answer：First, I need to determine the measures of the angles in triangle ( ABC ). The problem states that angle ( C ) is three times as large as angle ( A ), and angle ( B ) is twice as large as angle ( A ).Let ( x ) represent the measure of angle ( A ) in degrees. Then, angle ( C ) would be ( 3x ) degrees, and angle ( B ) would be ( 2x ) degrees.Since the sum of the interior angles in any triangle is ( 180^circ ), I can set up the equation:[x + 3x + 2x = 180^circ]Combining like terms, this simplifies to:[6x = 180^circ]Solving for ( x ), I divide both sides by 6:[x = 30^circ]Now, to find the measure of angle ( B ), I multiply ( x ) by 2:[angle B = 2x = 2 times 30^circ = 60^circ]Therefore, the measure of angle ( B ) is ( 60^circ ).

answer：First, I recognize that both the length and the width of the rectangular field are being reduced by 40%. This means each dimension is being multiplied by 60% of its original value.Next, I calculate the new area of the field by multiplying the reduced length and the reduced width. This results in the new area being 36% of the original area.Finally, to determine the percentage by which the area has diminished, I subtract the new area percentage from the original area percentage and find that the area has decreased by 64%.

answer：Alright, so I have this geometry problem here about three planes intersecting each other. Let me try to visualize it first. We have three planes, let's call them Plane A, Plane B, and Plane C. Each pair of these planes intersects along a line. So, Plane A and Plane B intersect along Line 1, Plane A and Plane C intersect along Line 2, and Plane B and Plane C intersect along Line 3.Now, the problem says that two of these three lines intersect at a point M. Let's say Line 1 and Line 2 intersect at point M. I need to prove that the third line, which is Line 3, also passes through point M.Hmm, okay. So, if Line 1 is the intersection of Plane A and Plane B, and Line 2 is the intersection of Plane A and Plane C, then both Line 1 and Line 2 lie on Plane A. Since they intersect at point M, that means M is a common point on both Line 1 and Line 2, and therefore, M must be on Plane A.Similarly, since Line 1 is the intersection of Plane A and Plane B, M is also on Plane B. And since Line 2 is the intersection of Plane A and Plane C, M is also on Plane C. So, M is a common point of all three planes: Plane A, Plane B, and Plane C.Now, if M is on all three planes, then it must lie on the intersection of any two of them. Specifically, since Line 3 is the intersection of Plane B and Plane C, M must lie on Line 3 as well. Therefore, Line 3 passes through point M.Wait, let me make sure I didn't skip any steps. So, M is on Plane A, Plane B, and Plane C. Since Line 3 is where Plane B and Plane C meet, any point common to both Plane B and Plane C must lie on Line 3. Since M is on both Plane B and Plane C, it must be on Line 3.Yeah, that makes sense. So, the key idea is that if a point is common to all three planes, it must lie on the intersection of any two of them, which in this case is Line 3. Therefore, Line 3 must pass through M.I think that's a solid reasoning. Let me try to put it in a more formal way to make sure it's clear.First, define the three planes and their lines of intersection. Then, show that the point M, being the intersection of two lines, is common to all three planes. Finally, conclude that M must lie on the third line of intersection because it's common to the two planes that define that line.Yeah, that should do it.

