💡Alright, so I have this problem here about functions and calculus. Let me try to understand it step by step.First, the function given is ( f(x) = x + a ln x ). I need to find the value of the real number ( a ). The problem says that the tangent line to the function at ( x = 1 ) is perpendicular to the line ( x + 2y = 0 ). Hmm, okay. I remember that if two lines are perpendicular, the product of their slopes is ( -1 ).So, let me find the slope of the given line ( x + 2y = 0 ). I can rewrite this in slope-intercept form ( y = mx + b ). Subtracting ( x ) from both sides gives ( 2y = -x ), so dividing both sides by 2, we get ( y = -frac{1}{2}x ). Therefore, the slope of this line is ( -frac{1}{2} ).Since the tangent line at ( x = 1 ) is perpendicular to this line, its slope must be the negative reciprocal of ( -frac{1}{2} ). The negative reciprocal of ( -frac{1}{2} ) is ( 2 ). So, the slope of the tangent line at ( x = 1 ) is ( 2 ).Now, to find the slope of the tangent line to ( f(x) ) at ( x = 1 ), I need to compute the derivative ( f'(x) ). The function is ( f(x) = x + a ln x ), so the derivative is ( f'(x) = 1 + frac{a}{x} ). Evaluating this derivative at ( x = 1 ), we get ( f'(1) = 1 + frac{a}{1} = 1 + a ). We already know that this slope must be ( 2 ), so:( 1 + a = 2 )Solving for ( a ), we subtract 1 from both sides:( a = 1 )Okay, so that's part (1) done. ( a = 1 ).Moving on to part (2). The function ( g(x) ) is given by ( g(x) = f(x) + frac{1}{2}x^2 - b x ). Since we found ( a = 1 ), let's substitute that into ( f(x) ):( f(x) = x + ln x )Therefore, ( g(x) = x + ln x + frac{1}{2}x^2 - b x ). Let's simplify this:Combine like terms: ( x - b x = (1 - b)x ). So,( g(x) = ln x + frac{1}{2}x^2 + (1 - b)x )Now, the problem mentions that ( x_1 ) and ( x_2 ) (with ( x_1 < x_2 )) are the two extreme points of ( g(x) ). Extreme points occur where the derivative is zero, so let's find ( g'(x) ).First, compute the derivative of ( g(x) ):( g'(x) = frac{d}{dx}[ln x] + frac{d}{dx}[frac{1}{2}x^2] + frac{d}{dx}[(1 - b)x] )Calculating each term:- ( frac{d}{dx}[ln x] = frac{1}{x} )- ( frac{d}{dx}[frac{1}{2}x^2] = x )- ( frac{d}{dx}[(1 - b)x] = 1 - b )So, putting it all together:( g'(x) = frac{1}{x} + x + (1 - b) )Let me write this as:( g'(x) = x + frac{1}{x} + (1 - b) )To find the critical points, set ( g'(x) = 0 ):( x + frac{1}{x} + (1 - b) = 0 )Multiply both sides by ( x ) to eliminate the denominator:( x^2 + 1 + (1 - b)x = 0 )So, we have the quadratic equation:( x^2 + (1 - b)x + 1 = 0 )Let me write this as:( x^2 - (b - 1)x + 1 = 0 )So, the quadratic equation is ( x^2 - (b - 1)x + 1 = 0 ). Let's denote this as ( x^2 - (b - 1)x + 1 = 0 ).The solutions to this quadratic will give us the critical points ( x_1 ) and ( x_2 ). Since it's a quadratic, the solutions are:( x = frac{(b - 1) pm sqrt{(b - 1)^2 - 4 cdot 1 cdot 1}}{2} )Simplify the discriminant:( D = (b - 1)^2 - 4 = b^2 - 2b + 1 - 4 = b^2 - 2b - 3 )For real solutions, the discriminant must be positive:( b^2 - 2b - 3 > 0 )Factorizing:( (b - 3)(b + 1) > 0 )So, the inequality holds when ( b > 3 ) or ( b < -1 ). However, since ( b geq frac{13}{3} ) (which is approximately 4.333), we are in the case where ( b > 3 ), so the quadratic has two real roots ( x_1 ) and ( x_2 ).Given that ( x_1 < x_2 ), we can denote ( x_1 ) as the smaller root and ( x_2 ) as the larger root.We are told to let ( t = frac{x_1}{x_2} ). So, ( t ) is the ratio of the smaller root to the larger root.The problem has two parts:① Find the range of values of ( t ) when ( b geq frac{13}{3} ).② Find the minimum value of ( g(x_1) - g(x_2) ).Let's tackle part ① first.First, let's recall that for a quadratic equation ( x^2 - (b - 1)x + 1 = 0 ), the sum of the roots is ( x_1 + x_2 = b - 1 ) and the product of the roots is ( x_1 x_2 = 1 ).Given that ( t = frac{x_1}{x_2} ), let's express ( x_1 = t x_2 ). Since ( x_1 x_2 = 1 ), substituting ( x_1 = t x_2 ) gives:( t x_2 cdot x_2 = 1 )( t x_2^2 = 1 )So, ( x_2^2 = frac{1}{t} ), which implies ( x_2 = frac{1}{sqrt{t}} ). Since ( x_2 > x_1 > 0 ), ( x_2 ) is positive, so we take the positive root.Similarly, ( x_1 = t x_2 = t cdot frac{1}{sqrt{t}} = sqrt{t} ).Now, the sum of the roots is:( x_1 + x_2 = sqrt{t} + frac{1}{sqrt{t}} = b - 1 )Let me denote ( s = sqrt{t} ). Then, ( x_1 = s ) and ( x_2 = frac{1}{s} ). So, the sum becomes:( s + frac{1}{s} = b - 1 )Let me write this as:( s + frac{1}{s} = b - 1 )Multiplying both sides by ( s ):( s^2 + 1 = (b - 1)s )Rearranging:( s^2 - (b - 1)s + 1 = 0 )Wait, this is the same quadratic equation as before, but in terms of ( s ). Hmm, interesting.But perhaps another approach is better. Let's express ( (x_1 + x_2)^2 ) in terms of ( t ).We have:( (x_1 + x_2)^2 = x_1^2 + 2 x_1 x_2 + x_2^2 )But since ( x_1 x_2 = 1 ), this becomes:( (x_1 + x_2)^2 = x_1^2 + 2 + x_2^2 )Also, ( x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2 x_1 x_2 = (b - 1)^2 - 2 )Wait, perhaps I should express ( frac{x_1}{x_2} = t ), so ( x_1 = t x_2 ). Then, ( x_1 + x_2 = t x_2 + x_2 = (t + 1) x_2 = b - 1 ).But we also have ( x_1 x_2 = 1 ), so ( t x_2^2 = 1 ), so ( x_2 = frac{1}{sqrt{t}} ).Substituting back into ( (t + 1) x_2 = b - 1 ):( (t + 1) cdot frac{1}{sqrt{t}} = b - 1 )Simplify:( frac{t + 1}{sqrt{t}} = b - 1 )Let me write ( frac{t + 1}{sqrt{t}} ) as ( sqrt{t} + frac{1}{sqrt{t}} ), which is similar to the earlier expression.So,( sqrt{t} + frac{1}{sqrt{t}} = b - 1 )Let me denote ( u = sqrt{t} ). Then, ( u + frac{1}{u} = b - 1 ).Multiplying both sides by ( u ):( u^2 + 1 = (b - 1) u )Rearranging:( u^2 - (b - 1) u + 1 = 0 )This is a quadratic in ( u ). The discriminant is:( D = (b - 1)^2 - 4 cdot 1 cdot 1 = (b - 1)^2 - 4 )Since ( b geq frac{13}{3} ), which is approximately 4.333, ( (b - 1) geq frac{13}{3} - 1 = frac{10}{3} approx 3.333 ). So, ( (b - 1)^2 geq (frac{10}{3})^2 = frac{100}{9} approx 11.111 ). Therefore, ( D geq frac{100}{9} - 4 = frac{100}{9} - frac{36}{9} = frac{64}{9} ), which is positive, so real solutions exist.The solutions for ( u ) are:( u = frac{(b - 1) pm sqrt{(b - 1)^2 - 4}}{2} )Since ( u = sqrt{t} ) must be positive, both solutions are positive because ( (b - 1) ) is positive and the square root term is less than ( (b - 1) ) (since ( (b - 1)^2 - 4 < (b - 1)^2 )), so both solutions are positive.But since ( t = frac{x_1}{x_2} ) and ( x_1 < x_2 ), ( t < 1 ). Therefore, ( u = sqrt{t} < 1 ). So, we need to find which of the two solutions for ( u ) is less than 1.Let me denote the two solutions as:( u_1 = frac{(b - 1) - sqrt{(b - 1)^2 - 4}}{2} )and( u_2 = frac{(b - 1) + sqrt{(b - 1)^2 - 4}}{2} )Since ( u_1 < u_2 ), and we need ( u = sqrt{t} < 1 ), let's check which of ( u_1 ) or ( u_2 ) is less than 1.Let me compute ( u_1 ):( u_1 = frac{(b - 1) - sqrt{(b - 1)^2 - 4}}{2} )Let me denote ( c = b - 1 ), so ( c geq frac{13}{3} - 1 = frac{10}{3} approx 3.333 ).So, ( u_1 = frac{c - sqrt{c^2 - 4}}{2} )I need to see if ( u_1 < 1 ).Let me compute ( u_1 ) when ( c = frac{10}{3} ):( u_1 = frac{frac{10}{3} - sqrt{(frac{10}{3})^2 - 4}}{2} = frac{frac{10}{3} - sqrt{frac{100}{9} - frac{36}{9}}}{2} = frac{frac{10}{3} - sqrt{frac{64}{9}}}{2} = frac{frac{10}{3} - frac{8}{3}}{2} = frac{frac{2}{3}}{2} = frac{1}{3} )So, when ( c = frac{10}{3} ), ( u_1 = frac{1}{3} ), which is less than 1.As ( c ) increases beyond ( frac{10}{3} ), let's see what happens to ( u_1 ).Compute ( u_1 ) as ( c ) approaches infinity:( u_1 approx frac{c - c}{2} = 0 )So, ( u_1 ) approaches 0 as ( c ) becomes large. Therefore, ( u_1 ) is always less than 1 for ( c geq frac{10}{3} ).Similarly, ( u_2 = frac{c + sqrt{c^2 - 4}}{2} ). Let's see what this is when ( c = frac{10}{3} ):( u_2 = frac{frac{10}{3} + frac{8}{3}}{2} = frac{frac{18}{3}}{2} = frac{6}{2} = 3 )Which is greater than 1. As ( c ) increases, ( u_2 ) increases as well.Therefore, since ( u = sqrt{t} ) must be less than 1, we have ( u = u_1 ), so:( sqrt{t} = frac{(b - 1) - sqrt{(b - 1)^2 - 4}}{2} )Therefore,( t = left( frac{(b - 1) - sqrt{(b - 1)^2 - 4}}{2} right)^2 )But we need to find the range of ( t ) as ( b ) varies from ( frac{13}{3} ) to infinity.When ( b = frac{13}{3} ), ( c = frac{10}{3} ), so:( t = left( frac{frac{10}{3} - frac{8}{3}}{2} right)^2 = left( frac{frac{2}{3}}{2} right)^2 = left( frac{1}{3} right)^2 = frac{1}{9} )As ( b ) increases beyond ( frac{13}{3} ), ( c = b - 1 ) increases, so ( u_1 ) decreases towards 0, so ( t = u_1^2 ) also decreases towards 0.Therefore, the range of ( t ) is ( 0 < t leq frac{1}{9} ).So, for part ①, the range of ( t ) is ( (0, frac{1}{9}] ).Now, moving on to part ②: Find the minimum value of ( g(x_1) - g(x_2) ).First, let's recall that ( x_1 ) and ( x_2 ) are critical points, so ( g'(x_1) = 0 ) and ( g'(x_2) = 0 ). We need to compute ( g(x_1) - g(x_2) ).Let me