💡Okay, so I have this problem about functions and their inverses, and it's divided into three parts. Let me try to tackle each part step by step.Starting with part (1): It says that the graph of y = f(x) is symmetric to the graph of y = a^x - 1 about the line y = x. Hmm, symmetry about y = x usually means that they are inverse functions of each other. So, if y = f(x) is the inverse of y = a^x - 1, then f(x) should be the inverse function of a^x - 1.Alright, let's find the inverse of y = a^x - 1. To find the inverse, I need to swap x and y and solve for y. So, starting with:y = a^x - 1Swapping x and y:x = a^y - 1Now, solve for y:x + 1 = a^yTake the logarithm base a of both sides:log_a(x + 1) = ySo, the inverse function is y = log_a(x + 1). Therefore, f(x) = log_a(x + 1). That seems straightforward.Moving on to part (2): It says that the range of f(x) on the interval [m, n] (where m > -1) is [log_a(p/m), log_a(p/n)]. I need to find the range of the real number p.Since f(x) = log_a(x + 1), and a > 1, this function is strictly increasing. So, on the interval [m, n], the minimum value is f(m) and the maximum is f(n). Therefore, f(m) = log_a(m + 1) and f(n) = log_a(n + 1).According to the problem, the range is [log_a(p/m), log_a(p/n)]. Since f(x) is increasing, the minimum is log_a(p/m) and the maximum is log_a(p/n). Therefore, we can set up the equations:log_a(m + 1) = log_a(p/m)log_a(n + 1) = log_a(p/n)Since the logarithm function is injective (one-to-one), we can equate the arguments:m + 1 = p/mn + 1 = p/nSo, we have two equations:1. m + 1 = p/m2. n + 1 = p/nLet me rearrange these equations:From equation 1: m(m + 1) = p => p = m^2 + mFrom equation 2: n(n + 1) = p => p = n^2 + nSo, both m and n satisfy the equation x^2 + x - p = 0. That means m and n are roots of the quadratic equation x^2 + x - p = 0.Since m and n are distinct points in the interval [m, n], and m > -1, we need to ensure that the quadratic has two distinct real roots in the domain x > -1.First, let's find the discriminant of the quadratic equation x^2 + x - p = 0.Discriminant D = 1^2 - 4*1*(-p) = 1 + 4pFor the quadratic to have two distinct real roots, D must be positive:1 + 4p > 0 => p > -1/4Also, since m and n are greater than -1, we need to ensure that the roots are greater than -1.Let me consider the quadratic equation x^2 + x - p = 0. The roots are:x = [-1 ± sqrt(1 + 4p)] / 2We need both roots to be greater than -1.Let's denote the roots as:x1 = [-1 + sqrt(1 + 4p)] / 2x2 = [-1 - sqrt(1 + 4p)] / 2We need x1 > -1 and x2 > -1.First, let's check x2:x2 = [-1 - sqrt(1 + 4p)] / 2We need x2 > -1:[-1 - sqrt(1 + 4p)] / 2 > -1Multiply both sides by 2:-1 - sqrt(1 + 4p) > -2Add 1 to both sides:-sqrt(1 + 4p) > -1Multiply both sides by -1 (inequality sign flips):sqrt(1 + 4p) < 1Square both sides:1 + 4p < 1 => 4p < 0 => p < 0So, p must be less than 0.We already have p > -1/4 from the discriminant condition.Therefore, combining both results, p must satisfy:-1/4 < p < 0So, the range of p is (-1/4, 0).Now, moving on to part (3): We have g(x) = log_a(x^2 - 3x + 3), and F(x) = a^{f(x) - g(x)}. We need to find the range of w such that w ≥ F(x) for all x in (-1, +∞).First, let's express F(x) in terms of f(x) and g(x):F(x) = a^{f(x) - g(x)} = a^{log_a(x + 1) - log_a(x^2 - 3x + 3)}Using logarithm properties, a^{log_a(A) - log_a(B)} = A / B.Therefore, F(x) = (x + 1) / (x^2 - 3x + 3)So, F(x) = (x + 1)/(x^2 - 3x + 3)We need to find the maximum value of F(x) on the interval (-1, +∞) because w must be greater than or equal to F(x) for all x in that interval. Therefore, w must be at least the maximum value of F(x).So, let's analyze F(x):F(x) = (x + 1)/(x^2 - 3x + 3)First, let's find the critical points by taking the derivative and setting it equal to zero.Let me denote numerator as N = x + 1 and denominator as D = x^2 - 3x + 3.Then, F(x) = N/DThe derivative F’(x) is (N’ D - N D’) / D^2Compute N’ = 1Compute D’ = 2x - 3So,F’(x) = [1*(x^2 - 3x + 3) - (x + 1)*(2x - 3)] / (x^2 - 3x + 3)^2Let me expand the numerator:1*(x^2 - 3x + 3) = x^2 - 3x + 3(x + 1)*(2x - 3) = 2x^2 - 3x + 2x - 3 = 2x^2 - x - 3Subtracting these:(x^2 - 3x + 3) - (2x^2 - x - 3) = x^2 - 3x + 3 - 2x^2 + x + 3 = -x^2 - 2x + 6So, F’(x) = (-x^2 - 2x + 6) / (x^2 - 3x + 3)^2Set numerator equal to zero to find critical points:-x^2 - 2x + 6 = 0Multiply both sides by -1:x^2 + 2x - 6 = 0Solutions:x = [-2 ± sqrt(4 + 24)] / 2 = [-2 ± sqrt(28)] / 2 = [-2 ± 2*sqrt(7)] / 2 = -1 ± sqrt(7)So, critical points at x = -1 + sqrt(7) and x = -1 - sqrt(7)But our domain is x > -1, so x = -1 - sqrt(7) is approximately -1 - 2.6458 ≈ -3.6458, which is less than -1, so we can ignore it.Therefore, the only critical