answer：Okay, so I have this problem about a quadratic function ( f(x) = ax^2 + bx + c ) with some conditions, and I need to find the minimum value of the product ( f(x_1)f(x_2)f(x_3) ) given certain constraints on ( x_1, x_2, x_3 ). Hmm, let me try to break this down step by step.First, the quadratic function has three conditions:1. ( f(-1) = 0 ). That means when I plug in ( x = -1 ) into the function, it equals zero. So, ( a(-1)^2 + b(-1) + c = 0 ), which simplifies to ( a - b + c = 0 ).2. For all real numbers ( x ), ( f(x) geq x ). So, the quadratic function is always above or equal to the line ( y = x ).3. For all ( x ) in the interval ( (0, 2) ), ( f(x) leq frac{(x + 1)^2}{4} ). That means within this interval, the quadratic function is below or equal to this specific parabola.Additionally, I have three variables ( x_1, x_2, x_3 ) in the interval ( (0, 2) ) such that ( frac{1}{x_1} + frac{1}{x_2} + frac{1}{x_3} = 3 ). I need to find the minimum of the product ( f(x_1)f(x_2)f(x_3) ).Alright, let's tackle the quadratic function first. From condition 1, I have ( a - b + c = 0 ). Maybe I can find another equation by using condition 2 or 3.Condition 2 says ( f(x) geq x ) for all ( x ). Let me rewrite this inequality:[ ax^2 + bx + c geq x ][ ax^2 + (b - 1)x + c geq 0 ]Since this must hold for all real ( x ), the quadratic ( ax^2 + (b - 1)x + c ) must be non-negative everywhere. That means it must either touch the x-axis or lie entirely above it. Therefore, its discriminant must be less than or equal to zero.The discriminant ( D ) of ( ax^2 + (b - 1)x + c ) is:[ D = (b - 1)^2 - 4ac leq 0 ]So, ( (b - 1)^2 leq 4ac ).Now, condition 3 says that for ( x in (0, 2) ), ( f(x) leq frac{(x + 1)^2}{4} ). Let me compute ( frac{(x + 1)^2}{4} ):[ frac{(x + 1)^2}{4} = frac{x^2 + 2x + 1}{4} = frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4} ]So, for ( x in (0, 2) ),[ ax^2 + bx + c leq frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4} ]Which can be rewritten as:[ (a - frac{1}{4})x^2 + (b - frac{1}{2})x + (c - frac{1}{4}) leq 0 ]This inequality must hold for all ( x ) in ( (0, 2) ). Hmm, so the quadratic ( (a - frac{1}{4})x^2 + (b - frac{1}{2})x + (c - frac{1}{4}) ) is non-positive on ( (0, 2) ). That suggests that the quadratic might be tangent to the x-axis or lies below it in this interval.But since ( f(x) geq x ) for all ( x ), and ( f(x) leq frac{(x + 1)^2}{4} ) in ( (0, 2) ), perhaps ( f(x) ) is equal to ( frac{(x + 1)^2}{4} ) at some points in ( (0, 2) ). Maybe at ( x = 1 )?Let me check ( x = 1 ). From condition 2, ( f(1) geq 1 ). From condition 3, ( f(1) leq frac{(1 + 1)^2}{4} = 1 ). So, ( f(1) = 1 ).So, ( f(1) = 1 ). Let's write that equation:[ a(1)^2 + b(1) + c = 1 ][ a + b + c = 1 ]Now, I have two equations:1. ( a - b + c = 0 ) (from ( f(-1) = 0 ))2. ( a + b + c = 1 ) (from ( f(1) = 1 ))Let me subtract equation 1 from equation 2:[ (a + b + c) - (a - b + c) = 1 - 0 ][ 2b = 1 ][ b = frac{1}{2} ]So, ( b = frac{1}{2} ). Plugging this back into equation 1:[ a - frac{1}{2} + c = 0 ][ a + c = frac{1}{2} ]So, ( a + c = frac{1}{2} ). Now, let's recall the discriminant condition from earlier:[ (b - 1)^2 leq 4ac ]Since ( b = frac{1}{2} ),[ (frac{1}{2} - 1)^2 leq 4ac ][ (-frac{1}{2})^2 leq 4ac ][ frac{1}{4} leq 4ac ][ ac geq frac{1}{16} ]But from ( a + c = frac{1}{2} ), by AM-GM inequality:[ sqrt{ac} leq frac{a + c}{2} = frac{1}{4} ][ ac leq frac{1}{16} ]So, combining ( ac geq frac{1}{16} ) and ( ac leq frac{1}{16} ), we get ( ac = frac{1}{16} ). Therefore, ( a = c = frac{1}{4} ).So, the quadratic function is:[ f(x) = frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4} ]Which can be written as:[ f(x) = frac{1}{4}(x^2 + 2x + 1) = frac{1}{4}(x + 1)^2 ]Alright, so ( f(x) = frac{1}{4}(x + 1)^2 ). Now, I need to find the minimum of ( f(x_1)f(x_2)f(x_3) ) where ( x_1, x_2, x_3 in (0, 2) ) and ( frac{1}{x_1} + frac{1}{x_2} + frac{1}{x_3} = 3 ).Let me express ( f(x_i) ) as:[ f(x_i) = frac{1}{4}(x_i + 1)^2 ]So, the product is:[ f(x_1)f(x_2)f(x_3) = left(frac{1}{4}right)^3 (x_1 + 1)^2(x_2 + 1)^2(x_3 + 1)^2 = frac{1}{64} left[(x_1 + 1)(x_2 + 1)(x_3 + 1)right]^2 ]Therefore, to minimize ( f(x_1)f(x_2)f(x_3) ), I need to minimize ( left[(x_1 + 1)(x_2 + 1)(x_3 + 1)right]^2 ), which is equivalent to minimizing ( (x_1 + 1)(x_2 + 1)(x_3 + 1) ).So, the problem reduces to minimizing ( (x_1 + 1)(x_2 + 1)(x_3 + 1) ) given that ( frac{1}{x_1} + frac{1}{x_2} + frac{1}{x_3} = 3 ) and ( x_i in (0, 2) ).Hmm, this seems like an optimization problem with constraints. Maybe I can use Lagrange multipliers or some inequality to solve it.Alternatively, since the variables are symmetric, perhaps the minimum occurs when all ( x_i ) are equal. Let me test that.Assume ( x_1 = x_2 = x_3 = x ). Then, the constraint becomes:[ frac{3}{x} = 3 ][ frac{1}{x} = 1 ][ x = 1 ]So, if ( x_1 = x_2 = x_3 = 1 ), then:[ (1 + 1)^3 = 8 ]So, ( (x_1 + 1)(x_2 + 1)(x_3 + 1) = 8 ), and the product ( f(x_1)f(x_2)f(x_3) = frac{1}{64} times 8^2 = frac{64}{64} = 1 ).Is this the minimum? Let me see if I can find a configuration where the product is smaller.Suppose one of the ( x_i ) is close to 0, say ( x_1 ) approaches 0. Then, ( frac{1}{x_1} ) approaches infinity, which would violate the constraint ( frac{1}{x_1} + frac{1}{x_2} + frac{1}{x_3} = 3 ). So, none of the ( x_i ) can be too close to 0.Alternatively, suppose one of the ( x_i ) is close to 2. Let me set ( x_1 = 2 ), then ( frac{1}{x_1} = frac{1}{2} ). Then, the remaining ( frac{1}{x_2} + frac{1}{x_3} = 3 - frac{1}{2} = frac{5}{2} ). Let me set ( x_2 = x_3 ), so ( frac{2}{x_2} = frac{5}{2} ), which gives ( x_2 = frac{4}{5} ).Then, ( (x_1 + 1)(x_2 + 1)(x_3 + 1) = (2 + 1)left(frac{4}{5} + 1right)^2 = 3 times left(frac{9}{5}right)^2 = 3 times frac{81}{25} = frac{243}{25} approx 9.72 ), which is larger than 8. So, the product is larger in this case.What if two variables are equal and one is different? Let me try ( x_1 = x_2 ) and ( x_3 ) different.Let ( x_1 = x_2 = a ) and ( x_3 = b ). Then, the constraint becomes:[ frac{2}{a} + frac{1}{b} = 3 ]Let me express ( b ) in terms of ( a ):[ frac{1}{b} = 3 - frac{2}{a} ][ b = frac{1}{3 - frac{2}{a}} = frac{a}{3a - 2} ]Now, ( x_3 = frac{a}{3a - 2} ). Since ( x_3 in (0, 2) ), the denominator must be positive:[ 3a - 2 > 0 ][ a > frac{2}{3} ]Also, ( x_3 < 2 ):[ frac{a}{3a - 2} < 2 ][ a < 2(3a - 2) ][ a < 6a - 4 ][ -5a < -4 ][ a > frac{4}{5} ]So, ( a in (frac{4}{5}, 2) ).Now, the product ( (a + 1)^2 left( frac{a}{3a - 2} + 1 right) ).Simplify ( frac{a}{3a - 2} + 1 ):[ frac{a}{3a - 2} + 1 = frac{a + 3a - 2}{3a - 2} = frac{4a - 2}{3a - 2} ]So, the product becomes:[ (a + 1)^2 times frac{4a - 2}{3a - 2} ]Let me denote this as ( P(a) ):[ P(a) = (a + 1)^2 times frac{4a - 2}{3a - 2} ]I need to find the minimum of ( P(a) ) for ( a in (frac{4}{5}, 2) ).To find the minimum, I can take the derivative of ( P(a) ) with respect to ( a ) and set it to zero.First, let me write ( P(a) ) as:[ P(a) = (a + 1)^2 times frac{4a - 2}{3a - 2} ]Let me compute the derivative ( P'(a) ).Let me denote ( u = (a + 1)^2 ) and ( v = frac{4a - 2}{3a - 2} ). Then, ( P(a) = u times v ).Using the product rule:[ P'(a) = u'v + uv' ]First, compute ( u' ):[ u = (a + 1)^2 ][ u' = 2(a + 1) ]Next, compute ( v' ):[ v = frac{4a - 2}{3a - 2} ]Using the quotient rule:[ v' = frac{(4)(3a - 2) - (4a - 2)(3)}{(3a - 2)^2} ][ v' = frac{12a - 8 - 12a + 6}{(3a - 2)^2} ][ v' = frac{-2}{(3a - 2)^2} ]Now, put it all together:[ P'(a) = 2(a + 1) times frac{4a - 2}{3a - 2} + (a + 1)^2 times left( frac{-2}{(3a - 2)^2} right) ]Let me factor out ( 2(a + 1) ):[ P'(a) = 2(a + 1) left[ frac{4a - 2}{3a - 2} - frac{(a + 1)}{(3a - 2)^2} right] ]Let me combine the terms inside the brackets:Find a common denominator, which is ( (3a - 2)^2 ):[ frac{(4a - 2)(3a - 2) - (a + 1)}{(3a - 2)^2} ]Compute the numerator:[ (4a - 2)(3a - 2) - (a + 1) ][ = 12a^2 - 8a - 6a + 4 - a - 1 ][ = 12a^2 - 15a + 3 ]So, the derivative becomes:[ P'(a) = 2(a + 1) times frac{12a^2 - 15a + 3}{(3a - 2)^2} ]Set ( P'(a) = 0 ):[ 2(a + 1) times frac{12a^2 - 15a + 3}{(3a - 2)^2} = 0 ]Since ( 2(a + 1) ) is never zero for ( a > frac{4}{5} ), and the denominator ( (3a - 2)^2 ) is always positive, we set the numerator equal to zero:[ 12a^2 - 15a + 3 = 0 ]Solve the quadratic equation:[ 12a^2 - 15a + 3 = 0 ]Divide all terms by 3:[ 4a^2 - 5a + 1 = 0 ]Use quadratic formula:[ a = frac{5 pm sqrt{25 - 16}}{8} ][ a = frac{5 pm 3}{8} ]So, ( a = 1 ) or ( a = frac{1}{4} ).But ( a in (frac{4}{5}, 2) ), so ( a = 1 ) is the critical point.Now, check the second derivative or test intervals to confirm if this is a minimum.But since we have only one critical point in the interval, and the function tends to infinity as ( a ) approaches ( frac{4}{5} ) from the right and as ( a ) approaches 2 from the left, it's likely that ( a = 1 ) gives the minimum.So, when ( a = 1 ), ( b = frac{1}{3 - 2} = 1 ). So, all ( x_i = 1 ), which is the symmetric case.Therefore, the minimum of ( (x_1 + 1)(x_2 + 1)(x_3 + 1) ) is 8, achieved when ( x_1 = x_2 = x_3 = 1 ).Thus, the minimum value of ( f(x_1)f(x_2)f(x_3) ) is:[ frac{1}{64} times 8^2 = frac{64}{64} = 1 ]So, the minimum value is 1.**Final Answer**The minimum value of ( f(x_1) f(x_2) f(x_3) ) is boxed{1}.