write down ( g(x) ):( g(x) = ln x + frac{1}{2}x^2 + (1 - b)x )So, ( g(x_1) - g(x_2) = ln x_1 + frac{1}{2}x_1^2 + (1 - b)x_1 - left( ln x_2 + frac{1}{2}x_2^2 + (1 - b)x_2 right) )Simplify:( g(x_1) - g(x_2) = ln left( frac{x_1}{x_2} right) + frac{1}{2}(x_1^2 - x_2^2) + (1 - b)(x_1 - x_2) )We know that ( t = frac{x_1}{x_2} ), so ( ln left( frac{x_1}{x_2} right) = ln t ).Also, ( x_1 = t x_2 ), so ( x_1^2 = t^2 x_2^2 ). Therefore, ( x_1^2 - x_2^2 = t^2 x_2^2 - x_2^2 = x_2^2 (t^2 - 1) ).Similarly, ( x_1 - x_2 = t x_2 - x_2 = x_2 (t - 1) ).So, substituting back into ( g(x_1) - g(x_2) ):( g(x_1) - g(x_2) = ln t + frac{1}{2} x_2^2 (t^2 - 1) + (1 - b) x_2 (t - 1) )Now, we need to express everything in terms of ( t ) and ( b ). We already have relationships between ( x_2 ) and ( t ), and ( b ).From earlier, we have:( x_1 x_2 = 1 ) and ( x_1 = t x_2 ), so ( t x_2^2 = 1 ) which gives ( x_2^2 = frac{1}{t} ). Therefore, ( x_2 = frac{1}{sqrt{t}} ).Also, from the sum of roots:( x_1 + x_2 = b - 1 )Substituting ( x_1 = t x_2 ):( t x_2 + x_2 = (t + 1) x_2 = b - 1 )So, ( x_2 = frac{b - 1}{t + 1} )But we also have ( x_2 = frac{1}{sqrt{t}} ), so:( frac{1}{sqrt{t}} = frac{b - 1}{t + 1} )Cross-multiplying:( t + 1 = (b - 1) sqrt{t} )Let me square both sides to eliminate the square root:( (t + 1)^2 = (b - 1)^2 t )Expanding the left side:( t^2 + 2t + 1 = (b - 1)^2 t )Rearranging:( t^2 + 2t + 1 - (b - 1)^2 t = 0 )( t^2 + [2 - (b - 1)^2] t + 1 = 0 )This is a quadratic equation in ( t ). However, since we already have ( t ) expressed in terms of ( b ), perhaps it's better to express ( b ) in terms of ( t ).From ( sqrt{t} + frac{1}{sqrt{t}} = b - 1 ), we can write:( b = 1 + sqrt{t} + frac{1}{sqrt{t}} )Let me denote ( s = sqrt{t} ), so ( s + frac{1}{s} = b - 1 ), which we already used earlier.So, ( b = 1 + s + frac{1}{s} )Now, let's go back to the expression for ( g(x_1) - g(x_2) ):( g(x_1) - g(x_2) = ln t + frac{1}{2} x_2^2 (t^2 - 1) + (1 - b) x_2 (t - 1) )We have ( x_2 = frac{1}{sqrt{t}} ), so ( x_2^2 = frac{1}{t} ). Therefore:( frac{1}{2} x_2^2 (t^2 - 1) = frac{1}{2} cdot frac{1}{t} (t^2 - 1) = frac{1}{2} cdot left( t - frac{1}{t} right) )Similarly, ( (1 - b) x_2 (t - 1) ). Let's compute ( 1 - b ):( 1 - b = 1 - left( 1 + s + frac{1}{s} right) = - left( s + frac{1}{s} right) )But ( s = sqrt{t} ), so ( s + frac{1}{s} = sqrt{t} + frac{1}{sqrt{t}} ). Therefore,( 1 - b = - left( sqrt{t} + frac{1}{sqrt{t}} right) )Now, ( x_2 = frac{1}{sqrt{t}} ), so:( (1 - b) x_2 (t - 1) = - left( sqrt{t} + frac{1}{sqrt{t}} right) cdot frac{1}{sqrt{t}} cdot (t - 1) )Simplify:First, multiply ( sqrt{t} + frac{1}{sqrt{t}} ) by ( frac{1}{sqrt{t}} ):( left( sqrt{t} + frac{1}{sqrt{t}} right) cdot frac{1}{sqrt{t}} = frac{sqrt{t}}{sqrt{t}} + frac{1}{t} = 1 + frac{1}{t} )Therefore,( (1 - b) x_2 (t - 1) = - left( 1 + frac{1}{t} right) (t - 1) )Expanding this:( - left( (1)(t - 1) + frac{1}{t}(t - 1) right) = - left( t - 1 + 1 - frac{1}{t} right) = - left( t - frac{1}{t} right) )So, putting it all together, ( g(x_1) - g(x_2) ) becomes:( ln t + frac{1}{2} left( t - frac{1}{t} right) - left( t - frac{1}{t} right) )Simplify the terms:Combine the terms involving ( t - frac{1}{t} ):( frac{1}{2} left( t - frac{1}{t} right) - left( t - frac{1}{t} right) = - frac{1}{2} left( t - frac{1}{t} right) )Therefore,( g(x_1) - g(x_2) = ln t - frac{1}{2} left( t - frac{1}{t} right) )Let me write this as:( g(x_1) - g(x_2) = ln t - frac{1}{2} t + frac{1}{2 t} )Now, we need to find the minimum value of this expression with respect to ( t ) in the interval ( (0, frac{1}{9}] ).Let me denote this function as:( h(t) = ln t - frac{1}{2} t + frac{1}{2 t} )We need to find the minimum of ( h(t) ) for ( t in (0, frac{1}{9}] ).To find the minimum, let's compute the derivative of ( h(t) ) with respect to ( t ):( h'(t) = frac{1}{t} - frac{1}{2} - frac{1}{2 t^2} )Simplify:( h'(t) = frac{1}{t} - frac{1}{2} - frac{1}{2 t^2} )Let me combine the terms:( h'(t) = frac{2 t - t^2 - 1}{2 t^2} )Wait, let me compute it step by step:First, ( frac{1}{t} = frac{2}{2 t} ), so:( h'(t) = frac{2}{2 t} - frac{1}{2} - frac{1}{2 t^2} )Combine over a common denominator of ( 2 t^2 ):( h'(t) = frac{2 t}{2 t^2} - frac{t^2}{2 t^2} - frac{1}{2 t^2} = frac{2 t - t^2 - 1}{2 t^2} )So,( h'(t) = frac{ - t^2 + 2 t - 1 }{2 t^2 } = frac{ - (t^2 - 2 t + 1) }{2 t^2 } = frac{ - (t - 1)^2 }{2 t^2 } )Therefore,( h'(t) = - frac{(t - 1)^2}{2 t^2} )Notice that ( (t - 1)^2 ) is always non-negative, and ( 2 t^2 ) is positive for ( t > 0 ). Therefore, ( h'(t) ) is always non-positive. It is zero only when ( t = 1 ), but since ( t leq frac{1}{9} ), ( h'(t) ) is strictly negative in the interval ( (0, frac{1}{9}] ).This means that ( h(t) ) is strictly decreasing on ( (0, frac{1}{9}] ). Therefore, the minimum value of ( h(t) ) occurs at the right endpoint of the interval, which is ( t = frac{1}{9} ).So, the minimum value is:( hleft( frac{1}{9} right) = ln left( frac{1}{9} right) - frac{1}{2} cdot frac{1}{9} + frac{1}{2 cdot frac{1}{9}} )Simplify each term:1. ( ln left( frac{1}{9} right) = - ln 9 = - 2 ln 3 )2. ( - frac{1}{2} cdot frac{1}{9} = - frac{1}{18} )3. ( frac{1}{2 cdot frac{1}{9}} = frac{9}{2} )Putting it all together:( hleft( frac{1}{9} right) = - 2 ln 3 - frac{1}{18} + frac{9}{2} )Convert ( frac{9}{2} ) to eighteenths to combine with ( - frac{1}{18} ):( frac{9}{2} = frac{81}{18} )So,( hleft( frac{1}{9} right) = - 2 ln 3 - frac{1}{18} + frac{81}{18} = - 2 ln 3 + frac{80}{18} )Simplify ( frac{80}{18} ):( frac{80}{18} = frac{40}{9} )Therefore,( hleft( frac{1}{9} right) = frac{40}{9} - 2 ln 3 )So, the minimum value of ( g(x_1) - g(x_2) ) is ( frac{40}{9} - 2 ln 3 ).**Final Answer**(1) The value of ( a ) is (boxed{1}).(2) ① The range of ( t ) is (boxed{left(0, dfrac{1}{9}right]}).② The minimum value of ( g(x_1) - g(x_2) ) is (boxed{dfrac{40}{9} - 2ln 3}).