point in the domain is x = -1 + sqrt(7) ≈ -1 + 2.6458 ≈ 1.6458.Now, let's check the behavior of F(x) around this critical point and the endpoints.First, let's compute F(x) at x = -1 + sqrt(7):Compute numerator: x + 1 = (-1 + sqrt(7)) + 1 = sqrt(7)Denominator: x^2 - 3x + 3Compute x^2: (-1 + sqrt(7))^2 = 1 - 2*sqrt(7) + 7 = 8 - 2*sqrt(7)Compute -3x: -3*(-1 + sqrt(7)) = 3 - 3*sqrt(7)So, denominator = (8 - 2*sqrt(7)) + (3 - 3*sqrt(7)) + 3 = 8 + 3 + 3 - 2*sqrt(7) - 3*sqrt(7) = 14 - 5*sqrt(7)Therefore, F(x) at x = -1 + sqrt(7) is sqrt(7)/(14 - 5*sqrt(7))Let me rationalize the denominator:Multiply numerator and denominator by (14 + 5*sqrt(7)):Numerator: sqrt(7)*(14 + 5*sqrt(7)) = 14*sqrt(7) + 5*7 = 14*sqrt(7) + 35Denominator: (14)^2 - (5*sqrt(7))^2 = 196 - 25*7 = 196 - 175 = 21So, F(x) = (14*sqrt(7) + 35)/21 = (14*sqrt(7))/21 + 35/21 = (2*sqrt(7))/3 + 5/3 = (2*sqrt(7) + 5)/3So, the maximum value of F(x) is (2*sqrt(7) + 5)/3 ≈ (5.2915 + 5)/3 ≈ 10.2915/3 ≈ 3.4305Now, let's check the behavior as x approaches infinity:F(x) = (x + 1)/(x^2 - 3x + 3) ≈ x / x^2 = 1/x, which approaches 0 as x approaches infinity.Also, at x approaching -1 from the right:x approaches -1, so numerator approaches 0, denominator approaches (-1)^2 - 3*(-1) + 3 = 1 + 3 + 3 = 7. So, F(x) approaches 0.Therefore, the maximum value of F(x) is indeed at x = -1 + sqrt(7), which is (2*sqrt(7) + 5)/3.Hence, to satisfy w ≥ F(x) for all x in (-1, +∞), w must be at least (2*sqrt(7) + 5)/3.So, the range of w is [(2*sqrt(7) + 5)/3, +∞).**Final Answer**(1) boxed{log_a (x + 1)}(2) boxed{left( -dfrac{1}{4}, 0 right)}(3) boxed{left[ dfrac{2sqrt{7} + 5}{3}, +infty right)}

💡First, determine how many stuffies Jean keeps by calculating one-third of the total.Next, subtract the number of stuffies Jean keeps from the total to find out how many she gives away.Then, divide the number of stuffies Janet received by the total number Jean gave away to find the fraction.Finally, simplify the fraction to get the answer.

💡Okay, so I have this equation to solve: [sqrt{2 - x} + sqrt{2 + 2x} = sqrt{frac{x^4 + 1}{x^2 + 1}} + frac{x + 3}{x + 1}]And I need to find all real numbers ( x ) such that ( -1 < x leq 2 ) satisfy this equation. Hmm, that looks a bit complicated, but let me try to break it down step by step.First, I notice that both sides of the equation have square roots and rational expressions. Maybe I can simplify each side separately or find some way to relate them.Starting with the left-hand side (LHS):[sqrt{2 - x} + sqrt{2 + 2x}]I wonder if I can combine these two square roots or find some relationship between them. Maybe I can square both sides to eliminate the square roots, but that might complicate things. Let me see if there's another approach.Looking at the right-hand side (RHS):[sqrt{frac{x^4 + 1}{x^2 + 1}} + frac{x + 3}{x + 1}]This also looks a bit tricky. The first term is a square root of a fraction, and the second term is a rational expression. Maybe I can simplify the fraction inside the square root.Let me try simplifying the fraction inside the square root on the RHS:[frac{x^4 + 1}{x^2 + 1}]Hmm, I recall that ( x^4 + 1 ) can be written as ( (x^2)^2 + 1 ). Maybe I can factor it or find a way to simplify it with the denominator ( x^2 + 1 ). Let me try polynomial division or see if it's a multiple.Wait, ( x^4 + 1 ) divided by ( x^2 + 1 ) is ( x^2 - 1 + frac{2}{x^2 + 1} ). Let me check:[(x^2 + 1)(x^2 - 1) = x^4 - 1]So, ( x^4 + 1 = (x^2 + 1)(x^2 - 1) + 2 ). Therefore,[frac{x^4 + 1}{x^2 + 1} = x^2 - 1 + frac{2}{x^2 + 1}]So, the square root becomes:[sqrt{x^2 - 1 + frac{2}{x^2 + 1}}]Hmm, not sure if that helps much. Maybe I can consider another approach.Let me think about the domain of the equation. Since we have square roots, the expressions inside must be non-negative.For LHS:1. ( 2 - x geq 0 ) implies ( x leq 2 )2. ( 2 + 2x geq 0 ) implies ( x geq -1 )So, combining these, ( -1 leq x leq 2 ). But the problem states ( -1 < x leq 2 ), so we're good.For RHS:1. ( frac{x^4 + 1}{x^2 + 1} geq 0 ) which is always true since both numerator and denominator are positive for all real ( x ).2. ( frac{x + 3}{x + 1} ) is defined when ( x neq -1 ), which is already excluded.So, the domain is ( -1 < x leq 2 ), as given.Now, maybe I can try plugging in some values of ( x ) within this interval to see if they satisfy the equation.Let me try ( x = 1 ):LHS:[sqrt{2 - 1} + sqrt{2 + 2(1)} = sqrt{1} + sqrt{4} = 1 + 2 = 3]RHS:[sqrt{frac{1^4 + 1}{1^2 + 1}} + frac{1 + 3}{1 + 1} = sqrt{frac{2}{2}} + frac{4}{2} = sqrt{1} + 2 = 1 + 2 = 3]So, ( x = 1 ) works. That's a good sign.Let me try another value, say ( x = 0 ):LHS:[sqrt{2 - 0} + sqrt{2 + 0} = sqrt{2} + sqrt{2} = 