💡First, I need to determine the duration of the hunting season. Since it lasts for one quarter of the year and there are 12 months in a year, the hunting season is 3 months long.Next, I'll calculate how many times Jack goes hunting during the season. If he hunts 6 times each month, then over 3 months, he hunts 6 times/month * 3 months = 18 times.Now, I'll find out how many deer Jack catches in total during the season. Since he catches 2 deer each time he hunts, the total number of deer caught is 2 deer * 18 times = 36 deer.Jack keeps half the weight of the deer he catches each year. Given that he keeps 10,800 pounds of deer, the total weight of the deer he catches in a year is 10,800 pounds * 2 = 21,600 pounds.Finally, to find out how much each deer weighs, I'll divide the total weight of the deer by the number of deer caught. So, each deer weighs 21,600 pounds / 36 deer = 600 pounds.

💡Okay, so I have this hyperbola equation: ( E: frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ) where ( a > 0 ) and ( b > 0 ). The problem says that the focus of the hyperbola is symmetric with respect to the asymptote line and lies on the hyperbola ( E ). I need to find the eccentricity of the hyperbola ( E ).First, I remember that for a hyperbola of the form ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ), the foci are located at ( (pm c, 0) ) where ( c^2 = a^2 + b^2 ). The eccentricity ( e ) is given by ( e = frac{c}{a} ), so that's what I need to find.Next, the asymptotes of this hyperbola are the lines ( y = pm frac{b}{a}x ). The problem mentions that the focus is symmetric with respect to the asymptote line. Hmm, symmetry with respect to a line usually means that if you reflect the point over that line, you get another point. But it also says that the focus lies on the hyperbola. Wait, the focus is already on the hyperbola? But normally, the foci are inside the hyperbola, not on it. That seems confusing.Wait, maybe I misread. It says the focus is symmetric with respect to the asymptote line and lies on the hyperbola. So, perhaps the reflection of the focus over the asymptote lies on the hyperbola. That makes more sense. So, if I take one focus, say ( (c, 0) ), and reflect it over one of the asymptotes, say ( y = frac{b}{a}x ), the reflected point should lie on the hyperbola.Okay, so let me try to visualize this. The hyperbola has two foci at ( (c, 0) ) and ( (-c, 0) ). The asymptotes are ( y = frac{b}{a}x ) and ( y = -frac{b}{a}x ). If I reflect the focus ( (c, 0) ) over the asymptote ( y = frac{b}{a}x ), the image point should lie on the hyperbola.To find the reflection of a point over a line, I can use the formula for reflection over a line. The formula for reflecting a point ( (x, y) ) over the line ( ax + by + c = 0 ) is a bit complicated, but in this case, the line is ( y = frac{b}{a}x ), which can be rewritten as ( frac{b}{a}x - y = 0 ). So, ( a' = frac{b}{a} ), ( b' = -1 ), and ( c' = 0 ).The reflection formula is:[x' = frac{x(b'^2 - a'^2) - 2a'y b'}{a'^2 + b'^2}][y' = frac{y(a'^2 - b'^2) - 2a'x b'}{a'^2 + b'^2}]Wait, maybe it's easier to use another method. Let me recall that the reflection of a point over a line can be found by finding the perpendicular projection of the point onto the line and then moving the same distance on the other side.So, let's denote the focus as ( F(c, 0) ) and the asymptote as ( L: y = frac{b}{a}x ). To find the reflection ( F' ) of ( F ) over ( L ), I can follow these steps:1. Find the equation of the line perpendicular to ( L ) that passes through ( F ).2. Find the intersection point ( M ) of this perpendicular line with ( L ).3. The reflection point ( F' ) is such that ( M ) is the midpoint between ( F ) and ( F' ).Let's do that step by step.First, the slope of ( L ) is ( frac{b}{a} ), so the slope of the perpendicular line is ( -frac{a}{b} ).The equation of the perpendicular line passing through ( F(c, 0) ) is:[y - 0 = -frac{a}{b}(x - c)][y = -frac{a}{b}x + frac{a c}{b}]Now, find the intersection point ( M ) between this line and ( L: y = frac{b}{a}x ).Set the two equations equal:[-frac{a}{b}x + frac{a c}{b} = frac{b}{a}x]Multiply both sides by ( ab ) to eliminate denominators:[-a^2 x + a^2 c = b^2 x][-a^2 x - b^2 x = -a^2 c][-(a^2 + b^2)x = -a^2 c][x = frac{a^2 c}{a^2 + b^2}]Now, plug this back into ( L ) to find ( y ):[y = frac{b}{a} cdot frac{a^2 c}{a^2 + b^2} = frac{a b c}{a^2 + b^2}]So, the midpoint ( M ) is ( left( frac{a^2 c}{a^2 + b^2}, frac{a b c}{a^2 + b^2} right) ).Since ( M ) is the midpoint between ( F(c, 0) ) and ( F'(x', y') ), we can write:[frac{c + x'}{2} = frac{a^2 c}{a^2 + b^2}][frac{0 + y'}{2} = frac{a b c}{a^2 + b^2}]Solving for ( x' ) and ( y' ):[c + x' = frac{2 a^2 c}{a^2 + b^2}][x' = frac{2 a^2 c}{a^2 + b^2} - c = frac{2 a^2 c - c(a^2 + b^2)}{a^2 + b^2} = frac{2 a^2 c - a^2 c - b^2 c}{a^2 + b^2} = frac{a^2 c - b^2 c}{a^2 + b^2} = frac{c(a^2 - b^2)}{a^2 + b^2}][y' = 2 cdot frac{a b c}{a^2 + b^2} = frac{2 a b c}{a^2 + b^2}]So, the reflection point ( F' ) is ( left( frac{c(a^2 - b^2)}{a^2 + b^2}, frac{2 a b c}{a^2 + b^2} right) ).Now, according to the problem, this point ( F' ) lies on the hyperbola ( E ). So, plugging ( F' ) into the hyperbola equation:[frac{left( frac{c(a^2 - b^2)}{a^2 + b^2} right)^2}{a^2} - frac{left( frac{2 a b c}{a^2 + b^2} right)^2}{b^2} = 1]Let me simplify this equation step by step.First, compute the x-term:[frac{left( frac{c(a^2 - b^2)}{a^2 + b^2} right)^2}{a^2} = frac{c^2 (a^2 - b^2)^2}{a^2 (a^2 + b^2)^2}]Next, compute the y-term:[frac{left( frac{2 a b c}{a^2 + b^2} right)^2}{b^2} = frac{4 a^2 b^2 c^2}{(a^2 + b^2)^2 b^2} = frac{4 a^2 c^2}{(a^2 + b^2)^2}]So, putting it all together:[frac{c^2 (a^2 - b^2)^2}{a^2 (a^2 + b^2)^2} - frac{4 a^2 c^2}{(a^2 + b^2)^2} = 1]Let me factor out ( frac{c^2}{(a^2 + b^2)^2} ):[frac{c^2}{(a^2 + b^2)^2} left( frac{(a^2 - b^2)^2}{a^2} - 4 a^2 right) = 1]Now, let's compute the expression inside the parentheses:[frac{(a^2 - b^2)^2}{a^2} - 4 a^2 = frac{a^4 - 2 a^2 b^2 + b^4}{a^2} - 4 a^2 = frac{a^4 - 2 a^2 b^2 + b^4 - 4 a^4}{a^2} = frac{-3 