💡Okay, so I have this problem where I need to find the range of values for 'a' such that the function f(x) = x² + 2a x + 1 has no fixed points. Hmm, fixed points are where f(x) equals x, right? So, I guess I need to set up the equation f(x) = x and then figure out when that equation has no real solutions.Let me write that down. If f(x) = x² + 2a x + 1, then setting f(x) equal to x gives:x² + 2a x + 1 = xAlright, now I need to rearrange this equation to standard quadratic form. So, subtract x from both sides:x² + 2a x + 1 - x = 0Simplify the terms with x:x² + (2a - 1)x + 1 = 0Okay, so now I have a quadratic equation: x² + (2a - 1)x + 1 = 0. For this quadratic to have no real roots, the discriminant must be negative. I remember that for a quadratic equation ax² + bx + c = 0, the discriminant is b² - 4ac. If that's less than zero, there are no real solutions.So, let's compute the discriminant for our equation. Here, the coefficients are:a = 1 (coefficient of x²)b = (2a - 1) (coefficient of x)c = 1 (constant term)So, discriminant D = b² - 4ac = (2a - 1)² - 4 * 1 * 1Let me calculate that:(2a - 1)² = 4a² - 4a + 1Then, subtract 4:4a² - 4a + 1 - 4 = 4a² - 4a - 3So, the discriminant is 4a² - 4a - 3. We need this to be less than zero:4a² - 4a - 3 < 0Hmm, now I have a quadratic inequality. To solve this, I should probably find the roots of the quadratic equation 4a² - 4a - 3 = 0 and then determine the intervals where the quadratic is negative.Let me solve 4a² - 4a - 3 = 0. Using the quadratic formula:a = [4 ± sqrt(16 + 48)] / 8Wait, let me compute the discriminant for this equation:Discriminant D' = (-4)² - 4 * 4 * (-3) = 16 + 48 = 64Oh, that's nice because sqrt(64) is 8.So, a = [4 ± 8] / 8Calculating both roots:First root: (4 + 8)/8 = 12/8 = 3/2 = 1.5Second root: (4 - 8)/8 = (-4)/8 = -0.5So, the roots are at a = -0.5 and a = 1.5.Since the coefficient of a² is positive (4), the parabola opens upwards. That means the quadratic expression 4a² - 4a - 3 is negative between the two roots.Therefore, the inequality 4a² - 4a - 3 < 0 holds when a is between -0.5 and 1.5.So, the range of values for 'a' is (-0.5, 1.5). In fraction terms, that's (-1/2, 3/2).Let me just double-check my steps to make sure I didn't make any mistakes.1. Started by setting f(x) = x, which gives x² + 2a x + 1 = x.2. Rearranged to x² + (2a - 1)x + 1 = 0.3. Calculated discriminant D = (2a - 1)² - 4 = 4a² - 4a - 3.4. Set discriminant < 0: 4a² - 4a - 3 < 0.5. Solved quadratic equation 4a² - 4a - 3 = 0, found roots at a = -0.5 and a = 1.5.6. Since parabola opens upwards, inequality holds between roots: -0.5 < a < 1.5.Everything seems to check out. So, I think that's the correct range for 'a'.**Final Answer**The range of values for ( a ) is boxed{left( -frac{1}{2}, frac{3}{2} right)}.