2sqrt{2} approx 2.828]RHS:[sqrt{frac{0 + 1}{0 + 1}} + frac{0 + 3}{0 + 1} = sqrt{1} + 3 = 1 + 3 = 4]So, LHS ≈ 2.828 and RHS = 4. Not equal. So, ( x = 0 ) doesn't work.How about ( x = 2 ):LHS:[sqrt{2 - 2} + sqrt{2 + 4} = sqrt{0} + sqrt{6} = 0 + sqrt{6} approx 2.449]RHS:[sqrt{frac{16 + 1}{4 + 1}} + frac{2 + 3}{2 + 1} = sqrt{frac{17}{5}} + frac{5}{3} approx sqrt{3.4} + 1.666 approx 1.843 + 1.666 approx 3.509]So, LHS ≈ 2.449 and RHS ≈ 3.509. Not equal.How about ( x = -0.5 ):LHS:[sqrt{2 - (-0.5)} + sqrt{2 + 2(-0.5)} = sqrt{2.5} + sqrt{1} approx 1.581 + 1 = 2.581]RHS:[sqrt{frac{(-0.5)^4 + 1}{(-0.5)^2 + 1}} + frac{-0.5 + 3}{-0.5 + 1} = sqrt{frac{0.0625 + 1}{0.25 + 1}} + frac{2.5}{0.5} = sqrt{frac{1.0625}{1.25}} + 5 approx sqrt{0.85} + 5 approx 0.922 + 5 = 5.922]So, LHS ≈ 2.581 and RHS ≈ 5.922. Not equal.Hmm, so far only ( x = 1 ) works. Maybe it's the only solution. But I should check another value, say ( x = 0.5 ):LHS:[sqrt{2 - 0.5} + sqrt{2 + 2(0.5)} = sqrt{1.5} + sqrt{3} approx 1.225 + 1.732 approx 2.957]RHS:[sqrt{frac{(0.5)^4 + 1}{(0.5)^2 + 1}} + frac{0.5 + 3}{0.5 + 1} = sqrt{frac{0.0625 + 1}{0.25 + 1}} + frac{3.5}{1.5} approx sqrt{frac{1.0625}{1.25}} + 2.333 approx sqrt{0.85} + 2.333 approx 0.922 + 2.333 approx 3.255]So, LHS ≈ 2.957 and RHS ≈ 3.255. Not equal.How about ( x = 1.5 ):LHS:[sqrt{2 - 1.5} + sqrt{2 + 2(1.5)} = sqrt{0.5} + sqrt{5} approx 0.707 + 2.236 approx 2.943]RHS:[sqrt{frac{(1.5)^4 + 1}{(1.5)^2 + 1}} + frac{1.5 + 3}{1.5 + 1} = sqrt{frac{5.0625 + 1}{2.25 + 1}} + frac{4.5}{2.5} = sqrt{frac{6.0625}{3.25}} + 1.8 approx sqrt{1.865} + 1.8 approx 1.366 + 1.8 approx 3.166]So, LHS ≈ 2.943 and RHS ≈ 3.166. Not equal.Hmm, seems like ( x = 1 ) is the only one so far. Maybe I can try to analyze the functions more carefully.Let me define:( f(x) = sqrt{2 - x} + sqrt{2 + 2x} )( g(x) = sqrt{frac{x^4 + 1}{x^2 + 1}} + frac{x + 3}{x + 1} )I need to find ( x ) such that ( f(x) = g(x) ).I can analyze the behavior of ( f(x) ) and ( g(x) ) over the interval ( (-1, 2] ).First, let's see the behavior of ( f(x) ):- At ( x = -1 ): Not defined (since denominator in ( g(x) ) is zero, but ( x > -1 )).- At ( x = 1 ): ( f(1) = 3 )- At ( x = 2 ): ( f(2) approx 2.449 )Let me compute the derivative of ( f(x) ) to see if it's increasing or decreasing.( f(x) = sqrt{2 - x} + sqrt{2 + 2x} )Derivative:( f'(x) = frac{-1}{2sqrt{2 - x}} + frac{2}{2sqrt{2 + 2x}} = frac{-1}{2sqrt{2 - x}} + frac{1}{sqrt{2 + 2x}} )Simplify:( f'(x) = frac{-1}{2sqrt{2 - x}} + frac{1}{sqrt{2(1 + x)}} )Hmm, let's see when ( f'(x) = 0 ):[frac{-1}{2sqrt{2 - x}} + frac{1}{sqrt{2(1 + x)}} = 0][frac{1}{sqrt{2(1 + x)}} = frac{1}{2sqrt{2 - x}}]Cross-multiplying:[2sqrt{2 - x} = sqrt{2(1 + x)}]Square both sides:[4(2 - x) = 2(1 + x)][8 - 4x = 2 + 2x][8 - 2 = 4x + 2x][6 = 6x][x = 1]So, ( f(x) ) has a critical point at ( x = 1 ). Let's check the second derivative or test intervals to see if it's a maximum or minimum.For ( x < 1 ), say ( x = 0 ):( f'(0) = frac{-1}{2sqrt{2}} + frac{1}{sqrt{2}} = frac{-1 + 2}{2sqrt{2}} = frac{1}{2sqrt{2}} > 0 )So, increasing before ( x = 1 ).For ( x > 1 ), say ( x = 1.5 ):( f'(1.5) = frac{-1}{2sqrt{0.5}} + frac{1}{sqrt{5}} approx frac{-1}{1.414} + 0.447 approx -0.707 + 0.447 approx -0.26 )So, decreasing after ( x = 1 ).Thus, ( f(x) ) has a maximum at ( x = 1 ), which is 3.Now, let's analyze ( g(x) ):( g(x) = sqrt{frac{x^4 + 1}{x^2 + 1}} + frac{x + 3}{x + 1} )Simplify ( sqrt{frac{x^4 + 1}{x^2 + 1}} ):As I tried earlier, ( x^4 + 1 = (x^2 + 1)(x^2 - 1) + 2 ), so:[sqrt{frac{x^4 + 1}{x^2 + 1}} = sqrt{x^2 - 1 + frac{2}{x^2 + 1}}]Not sure if that helps, but let's compute ( g(x) ) at some points:At ( x = 1 ):[sqrt{frac{1 + 1}{1 + 1}} + frac{4}{2} = sqrt{1} + 2 = 3]At ( x = 0 ):[sqrt{frac{0 + 1}{0 + 1}} + frac{3}{1} = 1 + 3 = 4]At ( x = 2 ):[sqrt{frac{16 + 1}{4 + 1}} + frac{5}{3} approx sqrt{3.4} + 1.666 approx 1.843 + 1.666 approx 3.509]At ( x = -0.5 ):[sqrt{frac{0.0625 + 1}{0.25 + 1}} + frac{2.5}{0.5} approx sqrt{0.85} + 5 approx 0.922 + 5 = 5.922]At ( x = 0.5 ):[sqrt{frac{0.0625 + 1}{0.25 + 1}} + frac{3.5}{1.5} approx sqrt{0.85} + 2.333 approx 0.922 + 2.333 approx 3.255]At ( x = 1.5 ):[sqrt{frac{5.0625 + 1}{2.25 + 1}} + frac{4.5}{2.5} approx sqrt{1.865} + 1.8 approx 1.366 + 1.8 approx 3.166]So, ( g(x) ) seems to decrease from ( x = -1 ) to some point and then maybe increase? Wait, at ( x = -0.5 ), it's 5.922, at ( x = 0 ), it's 4, at ( x = 0.5 ), it's ~3.255, at ( x = 1 ), it's 3, at ( x = 1.5 ), ~3.166, and at ( x = 2 ), ~3.509.So, it seems ( g(x) ) decreases from ( x = -1 ) to ( x = 1 ), reaching a minimum at ( x = 1 ), and then increases from ( x = 1 ) to ( x = 2 ).Wait, but at ( x = 1.5 ), it's ~3.166, which is higher than at ( x = 1 ) (which is 3). So, it's increasing after ( x = 1 ).Therefore, ( g(x) ) has a minimum at ( x = 1 ).So, both ( f(x) ) and ( g(x) ) have their extrema at ( x = 1 ), and at that point, they are equal to 3.Now, let's see the behavior around ( x = 1 ).For ( x ) slightly less than 1, say ( x = 0.9 ):Compute ( f(0.9) ):[sqrt{2 - 0.9} + sqrt{2 + 1.8} = sqrt{1.1} + sqrt{3.8} approx 1.049 + 1.949 approx 2.998]Compute ( g(0.9) ):[sqrt{frac{(0.9)^4 + 1}{(0.9)^2 + 1}} + frac{0.9 + 3}{0.9 + 1}]Calculate numerator: ( 0.9^4 = 0.6561 ), so ( 0.6561 + 1 = 1.6561 )Denominator: ( 0.81 + 1 = 1.81 )So, ( sqrt{frac{1.6561}{1.81}} approx sqrt{0.915} approx 0.957 )Then, ( frac{3.9}{1.9} approx 2.052 )So, ( g(0.9) approx 0.957 + 2.052 approx 3.009 )So, ( f(0.9) approx 2.998 ) and ( g(0.9) approx 3.009 ). So, ( f(x) < g(x) ) just below ( x = 1 ).Similarly, for ( x = 1.1 ):Compute ( f(1.1) ):[sqrt{2 - 1.1} + sqrt{2 + 2.2} = sqrt{0.9} + sqrt{4.2} approx 0.949 + 2.049 approx 3.0]Compute ( g(1.1) ):[sqrt{frac{(1.1)^4 + 1}{(1.1)^2 + 1}} + frac{1.1 + 3}{1.1 + 1}]Calculate numerator: ( 1.1^4 = 1.4641 ), so ( 1.4641 + 1 = 2.4641 )Denominator: ( 1.21 + 1 = 2.21 )So, ( sqrt{frac{2.4641}{2.21}} approx sqrt{1.115} approx 1.056 )Then, ( frac{4.1}{2.1} approx 1.952 )So, ( g(1.1) approx 1.056 + 1.952 approx 3.008 )So, ( f(1.1) approx 3.0 ) and ( g(1.1) approx 3.008 ). So, ( f(x) approx g(x) ) but slightly less.Wait, but at ( x = 1 ), both are exactly 3. So, around ( x = 1 ), ( f(x) ) is increasing before 1 and decreasing after 1, while ( g(x) ) is decreasing before 1 and increasing after 1.So, at ( x = 1 ), both functions reach 3, and around that point, ( f(x) ) is less than ( g(x) ) just below and just above ( x = 1 ). So, it's possible that ( x = 1 ) is the only solution where they intersect.But to be thorough, let me check another point, say ( x = 0.8 ):( f(0.8) = sqrt{2 - 0.8} + sqrt{2 + 1.6} = sqrt{1.2} + sqrt{3.6} approx 1.095 + 1.897 approx 2.992 )( g(0.8) = sqrt{frac{0.8^4 + 1}{0.8^2 + 1}} + frac{0.8 + 3}{0.8 + 1} )Compute numerator: ( 0.8^4 = 0.4096 ), so ( 0.4096 + 1 = 1.4096 )Denominator: ( 0.64 + 1 = 1.64 )So, ( sqrt{frac{1.4096}{1.64}} approx sqrt{0.859} approx 0.927 )Then, ( frac{3.8}{1.8} approx 2.111 )So, ( g(0.8) approx 0.927 + 2.111 approx 3.038 )Thus, ( f(0.8) approx 2.992 ) and ( g(0.8) approx 3.038 ). So, still ( f(x) < g(x) ).Similarly, at ( x = 1.2 ):( f(1.2) = sqrt{2 - 1.2} + sqrt{2 + 2.4} = sqrt{0.8} + sqrt{4.4} approx 0.894 + 2.098 approx 2.992 )( g(1.2) = sqrt{frac{1.2^4 + 1}{1.2^2 + 1}} + frac{1.2 + 3}{1.2 + 1} )Compute numerator: ( 1.2^4 = 2.0736 ), so ( 2.0736 + 1 = 3.0736 )Denominator: ( 1.44 + 1 = 2.44 )So, ( sqrt{frac{3.0736}{2.44}} approx sqrt{1.259} approx 1.122 )Then, ( frac{4.2}{2.2} approx 1.909 )So, ( g(1.2) approx 1.122 + 1.909 approx 3.031 )Thus, ( f(1.2) approx 2.992 ) and ( g(1.2) approx 3.031 ). So, again, ( f(x) < g(x) ).This suggests that ( x = 1 ) is the only point where ( f(x) = g(x) ). To confirm, let me check the derivatives around ( x = 1 ).We know that ( f(x) ) has a maximum at ( x = 1 ), so it's increasing before and decreasing after. ( g(x) ) has a minimum at ( x = 1 ), so it's decreasing before and increasing after.Therefore, at ( x = 1 ), ( f(x) ) is at its peak, and ( g(x) ) is at its trough. Since they meet exactly at 3, and around ( x = 1 ), ( f(x) ) is less than ( g(x) ), it's likely that ( x = 1 ) is the only solution.To be thorough, let me consider the possibility of another intersection point. Suppose there is another ( x ) where ( f(x) = g(x) ). Given the behavior of both functions, ( f(x) ) is increasing from ( x = -1 ) to ( x = 1 ), and ( g(x) ) is decreasing from ( x = -1 ) to ( x = 1 ). So, they can intersect at most once in that interval. Similarly, after ( x = 1 ), ( f(x) ) decreases while ( g(x) ) increases, so they can intersect at most once there as well. But since at ( x = 1 ), they both are equal, and the behavior around ( x = 1 ) shows that ( f(x) ) is less than ( g(x) ) on both sides, it's likely that ( x = 1 ) is the only solution.Therefore, after checking several points and analyzing the behavior of both functions, I conclude that the only solution is ( x = 1 ).

💡First, I recognize that the function ( y = frac{k}{x} ) is an inverse proportion function, where ( k ) is a constant and ( k neq 0 ).Given that the graph passes through the point ( (2, -1) ), I can substitute ( x = 2 ) and ( y = -1 ) into the equation to find ( k ).Substituting the values:[-1 = frac{k}{2}]To solve for ( k ), I multiply both sides of the equation by 2:[-1 cdot 2 = frac{k}{2} cdot 2]This simplifies to:[-2 = k]Therefore, the value of ( k ) is ( -2 ), which corresponds to option B.

