yes, that's another way to look at it, using substitution. So, that also leads to the same result.I think I've covered all the angles here. I can't find any mistakes in my reasoning, and the test cases all check out. So, I'm pretty sure that the function f(x) for x < 0 is x minus x squared.