a^4 - 2 a^2 b^2 + b^4}{a^2}]Wait, that seems complicated. Maybe I made a mistake in expanding ( (a^2 - b^2)^2 ). Let me double-check:( (a^2 - b^2)^2 = a^4 - 2 a^2 b^2 + b^4 ). Yes, that's correct.So, substituting back:[frac{c^2}{(a^2 + b^2)^2} cdot frac{-3 a^4 - 2 a^2 b^2 + b^4}{a^2} = 1]This seems messy. Maybe there's a better way to approach this problem.Wait, let me recall that ( c^2 = a^2 + b^2 ). Maybe substituting ( c^2 ) in terms of ( a ) and ( b ) can help.So, ( c^2 = a^2 + b^2 ). Let me substitute that into the equation.First, let's rewrite the equation:[frac{(a^2 + b^2) (a^2 - b^2)^2}{a^2 (a^2 + b^2)^2} - frac{4 a^2 (a^2 + b^2)}{(a^2 + b^2)^2} = 1]Simplify the first term:[frac{(a^2 - b^2)^2}{a^2 (a^2 + b^2)}]Second term:[frac{4 a^2}{(a^2 + b^2)}]So, the equation becomes:[frac{(a^2 - b^2)^2}{a^2 (a^2 + b^2)} - frac{4 a^2}{(a^2 + b^2)} = 1]Let me factor out ( frac{1}{a^2 + b^2} ):[frac{1}{a^2 + b^2} left( frac{(a^2 - b^2)^2}{a^2} - 4 a^2 right) = 1]Compute the expression inside the brackets:[frac{(a^2 - b^2)^2}{a^2} - 4 a^2 = frac{a^4 - 2 a^2 b^2 + b^4}{a^2} - 4 a^2 = frac{a^4 - 2 a^2 b^2 + b^4 - 4 a^4}{a^2} = frac{-3 a^4 - 2 a^2 b^2 + b^4}{a^2}]So, the equation becomes:[frac{-3 a^4 - 2 a^2 b^2 + b^4}{a^2 (a^2 + b^2)} = 1]Multiply both sides by ( a^2 (a^2 + b^2) ):[-3 a^4 - 2 a^2 b^2 + b^4 = a^2 (a^2 + b^2)]Expand the right side:[-3 a^4 - 2 a^2 b^2 + b^4 = a^4 + a^2 b^2]Bring all terms to the left side:[-3 a^4 - 2 a^2 b^2 + b^4 - a^4 - a^2 b^2 = 0][-4 a^4 - 3 a^2 b^2 + b^4 = 0]Hmm, this is a quartic equation in terms of ( a ) and ( b ). Maybe I can let ( k = frac{b}{a} ) to simplify it.Let ( k = frac{b}{a} ), so ( b = k a ). Substitute into the equation:[-4 a^4 - 3 a^2 (k a)^2 + (k a)^4 = 0][-4 a^4 - 3 a^4 k^2 + a^4 k^4 = 0]Factor out ( a^4 ):[a^4 (-4 - 3 k^2 + k^4) = 0]Since ( a > 0 ), ( a^4 neq 0 ), so:[-4 - 3 k^2 + k^4 = 0]Let me rewrite this as:[k^4 - 3 k^2 - 4 = 0]This is a quadratic in terms of ( k^2 ). Let me set ( m = k^2 ):[m^2 - 3 m - 4 = 0]Solve for ( m ):[m = frac{3 pm sqrt{9 + 16}}{2} = frac{3 pm sqrt{25}}{2} = frac{3 pm 5}{2}]So, ( m = frac{3 + 5}{2} = 4 ) or ( m = frac{3 - 5}{2} = -1 ).Since ( m = k^2 ) and ( k^2 ) cannot be negative, we discard ( m = -1 ). So, ( m = 4 ), which means ( k^2 = 4 ), so ( k = 2 ) (since ( k = frac{b}{a} > 0 )).Therefore, ( frac{b}{a} = 2 ), so ( b = 2 a ).Now, recall that ( c^2 = a^2 + b^2 ). Substitute ( b = 2 a ):[c^2 = a^2 + (2 a)^2 = a^2 + 4 a^2 = 5 a^2][c = sqrt{5} a]The eccentricity ( e ) is ( frac{c}{a} ):[e = frac{sqrt{5} a}{a} = sqrt{5}]So, the eccentricity is ( sqrt{5} ), which corresponds to option C.**Final Answer**boxed{C}