💡Okay, so I have this problem here with two propositions, p and q, and I need to figure out which of the given options is true. Let me take it step by step.First, let's understand proposition p. It says that the necessary and sufficient condition for the angle between vector a = (1, 2) and vector b = (2, k) to be acute is k > -1. Hmm, okay. I remember that the angle between two vectors is acute if their dot product is positive. So, maybe I should calculate the dot product of a and b.The dot product of a and b is (1)(2) + (2)(k) = 2 + 2k. For the angle to be acute, this dot product needs to be greater than zero. So, 2 + 2k > 0. Let me solve this inequality:2 + 2k > 0 Subtract 2 from both sides: 2k > -2 Divide both sides by 2: k > -1Okay, so that gives me k > -1. So, according to this, proposition p is saying that k > -1 is the necessary and sufficient condition. But wait, is that all? I remember that if two vectors are in the same direction, the angle is zero, which is technically acute, but what if they are collinear? Does that affect the condition?Let me think. If vectors a and b are collinear, then one is a scalar multiple of the other. So, if a = (1, 2) and b = (2, k), then for them to be collinear, there must be a scalar λ such that 2 = λ*1 and k = λ*2. So, λ = 2, which would make k = 4. So, when k = 4, vectors a and b are collinear.But wait, collinear vectors have an angle of 0 degrees, which is still considered acute. So, does that mean k = 4 is still acceptable? So, the condition k > -1 includes k = 4, which is fine because the angle is still acute. So, maybe proposition p is actually correct?Wait, but sometimes in math, necessary and sufficient conditions have to exclude certain cases. Is there a case where k > -1 doesn't guarantee an acute angle? Hmm, let me think. If k is greater than -1, the dot product is positive, so the angle is between 0 and 90 degrees, which is acute. So, maybe p is correct.But the initial thought was that p is false because when k = 4, the vectors are collinear. But collinear vectors still have an angle of 0 degrees, which is acute. So, maybe p is actually true. Hmm, this is confusing.Let me double-check. The dot product being positive means the angle is acute. If vectors are collinear in the same direction, the angle is 0, which is still acute. So, k > -1 is indeed the necessary and sufficient condition. So, proposition p is true.Wait, but in the initial solution, it was stated that p is false because when k = 4, vectors are collinear. But collinear vectors still have an acute angle. So, maybe the initial solution was wrong? Hmm, I need to clarify this.Let me look up the definition of an acute angle between vectors. An acute angle is one that is less than 90 degrees. So, 0 degrees is considered acute. Therefore, if vectors are collinear in the same direction, the angle is 0 degrees, which is still acute. So, k = 4 is acceptable, and thus, p is correct.Okay, moving on to proposition q. It says that the function f(x) defined as:f(x) = sin(x + π/3) when x ≤ 0 f(x) = cos(x + π/6) when x > 0is an even function. An even function satisfies f(-x) = f(x) for all x. So, I need to check if f(-x) equals f(x).Let me consider two cases: when x is positive and when x is negative.Case 1: x > 0 Then, f(x) = cos(x + π/6). Now, f(-x) would be f(-x) = sin(-x + π/3) because -x < 0. So, f(-x) = sin(-x + π/3). I need to see if this equals f(x) = cos(x + π/6).Let me simplify sin(-x + π/3). sin(-x + π/3) = sin(π/3 - x) Using the sine identity: sin(A - B) = sin A cos B - cos A sin B So, sin(π/3 - x) = sin(π/3)cos(x) - cos(π/3)sin(x) We know that sin(π/3) = √3/2 and cos(π/3) = 1/2 So, sin(π/3 - x) = (√3/2)cos(x) - (1/2)sin(x)Now, let's look at f(x) = cos(x + π/6). Using the cosine addition formula: cos(A + B) = cos A cos B - sin A sin B So, cos(x + π/6) = cos(x)cos(π/6) - sin(x)sin(π/6) cos(π/6) = √3/2 and sin(π/6) = 1/2 So, cos(x + π/6) = (√3/2)cos(x) - (1/2)sin(x)Hey, that's the same as sin(π/3 - x)! So, f(-x) = f(x) when x > 0.Case 2: x < 0 Then, f(x) = sin(x + π/3). Now, f(-x) would be f(-x) = cos(-x + π/6) because -x > 0. So, f(-x) = cos(-x + π/6). I need to see if this equals f(x) = sin(x + π/3).Let me simplify cos(-x + π/6). cos(-x + π/6) = cos(π/6 - x) Using the cosine identity: cos(A - B) = cos A cos B + sin A sin B So, cos(π/6 - x) = cos(π/6)cos(x) + sin(π/6)sin(x) cos(π/6) = √3/2 and sin(π/6) = 1/2 So, cos(π/6 - x) = (√3/2)cos(x) + (1/2)sin(x)Now, let's look at f(x) = sin(x + π/3). Using the sine addition formula: sin(A + B) = sin A cos B + cos A sin B So, sin(x + π/3) = sin(x)cos(π/3) + cos(x)sin(π/3) cos(π/3) = 1/2 and sin(π/3) = √3/2 So, sin(x + π/3) = (1/2)sin(x) + (√3/2)cos(x)Which is the same as cos(π/6 - x). So, f(-x) = f(x) when x < 0.What about x = 0? f(0) = sin(0 + π/3) = sin(π/3) = √3/2 f(-0) = f(0) = √3/2, so it's equal.Therefore, f(-x) = f(x) for all x, so f(x) is indeed an even function. So, proposition q is true.Wait, but in the initial solution, it was stated that q is true, so p is false, hence option B is correct. But according to my analysis, p is true and q is true, so p ∧ q would be true, which is option A.But in the initial solution, it was concluded that p is false because when k = 4, vectors are collinear. But as I thought earlier, collinear vectors still have an acute angle (0 degrees). So, maybe the initial solution was incorrect.Let me double-check the dot product condition. The dot product is 2 + 2k > 0, which gives k > -1. So, as long as k > -1, the angle is acute, regardless of whether they are collinear or not. So, p is correct.Therefore, both p and q are true, so p ∧ q is true, which is option A.Wait, but the initial solution said p is false because when k = 4, vectors are collinear, but that's still an acute angle. So, maybe the initial solution was wrong. Hmm.Alternatively, maybe the initial solution considered that collinear vectors are not considered to form an acute angle, but according to the definition, 0 degrees is acute. So, p is correct.Therefore, I think the correct answer should be A: p ∧ q.But the initial solution concluded B: (￢p) ∧ q. So, there must be something I'm missing.Wait, let me think again about proposition p. It says "the necessary and sufficient condition for the angle between vector a and vector b to be acute is k > -1". So, if k > -1, the angle is acute, and if the angle is acute, then k > -1.But is that the case? Let's see. If the angle is acute, then the dot product is positive, so k > -1. Conversely, if k > -1, the dot product is positive, so the angle is acute. So, it's a necessary and sufficient condition.But wait, what if k = -1? Then, the dot product is zero, so the angle is 90 degrees, which is not acute. So, k > -1 is indeed the necessary and sufficient condition. So, p is correct.So, why did the initial solution say p is false? Maybe because they thought collinear vectors are not considered acute, but according to the definition, 0 degrees is acute. So, maybe the initial solution was incorrect.Therefore, I think both p and q are true, so the correct answer is A: p ∧ q.But wait, let me check the initial solution again. It said:"When k = 4, a = 1/2 b, then a and b are collinear, thus the necessary and sufficient condition is k > -1 and k ≠ 4, so p is false."But as I thought, collinear vectors still have an acute angle, so k = 4 is acceptable. So, p is correct.Therefore, I think the initial solution was wrong, and the correct answer is A: p ∧ q.But wait, the initial solution concluded B: (￢p) ∧ q. So, maybe I'm missing something.Wait, maybe the definition of acute angle excludes 0 degrees? Let me check.According to the definition, an acute angle is one that is less than 90 degrees. So, 0 degrees is not considered an acute angle. Wait, is that correct?Wait, no, 0 degrees is considered an acute angle. Because acute angles are between 0 and 90 degrees, not including 90. So, 0 degrees is included.But wait, in some contexts, 0 degrees might not be considered an angle at all, but in vector terms, if two vectors are in the same direction, the angle between them is 0 degrees, which is acute.Hmm, I'm confused now. Let me check a reference.According to standard definitions, an acute angle is one that is less than 90 degrees but greater than 0 degrees. So, 0 degrees is not considered an acute angle. So, if two vectors are collinear in the same direction, the angle between them is 0 degrees, which is not acute.Therefore, in that case, when k = 4, the angle is 0 degrees, which is not acute. So, the condition should exclude k = 4.Therefore, the necessary and sufficient condition is k > -1 and k ≠ 4. So, proposition p is stating that the condition is k > -1, which is not entirely correct because it includes k = 4, which does not result in an acute angle.Therefore, proposition p is false.Okay, that makes sense now. So, p is false because it doesn't exclude k = 4, which results in a 0-degree angle, not acute.So, going back, p is false and q is true. Therefore, the correct composite proposition is (￢p) ∧ q, which is option B.So, the initial solution was correct after all. I had to clarify the definition of an acute angle, which excludes 0 degrees. Therefore, p is false because it includes k = 4, which doesn't satisfy the acute angle condition.Therefore, the correct answer is B: (￢p) ∧ q.