answer：Okay, so I have this problem here where I need to compute the value of the expression:[ frac{binom{1/2}{2014} cdot 4^{2014}}{binom{4028}{2014}}.]Hmm, binomial coefficients with a real number upper index. I remember that the binomial coefficient for a real number ( x ) and a nonnegative integer ( k ) is defined as:[ binom{x}{k} = frac{x(x - 1)(x - 2) dotsm (x - k + 1)}{k!}.]So, for ( binom{1/2}{2014} ), that would be:[ binom{1/2}{2014} = frac{(1/2)(1/2 - 1)(1/2 - 2) dotsm (1/2 - 2014 + 1)}{2014!}.]Let me write out the numerator terms to see the pattern:The first term is ( 1/2 ).The second term is ( 1/2 - 1 = -1/2 ).The third term is ( 1/2 - 2 = -3/2 ).Continuing this way, each subsequent term decreases by 1. So, the ( k )-th term is ( 1/2 - (k - 1) ).So, for ( k = 2014 ), the last term is ( 1/2 - 2013 = -4025/2 ).Therefore, the numerator is:[ (1/2)(-1/2)(-3/2) dotsm (-4025/2).]I notice that each term is negative except the first one. Since there are 2014 terms, and the first term is positive, the rest 2013 terms are negative. So, the product will be negative because an odd number of negative terms multiply together.Let me factor out the ( 1/2 ) from each term:[ left( frac{1}{2} right)^{2014} times (1)(-1)(-3) dotsm (-4025).]Wait, actually, each term is ( (1/2 - n) ) where ( n ) goes from 0 to 2013. So, each term is ( (1 - 2n)/2 ). So, the numerator can be written as:[ prod_{n=0}^{2013} left( frac{1 - 2n}{2} right) = frac{1}{2^{2014}} prod_{n=0}^{2013} (1 - 2n).]Simplifying the product ( prod_{n=0}^{2013} (1 - 2n) ):Let me write out the terms:For ( n = 0 ): ( 1 - 0 = 1 ).For ( n = 1 ): ( 1 - 2 = -1 ).For ( n = 2 ): ( 1 - 4 = -3 )....For ( n = 2013 ): ( 1 - 4026 = -4025 ).So, the product is:[ 1 times (-1) times (-3) times dotsm times (-4025).]Since there are 2014 terms, and starting from 1, each subsequent term is negative and odd. So, the product is:[ (-1)^{2013} times (1 times 3 times 5 times dotsm times 4025).]Because starting from the second term, each term is negative, so there are 2013 negative signs, hence ( (-1)^{2013} = -1 ).So, the numerator becomes:[ frac{1}{2^{2014}} times (-1) times (1 times 3 times 5 times dotsm times 4025).]Now, the denominator of ( binom{1/2}{2014} ) is ( 2014! ).Putting it all together:[ binom{1/2}{2014} = frac{ (-1) times (1 times 3 times 5 times dotsm times 4025) }{ 2^{2014} times 2014! }.]Hmm, I need to relate this product ( 1 times 3 times 5 times dotsm times 4025 ) to something more manageable. I recall that the product of the first ( n ) odd numbers is related to the double factorial.Yes, the double factorial of an odd number ( (2n - 1)!! = 1 times 3 times 5 times dotsm times (2n - 1) ).In this case, the product ( 1 times 3 times 5 times dotsm times 4025 ) is ( (4025)!! ). But 4025 is ( 2 times 2013 + 1 ), so ( n = 2013 ).Wait, actually, 4025 is ( 2 times 2012.5 ), which isn't an integer. Hmm, maybe I need to adjust.Wait, 4025 is the last term, which is ( 2 times 2013 - 1 = 4025 ). So, yes, ( (4025)!! = (2 times 2013 - 1)!! ).But I also know that ( (2n)!! = 2^n n! ) and ( (2n - 1)!! = frac{(2n)!}{2^n n!} ).So, ( (4025)!! = frac{4026!}{2^{2013} times 2013!} ).Wait, let me check that.Yes, for ( (2n - 1)!! = frac{(2n)!}{2^n n!} ). So, if ( 2n - 1 = 4025 ), then ( n = 2013 ). Therefore:[ (4025)!! = frac{4026!}{2^{2013} times 2013!}.]Great, so substituting back into the numerator:[ binom{1/2}{2014} = frac{ (-1) times frac{4026!}{2^{2013} times 2013!