💡Alright, so I've got this inequality to prove:frac{a_{1}}{a_{2}^{2}+cdots+a_{n}^{2}}+frac{a_{2}}{a_{1}^{2}+a_{3}^{2}+cdots+a_{n}^{2}}+cdots+frac{a_{n}}{a_{1}^{2}+cdots+a_{n-1}^{2}} geqslant frac{4}{a_{1}+a_{2}+cdots+a_{n}}.Hmm, okay. Let me try to unpack this step by step.First, all the ( a_i ) are non-negative real numbers. That's good because it means I don't have to worry about negative values complicating things. The left side of the inequality is a sum of fractions, each of which has one ( a_i ) in the numerator and the sum of the squares of all the other ( a_j ) in the denominator. The right side is 4 divided by the sum of all ( a_i ).I need to show that this sum of fractions is at least as big as ( frac{4}{a_1 + a_2 + cdots + a_n} ).Let me think about possible inequalities that could help here. The Cauchy-Schwarz inequality comes to mind because it often helps with sums involving fractions and squares. Maybe I can apply it here somehow.Let me recall the Cauchy-Schwarz inequality in the form:left( sum_{i=1}^{n} x_i y_i right)^2 leq left( sum_{i=1}^{n} x_i^2 right) left( sum_{i=1}^{n} y_i^2 right).But in this problem, I have a sum of fractions. Maybe I can rearrange the terms or find a way to express the sum in a form where Cauchy-Schwarz can be applied.Alternatively, maybe I can use the Titu's lemma, which is a specific case of the Cauchy-Schwarz inequality. Titu's lemma states that:frac{x_1^2}{a_1} + frac{x_2^2}{a_2} + cdots + frac{x_n^2}{a_n} geq frac{(x_1 + x_2 + cdots + x_n)^2}{a_1 + a_2 + cdots + a_n}.But in my case, the numerators are just ( a_i ), not ( a_i^2 ). So maybe Titu's lemma isn't directly applicable, but perhaps I can manipulate the expression to fit.Wait, let me think differently. Maybe I can consider each term ( frac{a_i}{sum_{j neq i} a_j^2} ) and see if I can bound it from below.Let me denote ( S = a_1 + a_2 + cdots + a_n ). Then, the right-hand side of the inequality is ( frac{4}{S} ).So, I need to show:sum_{i=1}^{n} frac{a_i}{sum_{j neq i} a_j^2} geq frac{4}{S}.Hmm, maybe I can use the Cauchy-Schwarz inequality in a clever way here. Let's consider the sum:sum_{i=1}^{n} frac{a_i}{sum_{j neq i} a_j^2}.If I think of this as a sum of fractions, perhaps I can pair each ( a_i ) with the denominator ( sum_{j neq i} a_j^2 ).Let me try applying the Cauchy-Schwarz inequality to this sum. Let me set up two sequences:Let ( x_i = sqrt{a_i} ) and ( y_i = frac{sqrt{a_i}}{sum_{j neq i} a_j^2} ).Then, by Cauchy-Schwarz,left( sum_{i=1}^{n} x_i y_i right)^2 leq left( sum_{i=1}^{n} x_i^2 right) left( sum_{i=1}^{n} y_i^2 right).But ( x_i y_i = frac{a_i}{sum_{j neq i} a_j^2} ), which is exactly the term in my sum. So,left( sum_{i=1}^{n} frac{a_i}{sum_{j neq i} a_j^2} right)^2 leq left( sum_{i=1}^{n} a_i right) left( sum_{i=1}^{n} frac{a_i}{left( sum_{j neq i} a_j^2 right)^2} right).Hmm, but this seems to give me an upper bound, not a lower bound. I need a lower bound, so maybe this approach isn't helpful.Let me try another angle. Maybe I can use the AM-GM inequality. The AM-GM inequality states that for non-negative real numbers,frac{x_1 + x_2 + cdots + x_n}{n} geq sqrt[n]{x_1 x_2 cdots x_n}.But I'm not sure how to apply this directly to the given sum.Wait, perhaps I can consider the denominators. Each denominator is ( sum_{j neq i} a_j^2 ). Let me denote ( D_i = sum_{j neq i} a_j^2 ). Then, the sum becomes:sum_{i=1}^{n} frac{a_i}{D_i}.I need to relate this sum to ( S = sum_{i=1}^{n} a_i ).Maybe I can use the Cauchy-Schwarz inequality in the following way. Consider:left( sum_{i=1}^{n} frac{a_i}{D_i} right) left( sum_{i=1}^{n} a_i D_i right) geq left( sum_{i=1}^{n} a_i right)^2.This is another form of the Cauchy-Schwarz inequality, where I pair ( frac{a_i}{D_i} ) with ( a_i D_i ).So,left( sum_{i=1}^{n} frac{a_i}{D_i} right) left( sum_{i=1}^{n} a_i D_i right) geq S^2.Therefore,sum_{i=1}^{n} frac{a_i}{D_i} geq frac{S^2}{sum_{i=1}^{n} a_i D_i}.So, if I can bound ( sum_{i=1}^{n} a_i D_i ) from above, then I can get a lower bound for the original sum.Let me compute ( sum_{i=1}^{n} a_i D_i ):sum_{i=1}^{n} a_i D_i = sum_{i=1}^{n} a_i left( sum_{j neq i} a_j^2 right) = sum_{i=1}^{n} sum_{j neq i} a_i a_j^2.This is a double sum where each term is ( a_i a_j^2 ) with ( i neq j ).Let me try to rearrange this sum. Notice that for each ( j ), ( a_j^2 ) is multiplied by ( a_i ) for all ( i neq j ). So,sum_{i=1}^{n} sum_{j neq i} a_i a_j^2 = sum_{j=1}^{n} a_j^2 sum_{i neq j} a_i = sum_{j=1}^{n} a_j^2 (S - a_j).So,sum_{i=1}^{n} a_i D_i = sum_{j=1}^{n} a_j^2 (S - a_j) = S sum_{j=1}^{n} a_j^2 - sum_{j=1}^{n} a_j^3.Let me denote ( T = sum_{j=1}^{n} a_j^2 ) and ( U = sum_{j=1}^{n} a_j^3 ). So,sum_{i=1}^{n} a_i D_i = S T - U.Now, going back to the inequality,sum_{i=1}^{n} frac{a_i}{D_i} geq frac{S^2}{S T - U}.So, I need to show that:frac{S^2}{S T - U} geq frac{4}{S}.Simplifying this,frac{S^2}{S T - U} geq frac{4}{S} implies S^3 geq 4 (S T - U).So,S^3 geq 4 S T - 4 U.Let me rearrange this:S^3 - 4 S T + 4 U geq 0.Hmm, I need to show that this expression is non-negative. Let me see if I can relate ( S ), ( T ), and ( U ) in some way.I know that ( T = sum_{j=1}^{n} a_j^2 ) and ( U = sum_{j=1}^{n} a_j^3 ). Also, ( S = sum_{j=1}^{n} a_j ).Perhaps I can use the Power Mean inequality or some other inequality to relate these sums.Wait, another idea: Maybe I can use the Cauchy-Schwarz inequality again on ( T ) and ( U ).Recall that:left( sum_{j=1}^{n} a_j^3 right) left( sum_{j=1}^{n} a_j right) geq left( sum_{j=1}^{n} a_j^2 right)^2.This is because of the Cauchy-Schwarz inequality applied to the sequences ( (a_j^{3/2}) ) and ( (a_j^{1/2}) ).So,U S geq T^2.Therefore,U geq frac{T^2}{S}.Let me substitute this into the expression ( S^3 - 4 S T + 4 U ):S^3 - 4 S T + 4 U geq S^3 - 4 S T + 4 cdot frac{T^2}{S} = S^3 - 4 S T + frac{4 T^2}{S}.Let me factor this expression:S^3 - 4 S T + frac{4 T^2}{S} = frac{S^4 - 4 S^2 T + 4 T^2}{S} = frac{(S^2 - 2 T)^2}{S}.Since ( S ) is the sum of non-negative real numbers, ( S geq 0 ). Therefore,frac{(S^2 - 2 T)^2}{S} geq 0.Thus,S^3 - 4 S T + 4 U geq 0.Which implies,frac{S^2}{S T - U} geq frac{4}{S}.Therefore,sum_{i=1}^{n} frac{a_i}{D_i} geq frac{4}{S}.Which is exactly what we needed to prove.So, putting it all together, by applying the Cauchy-Schwarz inequality twice, once to relate the sum of fractions to another sum, and then using another application to relate the sums ( S ), ( T ), and ( U ), we were able to establish the desired inequality.