💡Okay, so I have this problem with 2014 real numbers ( x_1, x_2, ldots, x_{2014} ) that satisfy a system of equations:[sum_{k=1}^{2014} frac{x_k}{n+k} = frac{1}{2n+1} quad text{for } n = 1, 2, ldots, 2014.]And I need to find the value of ( sum_{k=1}^{2014} frac{x_k}{2k+1} ).Hmm, this looks like a system of linear equations where each equation corresponds to a different ( n ). Each equation is a linear combination of the ( x_k ) terms with coefficients ( frac{1}{n+k} ). The right-hand side is ( frac{1}{2n+1} ).I remember that sometimes when dealing with systems of equations that have a specific structure, especially with fractions like ( frac{1}{n+k} ), it might be related to something like the Hilbert matrix or other structured matrices. But I'm not sure if that's directly applicable here.Another thought is to consider constructing a polynomial that can encapsulate these equations. Since we have 2014 equations, maybe a polynomial of degree 2014 could be involved. Let me think about how to construct such a polynomial.Let me define a polynomial ( f(x) ) as follows:[f(x) = left( prod_{i=1}^{2014} (x + i) right) left( (2x + 1) left( sum_{i=1}^{2014} frac{x_i}{x + i} right) - 1 right)]I chose this form because when ( x = n ) (where ( n ) is 1, 2, ..., 2014), the term ( prod_{i=1}^{2014} (x + i) ) becomes ( prod_{i=1}^{2014} (n + i) ), which is non-zero. The other part, ( (2x + 1) left( sum_{i=1}^{2014} frac{x_i}{x + i} right) - 1 ), when ( x = n ), becomes ( (2n + 1) left( frac{1}{2n + 1} right) - 1 = 1 - 1 = 0 ). So, ( f(n) = 0 ) for each ( n = 1, 2, ldots, 2014 ).That means ( f(x) ) has roots at ( x = 1, 2, ldots, 2014 ). Therefore, ( f(x) ) can be written as:[f(x) = c cdot prod_{i=1}^{2014} (x - i)]where ( c ) is a constant.Now, I need to find the value of ( c ). To do this, I can evaluate ( f(x) ) at a specific value of ( x ) that isn't one of the roots. Let me choose ( x = -frac{1}{2} ) because it might simplify some terms.Plugging ( x = -frac{1}{2} ) into ( f(x) ):[fleft( -frac{1}{2} right) = left( prod_{i=1}^{2014} left( -frac{1}{2} + i right) right) left( (2 cdot -frac{1}{2} + 1) left( sum_{i=1}^{2014} frac{x_i}{ -frac{1}{2} + i } right) - 1 right)]Simplify the terms:First, ( 2 cdot -frac{1}{2} + 1 = -1 + 1 = 0 ). So the term inside the brackets becomes:[0 cdot left( sum_{i=1}^{2014} frac{x_i}{ -frac{1}{2} + i } right) - 1 = -1]Therefore,[fleft( -frac{1}{2} right) = left( prod_{i=1}^{2014} left( i - frac{1}{2} right) right) cdot (-1)]But from the other expression of ( f(x) ), we have:[fleft( -frac{1}{2} right) = c cdot prod_{i=1}^{2014} left( -frac{1}{2} - i right) = c cdot (-1)^{2014} prod_{i=1}^{2014} left( i + frac{1}{2} right)]Since ( (-1)^{2014} = 1 ), this simplifies to:[c cdot prod_{i=1}^{2014} left( i + frac{1}{2} right)]Setting the two expressions for ( fleft( -frac{1}{2} right) ) equal:[c cdot prod_{i=1}^{2014} left( i + frac{1}{2} right) = - prod_{i=1}^{2014} left( i - frac{1}{2} right)]Therefore,[c = - frac{ prod_{i=1}^{2014} left( i - frac{1}{2} right) }{ prod_{i=1}^{2014} left( i + frac{1}{2} right) }]Hmm, this seems a bit complicated. Maybe I can simplify this ratio. Notice that:[prod_{i=1}^{2014} left( i - frac{1}{2} right) = prod_{i=1}^{2014} frac{2i - 1}{2} = frac{1}{2^{2014}} prod_{i=1}^{2014} (2i - 1)]Similarly,[prod_{i=1}^{2014} left( i + frac{1}{2} right) = prod_{i=1}^{2014} frac{2i + 1}{2} = frac{1}{2^{2014}} prod_{i=1}^{2014} (2i + 1)]So,[c = - frac{ frac{1}{2^{2014}} prod_{i=1}^{2014} (2i - 1) }{ frac{1}{2^{2014}} prod_{i=1}^{2014} (2i + 1) } = - frac{ prod_{i=1}^{2014} (2i - 1) }{ prod_{i=1}^{2014} (2i + 1) }]This simplifies to:[c = - frac{ (1 cdot 3 cdot 