💡First, I need to determine the cost price of the book based on the given profit percentages and the additional amount that would have resulted in a higher profit.Let ( C ) represent the cost price of the book.When the book is sold at a 10% profit, the selling price is:[text{Selling Price} = C + 0.10C = 1.10C]If the book had been sold for 140 more, the selling price would have been:[text{New Selling Price} = 1.10C + 140]This new selling price would result in a 15% profit:[1.10C + 140 = C + 0.15C = 1.15C]To find the cost price ( C ), I'll set up the equation:[1.10C + 140 = 1.15C]Subtracting ( 1.10C ) from both sides:[140 = 0.05C]Finally, solving for ( C ):[C = frac{140}{0.05} = 2800]Therefore, the cost price of the book is 2800.

💡Alright, so I have this problem here that says: There are ( n geq 2 ) people at a meeting. I need to show that there exist two people at the meeting who have the same number of friends among the persons at the meeting. It also mentions that friendship is mutual, meaning if A is a friend of B, then B is a friend of A, and nobody is their own friend.Hmm, okay. So, I need to prove that in any such gathering, there must be at least two people with the same number of friends. That sounds like it might be related to the pigeonhole principle, where if you have more pigeons than pigeonholes, at least two pigeons must share a hole. Maybe something similar applies here with the number of friends.Let me think. If there are ( n ) people, each person can potentially have anywhere from 0 to ( n-1 ) friends. Because you can't be friends with yourself, the maximum number of friends one person can have is ( n-1 ). So, the possible number of friends each person can have is in the range ( 0, 1, 2, ldots, n-1 ).Now, if we assume that everyone has a different number of friends, that would mean that each number from 0 to ( n-1 ) is represented exactly once among the group. So, one person has 0 friends, another has 1 friend, another has 2 friends, and so on, up to one person who has ( n-1 ) friends.Wait a minute, that seems problematic. If there's a person with ( n-1 ) friends, that means they are friends with everyone else at the meeting. But if there's also a person with 0 friends, that person isn't friends with anyone, including the person who supposedly is friends with everyone. That's a contradiction because the person with ( n-1 ) friends should be friends with the person who has 0 friends, but that's impossible.So, this assumption that everyone has a unique number of friends leads to a contradiction. Therefore, it must be that at least two people have the same number of friends.Let me try to formalize this a bit more. Suppose we have ( n ) people, and each person has a degree (number of friends) ( d_1, d_2, ldots, d_n ). The possible degrees range from 0 to ( n-1 ). If all degrees are distinct, then the degrees must be exactly the numbers ( 0, 1, 2, ldots, n-1 ) in some order.However, as I thought earlier, if one person has degree ( n-1 ), they must be connected to everyone else, including the person with degree 0. But the person with degree 0 cannot have any connections, which contradicts the fact that they are connected to the person with degree ( n-1 ). Therefore, our initial assumption that all degrees are distinct must be false.Hence, there must be at least two people with the same number of friends.I think that makes sense. It's like saying in any social network, you can't have someone who is friends with everyone and someone who is friends with no one at the same time. So, the degrees can't cover the entire range without overlapping, which means some degrees must repeat.Another way to think about it is using the pigeonhole principle. There are ( n ) people and ( n ) possible degrees (from 0 to ( n-1 )). But because of the contradiction when considering the extremes (0 and ( n-1 )), we can't have all degrees represented. Therefore, at least two people must share the same degree.Yeah, I think that solidifies the argument. So, in conclusion, in any gathering of ( n geq 2 ) people, there must be at least two individuals who have the same number of friends.