💡Okay, so I have this geometry problem here: In triangle ABC, the median AD makes an angle of 60 degrees with side BC. I need to find the value of |cot B - cot C|. Hmm, let me try to visualize this triangle first.Alright, triangle ABC. Let me label the vertices A, B, and C. AD is a median, so D must be the midpoint of BC. That means BD equals DC. Let me denote BD = DC = x. So, BC is 2x in length.Now, the median AD makes a 60-degree angle with BC. So, if I draw AD, it forms a 60-degree angle with BC. Let me think about how to model this. Maybe I can drop a perpendicular from A to BC, let's call the foot of this perpendicular P. So, AP is the altitude from A to BC.Since AD is a median, D is the midpoint, so BD = DC = x. Also, since P is the foot of the altitude, BP and PC can be different. Let me denote BP as y, so PC would be 2x - y.Now, in triangle ABP, we have a right triangle with legs AP and BP, and hypotenuse AB. Similarly, in triangle APC, another right triangle with legs AP and PC, and hypotenuse AC.Given that angle ADP is 60 degrees, since AD is making a 60-degree angle with BC. Wait, is that correct? Let me see. AD is the median, so AD connects A to D, which is the midpoint. So, angle between AD and BC is 60 degrees. So, at point D, the angle between AD and BC is 60 degrees. So, triangle ADP is a triangle with angle at D being 60 degrees.Wait, but P is the foot of the altitude from A to BC, so AP is perpendicular to BC. So, triangle APD is a right triangle at P, with angle at D being 60 degrees. So, in triangle APD, angle at D is 60 degrees, angle at P is 90 degrees, so angle at A must be 30 degrees.So, triangle APD is a 30-60-90 triangle. In such triangles, the sides are in the ratio 1 : √3 : 2. So, the side opposite 30 degrees is the shortest side, which would be AP, then PD is opposite 60 degrees, so it's AP * √3, and AD is the hypotenuse, which is twice AP.Wait, but in this case, AP is opposite the 60-degree angle, isn't it? Wait, no. Let me clarify. In triangle APD, angle at D is 60 degrees, angle at P is 90 degrees, so angle at A is 30 degrees. So, side opposite 30 degrees is PD, side opposite 60 degrees is AP, and side opposite 90 degrees is AD.So, PD is opposite 30 degrees, so PD = (AP) / √3. Wait, no, in a 30-60-90 triangle, the sides are 1 : √3 : 2, where 1 is opposite 30, √3 opposite 60, and 2 is the hypotenuse.So, in triangle APD, angle at D is 60 degrees, so side opposite is AP. So, AP = PD * √3. Wait, no, PD is adjacent to 60 degrees, so maybe I need to think in terms of adjacent and opposite sides.Wait, maybe I should assign variables to the sides. Let me denote AP as h, the height. Then, PD is adjacent to the 60-degree angle, so PD = h / tan(60°). Since tan(60°) is √3, PD = h / √3.But also, in triangle APD, PD is part of BC. Since D is the midpoint, BD = DC = x, so PD is the distance from P to D. So, PD = |BD - BP| or |BP - BD|, depending on where P is.Wait, BP is y, BD is x, so PD is |x - y|. So, PD = |x - y|. But from triangle APD, PD = h / √3. So, |x - y| = h / √3.Also, in triangle APD, AD is the hypotenuse. So, AD = 2 * (PD) because in a 30-60-90 triangle, the hypotenuse is twice the shorter side. Wait, PD is opposite 30 degrees, so PD is the shorter side, so AD = 2 * PD. So, AD = 2 * (h / √3) = 2h / √3.But I also know that AD is a median. There's a formula for the length of a median in terms of the sides of the triangle. The formula is:AD² = (2AB² + 2AC² - BC²) / 4But maybe that's too complicated. Alternatively, since I have coordinates, maybe I can assign coordinates to the triangle and solve it that way.Let me place point B at (0, 0), point C at (2x, 0), so that BC is along the x-axis. Then, point D, the midpoint, is at (x, 0). Point A is somewhere above the x-axis, at coordinates (p, q). Then, the median AD goes from A(p, q) to D(x, 0).Given that angle between AD and BC is 60 degrees. Since BC is along the x-axis, the angle between AD and the x-axis is 60 degrees. So, the slope of AD is tan(60°) = √3.So, the slope of AD is (0 - q) / (x - p) = -q / (x - p) = √3. So, -q / (x - p) = √3, which implies q = -√3 (x - p). So, q = √3 (p - x).So, point A is at (p, √3 (p - x)).Now, let's find the coordinates of P, the foot of the altitude from A to BC. Since BC is along the x-axis, the foot P will have the same x-coordinate as A, but y-coordinate 0. Wait, no, that's only if the