} }{ 2^{2014} times 2014! }.]Simplify the denominator:[ 2^{2014} times 2014! = 2^{2014} times 2014 times 2013!.]So, putting it all together:[ binom{1/2}{2014} = frac{ -4026! }{ 2^{2013} times 2013! times 2^{2014} times 2014 times 2013! }.]Wait, hold on, let me write it step by step:Numerator: ( -4026! )Denominator: ( 2^{2013} times 2013! times 2^{2014} times 2014 times 2013! )Combine the powers of 2: ( 2^{2013 + 2014} = 2^{4027} )Combine the factorials: ( (2013!)^2 times 2014 )So,[ binom{1/2}{2014} = frac{ -4026! }{ 2^{4027} times (2013!)^2 times 2014 }.]Simplify ( 4026! ) in terms of ( 2014 ):Wait, ( 4026 = 2 times 2013 ), so ( 4026! = (2 times 2013)! ).But I also know that ( (2n)! = 2^{2n} (n!)^2 times frac{(2n - 1)!!}{n!} ) or something like that? Wait, maybe I should use the relation for double factorials.Alternatively, perhaps express ( 4026! ) as ( (2 times 2013)! ) and use the formula for factorial of even numbers.Wait, another approach: perhaps express ( 4026! ) as ( (2 times 2013)! ) and use the relation:[ (2n)! = 2^{2n} n! times prod_{k=1}^{n} (2k - 1).]But I'm not sure if that helps directly.Alternatively, maybe express ( 4026! ) in terms of ( 2014! ) and ( 2013! ).Wait, ( 4026! = 4026 times 4025 times dotsm times 2014 times 2013! ).But that might not be directly helpful.Wait, let me think about the expression we have:[ binom{1/2}{2014} = frac{ -4026! }{ 2^{4027} times (2013!)^2 times 2014 }.]So, let's write ( 4026! = 4026 times 4025! ).But ( 4025! = (2 times 2012.5)! ), which isn't an integer, so that might not help.Alternatively, perhaps express ( 4026! = 2^{2013} times 2013! times text{something} ).Wait, actually, I recall that:[ (2n)! = 2^n n! times prod_{k=1}^{n} (2k - 1).]So, for ( n = 2013 ), we have:[ (4026)! = 2^{2013} times 2013! times prod_{k=1}^{2013} (2k - 1).]But ( prod_{k=1}^{2013} (2k - 1) = (4025)!! ), which we already expressed as ( frac{4026!}{2^{2013} times 2013!} ). Hmm, that seems circular.Wait, perhaps I should instead consider the central binomial coefficient or something related to ( binom{4028}{2014} ).Wait, let me recall that ( binom{4028}{2014} = frac{4028!}{2014! times 2014!} ).So, perhaps I can relate ( 4026! ) to ( 4028! ):[ 4028! = 4028 times 4027 times 4026!.]So, ( 4026! = frac{4028!}{4028 times 4027} ).Substituting back into the expression for ( binom{1/2}{2014} ):[ binom{1/2}{2014} = frac{ - frac{4028!}{4028 times 4027} }{ 2^{4027} times (2013!)^2 times 2014 }.]Simplify numerator and denominator:[ binom{1/2}{2014} = frac{ -4028! }{ 4028 times 4027 times 2^{4027} times (2013!)^2 times 2014 }.]Simplify ( 4028 times 2014 ):Note that ( 4028 = 2 times 2014 ), so ( 4028 times 2014 = 2 times 2014 times 2014 = 2 times (2014)^2 ).So, substituting back:[ binom{1/2}{2014} = frac{ -4028! }{ 2 times (2014)^2 times 4027 times 2^{4027} times (2013!)^2 }.]Hmm, let's see if we can express ( 4028! ) in terms of ( 2014! ) and ( 2013! ).Wait, ( 4028! = (2 times 2014)! ). There's a formula for ( (2n)! ) in terms of ( n! ) and double factorials, but I'm not sure if that helps here.Alternatively, perhaps express ( 4028! ) as ( 4028 times 4027 times 4026! ), but we already did that.Wait, maybe I can express ( 4028! ) as ( (2014 times 2)! ), which is ( (2n)! ) where ( n = 2014 ).And I know that ( (2n)! = 2^{2n} n! times prod_{k=1}^{n} (2k - 1) ).So, for ( n = 2014 ):[ 4028! = 2^{4028} times 2014! times prod_{k=1}^{2014} (2k - 1).]But ( prod_{k=1}^{2014} (2k - 1) = (4027)!! ), which is the double factorial.But I don't know if that helps directly. Alternatively, perhaps express ( 4028! ) in terms of ( (2014)! ) and ( (2014)! ) with some coefficients.Wait, perhaps using the relation for central binomial coefficients:[ binom{2n}{n} = frac{(2n)!}{(n!)^2}.]So, in our case, ( binom{4028}{2014} = frac{4028!}{(2014!)^2} ).So, ( 4028! = binom{4028}{2014} times (2014!)