💡Alright, so I've got these three probability questions about English football teams in a tournament. Let me try to wrap my head around them step by step. First, I need to understand the setup. There are 8 teams in total, and 4 of them are English. The tournament is structured in a way that teams are paired against each other in the first round, and the winners move on to the semifinals. The questions are about the probabilities of different pairing scenarios involving the English teams.Starting with part (a): Certain English fans wanted the English teams to avoid each other, allowing all four teams to reach the semifinals. So, I need to find the probability that none of the English teams are paired against each other in the first round.Okay, so to find this probability, I think I need to figure out the total number of possible pairings and then the number of favorable pairings where no two English teams are paired together.Total number of ways to pair 8 teams: I remember that the number of ways to pair up 2n teams is (2n)!)/(2^n n!). So, for 8 teams, it should be 8!/(2^4 * 4!) = 40320/(16 * 24) = 40320/384 = 105. So, there are 105 possible ways to pair up the 8 teams.Now, the number of favorable pairings where no two English teams are paired together. This means each English team is paired with a non-English team. There are 4 English teams and 4 non-English teams. So, essentially, we need to pair each English team with a non-English team.The number of ways to do this is 4! (since it's like assigning each English team to a non-English team). 4! = 24.Therefore, the probability that no two English teams are paired together is 24/105, which simplifies to 8/35 or approximately 0.2286.Moving on to part (b): Other English fans preferred that two pairs should each consist of two English teams, ensuring that at least two English teams would definitely make it to the semifinals. I need to find the probability of such pairings occurring.So, in this case, we want exactly two pairs of English teams. That means two pairs are English vs. English, and the other two pairs are non-English vs. non-English.First, let's find the number of ways to pair the English teams. There are 4 English teams, and we need to split them into two pairs. The number of ways to do this is (4!)/(2^2 * 2!) = 24/(4 * 2) = 3.Similarly, for the non-English teams, we have 4 non-English teams, and we need to split them into two pairs. The number of ways is also 3.So, the total number of favorable pairings is 3 (for English pairs) * 3 (for non-English pairs) = 9.Therefore, the probability is 9/105, which simplifies to 3/35 or approximately 0.0857.Wait, that seems low. Let me double-check. The total number of ways to pair the English teams into two pairs is indeed 3, and the same for non-English teams. So, 3*3=9 favorable pairings. Yes, that seems correct.Now, part (c): In the actual draw, two English teams were paired together, and the other two were matched with non-English opponents. What is the probability of such pairings occurring?So, this is a scenario where one pair is English vs. English, and the other two English teams are paired with non-English teams.First, let's find the number of ways this can happen.We need to choose 2 English teams out of 4 to be paired together. The number of ways to choose 2 teams from 4 is C(4,2) = 6.Once we've chosen the two English teams to be paired together, we have 2 English teams left and 4 non-English teams. Now, we need to pair each of the remaining 2 English teams with a non-English team.The number of ways to pair 2 English teams with 4 non-English teams is 4 * 3 = 12 (since the first English team can be paired with any of the 4 non-English teams, and the second English team can be paired with any of the remaining 3 non-English teams).Therefore, the total number of favorable pairings is 6 * 12 = 72.So, the probability is 72/105, which simplifies to 24/35 or approximately 0.6857.Wait, let me make sure. The total number of ways is 105, and favorable is 72, so 72/105 is indeed 24/35. That seems correct.So, summarizing:a) Probability that no two English teams are paired together: 24/105 ≈ 0.2286b) Probability that two pairs consist of two English teams: 9/105 ≈ 0.0857c) Probability that two English teams are paired together and the other two are matched with non-English opponents: 72/105 ≈ 0.6857I think that covers all the parts. I hope I didn't make any calculation errors. It's always good to double-check, but I believe these probabilities make sense given the total number of pairings.

💡First, I need to find the smallest 3-digit number that is divisible by 13. By calculating (13 times 8 = 104), I determine that 104 is the smallest 3-digit number divisible by 13.Next, I identify the largest 3-digit number divisible by 13. Since (13 times 77 = 1001), the largest number less than 1000 divisible by 13 is (13 times 76 = 988).Now, I have an arithmetic sequence starting at 104 and ending at 988 with a common difference of 13. To find the number of terms in this sequence, I use the formula for the nth term of an arithmetic sequence:[a_n = a_1 + (n-1)d]Plugging in the values:[988 = 104 + (n-1) times 13]Solving for (n):[988 - 104 = (n-1) times 13][884 = (n-1) times 13][n-1 = frac{884}{13} = 68][n = 68 + 1 = 69]Therefore, there are 69 three-digit numbers divisible by 13.