5 cdots 4027) }{ (3 cdot 5 cdot 7 cdots 4029) } = - frac{1}{4029}]Wait, let's see. The numerator is the product of odd numbers from 1 to 4027, and the denominator is the product of odd numbers from 3 to 4029. So, the numerator is missing the last term 4029, and the denominator is missing the first term 1. So, the ratio is ( frac{1}{4029} ). Therefore, ( c = - frac{1}{4029} ).So, now we have:[f(x) = - frac{1}{4029} prod_{i=1}^{2014} (x - i)]But we also have the original definition of ( f(x) ):[f(x) = left( prod_{i=1}^{2014} (x + i) right) left( (2x + 1) left( sum_{i=1}^{2014} frac{x_i}{x + i} right) - 1 right)]Setting these equal:[left( prod_{i=1}^{2014} (x + i) right) left( (2x + 1) left( sum_{i=1}^{2014} frac{x_i}{x + i} right) - 1 right) = - frac{1}{4029} prod_{i=1}^{2014} (x - i)]Now, I want to find ( sum_{k=1}^{2014} frac{x_k}{2k + 1} ). Notice that if I set ( x = frac{1}{2} ), then ( 2x + 1 = 2 cdot frac{1}{2} + 1 = 2 ). So, plugging ( x = frac{1}{2} ) into the equation:[left( prod_{i=1}^{2014} left( frac{1}{2} + i right) right) left( 2 left( sum_{i=1}^{2014} frac{x_i}{ frac{1}{2} + i } right) - 1 right) = - frac{1}{4029} prod_{i=1}^{2014} left( frac{1}{2} - i right)]Simplify the right-hand side:[- frac{1}{4029} prod_{i=1}^{2014} left( frac{1}{2} - i right) = - frac{1}{4029} prod_{i=1}^{2014} left( - left( i - frac{1}{2} right) right) = - frac{1}{4029} (-1)^{2014} prod_{i=1}^{2014} left( i - frac{1}{2} right)]Since ( (-1)^{2014} = 1 ), this becomes:[- frac{1}{4029} prod_{i=1}^{2014} left( i - frac{1}{2} right)]But from earlier, we know that:[prod_{i=1}^{2014} left( i - frac{1}{2} right) = frac{1}{2^{2014}} prod_{i=1}^{2014} (2i - 1)]And we also have:[prod_{i=1}^{2014} left( frac{1}{2} + i right) = frac{1}{2^{2014}} prod_{i=1}^{2014} (2i + 1)]So, plugging these into the equation:[left( frac{1}{2^{2014}} prod_{i=1}^{2014} (2i + 1) right) left( 2 left( sum_{i=1}^{2014} frac{x_i}{ frac{1}{2} + i } right) - 1 right) = - frac{1}{4029} cdot frac{1}{2^{2014}} prod_{i=1}^{2014} (2i - 1)]We can cancel ( frac{1}{2^{2014}} ) from both sides:[prod_{i=1}^{2014} (2i + 1) left( 2 left( sum_{i=1}^{2014} frac{x_i}{ frac{1}{2} + i } right) - 1 right) = - frac{1}{4029} prod_{i=1}^{2014} (2i - 1)]Let me denote ( S = sum_{i=1}^{2014} frac{x_i}{ frac{1}{2} + i } ). Then the equation becomes:[prod_{i=1}^{2014} (2i + 1) left( 2S - 1 right) = - frac{1}{4029} prod_{i=1}^{2014} (2i - 1)]We can solve for ( S ):[2S - 1 = - frac{1}{4029} cdot frac{ prod_{i=1}^{2014} (2i - 1) }{ prod_{i=1}^{2014} (2i + 1) }]From earlier, we found that:[frac{ prod_{i=1}^{2014} (2i - 1) }{ prod_{i=1}^{2014} (2i + 1) } = frac{1}{4029}]Therefore,[2S - 1 = - frac{1}{4029} cdot frac{1}{4029} = - frac{1}{4029^2}]So,[2S = 1 - frac{1}{4029^2}]Thus,[S = frac{1}{2} left( 1 - frac{1}{4029^2} right)]But ( S = sum_{i=1}^{2014} frac{x_i}{ frac{1}{2} + i } = sum_{i=1}^{2014} frac{x_i}{2i + 1} ), which is exactly the sum we need to find.Therefore,[sum_{k=1}^{2014} frac{x_k}{2k + 1} = frac{1}{2} left( 1 - frac{1}{4029^2} right)]Simplifying,[sum_{k=1}^{2014} frac{x_k}{2k + 1} = frac{1}{2} - frac{1}{2 cdot 4029^2}]But the problem asks for the value, so we can write it as:[sum_{k=1}^{2014} frac{x_k}{2k + 1} = frac{1}{2} left( 1 - frac{1}{4029^2} right)]Which is the same as:[sum_{k=1}^{2014} frac{x_k}{2k + 1} = frac{1}{2} - frac{1}{2 cdot 4029^2}]But to make it look cleaner, perhaps factor out the ( frac{1}{2} ):[sum_{k=1}^{2014} frac{x_k}{2k + 1} = frac{1}{2} left( 1 - frac{1}{4029^2} right)]So, that's the value we're looking for.**Final Answer**The value of ( sum_{k=1}^{2014} frac{x_k}{2k+1} ) is boxed{dfrac{1}{2} left(1 - dfrac{1}{4029^2}right)}.