^2 ).Substituting back into the expression for ( binom{1/2}{2014} ):[ binom{1/2}{2014} = frac{ - binom{4028}{2014} times (2014!)^2 }{ 2 times (2014)^2 times 4027 times 2^{4027} times (2013!)^2 }.]Simplify the ( (2014!)^2 ) and ( (2013!)^2 ):Note that ( 2014! = 2014 times 2013! ), so ( (2014!)^2 = (2014)^2 times (2013!)^2 ).Substituting this in:[ binom{1/2}{2014} = frac{ - binom{4028}{2014} times (2014)^2 times (2013!)^2 }{ 2 times (2014)^2 times 4027 times 2^{4027} times (2013!)^2 }.]Cancel out ( (2014)^2 ) and ( (2013!)^2 ):[ binom{1/2}{2014} = frac{ - binom{4028}{2014} }{ 2 times 4027 times 2^{4027} }.]So,[ binom{1/2}{2014} = frac{ - binom{4028}{2014} }{ 2^{4028} times 4027 }.]Wait, because ( 2 times 2^{4027} = 2^{4028} ).So,[ binom{1/2}{2014} = frac{ - binom{4028}{2014} }{ 2^{4028} times 4027 }.]Therefore, the original expression we need to compute is:[ frac{ binom{1/2}{2014} times 4^{2014} }{ binom{4028}{2014} }.]Substituting ( binom{1/2}{2014} ):[ frac{ left( frac{ - binom{4028}{2014} }{ 2^{4028} times 4027 } right) times 4^{2014} }{ binom{4028}{2014} }.]Simplify:The ( binom{4028}{2014} ) terms cancel out:[ frac{ -4^{2014} }{ 2^{4028} times 4027 }.]Note that ( 4^{2014} = (2^2)^{2014} = 2^{4028} ).So,[ frac{ -2^{4028} }{ 2^{4028} times 4027 } = frac{ -1 }{ 4027 }.]Therefore, the value of the expression is ( -frac{1}{4027} ).**Final Answer**boxed{-dfrac{1}{4027}}

answer：Okay, so I've got this problem here: "Given that the function y = mx + b is a decreasing function on ℝ, then ( ) A. m ≥ 0 B. m ≤ 0 C. m > 0 D. m < 0." Alright, let's try to figure this out step by step.First, I remember that the function y = mx + b is a linear function, which means it's a straight line when graphed. The 'm' in this equation represents the slope of the line, and 'b' is the y-intercept, where the line crosses the y-axis. So, the slope is crucial here because it tells us whether the function is increasing or decreasing.Now, the question says that the function is decreasing on ℝ, which means it's decreasing everywhere on the real number line. I recall that for a function to be decreasing, as x increases, y should decrease. In terms of the slope, that would mean the slope is negative because a negative slope causes the line to go downward from left to right.Let me think about this more carefully. If m is positive, then as x increases, y increases, right? So that would be an increasing function. But we need a decreasing function, so m must be negative. That makes sense because a negative slope would mean that as x increases, y decreases, which is exactly what we want for a decreasing function.Looking at the options, A says m ≥ 0, which includes zero and positive numbers. But if m is zero, the function is constant, not decreasing. So A can't be the right answer. B says m ≤ 0, which includes zero and negative numbers. Again, zero would make the function constant, not decreasing, so B might not be the best choice either.C says m > 0, which would mean the function is increasing, not decreasing, so that's definitely not correct. D says m < 0, which means the slope is negative, and as we established earlier, that makes the function decreasing. So D seems to be the correct answer.Wait, let me double-check. If m is negative, then y decreases as x increases. Yes, that's right. So, the function is decreasing if and only if m is less than zero. Therefore, D is the correct choice.I think I've got it. The key was remembering how the slope affects whether a linear function is increasing or decreasing. Positive slope means increasing, negative slope means decreasing, and zero slope means constant. Since the function is given as decreasing, m has to be negative. So, the answer is D.