altitude is vertical, but in this case, the altitude from A is perpendicular to BC, which is horizontal, so the altitude is vertical. So, yes, P is (p, 0).Wait, but in that case, PD is the distance from P(p, 0) to D(x, 0), which is |x - p|. So, PD = |x - p|.From earlier, we had PD = h / √3, where h is the height AP. Since AP is the vertical distance from A to BC, which is q. So, h = q = √3 (p - x). So, PD = |x - p| = |p - x| = |h / √3|.But h is positive, so PD = h / √3. Therefore, |x - p| = h / √3.But h = √3 (p - x), so substituting:|x - p| = (√3 (p - x)) / √3 = |p - x|.Which is consistent, since |x - p| = |p - x|. So, that checks out.Now, let's find the coordinates of A: (p, √3 (p - x)).Now, let's compute the lengths of AB and AC.Point B is at (0, 0), so AB is the distance from (p, √3 (p - x)) to (0, 0):AB² = p² + (√3 (p - x))² = p² + 3 (p - x)².Similarly, AC is the distance from (p, √3 (p - x)) to (2x, 0):AC² = (2x - p)² + (√3 (p - x))² = (2x - p)² + 3 (p - x)².Now, let's compute cot B and cot C.In triangle ABC, cot B is adjacent over opposite in triangle ABP. Wait, no, cot B is adjacent over opposite for angle B. In triangle ABC, angle B is at point B, so the sides adjacent and opposite are with respect to angle B.Wait, in triangle ABC, cot B = (length of adjacent side) / (length of opposite side). The adjacent side to angle B is AB, but actually, no. Wait, in triangle ABC, for angle B, the sides are:- Opposite side: AC- Adjacent sides: AB and BC.Wait, no, cotangent is adjacent over opposite in a right triangle, but in any triangle, cot B = (length of adjacent side) / (length of opposite side). Wait, maybe I should use the definition in terms of the triangle.Alternatively, cot B can be expressed using the sides of the triangle. In triangle ABC, cot B = (AB² + BC² - AC²) / (4 * area). Wait, that might be more complicated.Alternatively, since I have coordinates, maybe I can compute the slopes and use trigonometry.Wait, let's think about triangle ABP, which is a right triangle at P. So, in triangle ABP, angle at B is angle B of triangle ABC. So, in triangle ABP, cot B = adjacent / opposite = BP / AP.Similarly, in triangle APC, which is also a right triangle at P, cot C = adjacent / opposite = PC / AP.So, cot B = BP / AP = y / h, and cot C = PC / AP = (2x - y) / h.Wait, but earlier, I denoted BP as y, so PC is 2x - y.But wait, in my coordinate system, BP is the distance from B(0,0) to P(p,0), which is |p - 0| = p. So, BP = p, and PC is |2x - p|.But earlier, I had PD = |x - p|, which is |p - x|.Wait, but in the coordinate system, PD is |x - p|, which is the same as |p - x|.From earlier, PD = h / √3, and h = AP = q = √3 (p - x). So, PD = |p - x| = (√3 (p - x)) / √3 = |p - x|, which is consistent.So, now, in triangle ABP, cot B = BP / AP = p / h.Similarly, in triangle APC, cot C = PC / AP = (2x - p) / h.So, cot B = p / h, cot C = (2x - p) / h.Therefore, |cot B - cot C| = |(p / h) - ((2x - p)/h)| = |(p - (2x - p)) / h| = |(2p - 2x) / h| = |2(p - x)| / h.But from earlier, h = √3 (p - x). So, h = √3 (p - x). Therefore, |2(p - x)| / h = |2(p - x)| / (√3 (p - x)) ) = |2 / √3| = 2 / √3.Wait, but that's not matching the answer given earlier, which was 3. Hmm, maybe I made a mistake.Wait, let's go back. In the initial problem, the assistant's solution had |cot B - cot C| = 3, but in my calculation, I got 2 / √3. That's different. So, perhaps I made a mistake in the setup.Wait, in the coordinate system, I assumed that the foot of the altitude P is at (p, 0), but actually, in reality, P could be anywhere along BC, not necessarily at (p, 0). Wait, no, if I drop a perpendicular from A(p, q) to BC, which is along the x-axis, then the foot P must be (p, 0). So, that part is correct.Wait, but in the assistant's solution, they had AP = PD = (x + y)/2 * √3, but in my coordinate system, PD is |x - p|, and h = √3 (p - x). So, PD = |x - p| = |p - x| = h / √3.So, h = √3 PD.Wait, but in the assistant's solution, they had AP = PD = (x + y)/2 * √3, which seems different.Wait, maybe I need to express everything in terms of x and y, where y = BP, and x = BD = DC.So, in my coordinate system, BP = p, so y = p, and PC = 2x - p.So, PD = |x - p| = |x - y|.From triangle APD, which is a 30-60-90 triangle, we have PD = h / √3, and AD = 2h / √3.But also, AD is the median, so we can express AD in terms of the sides of the triangle.Alternatively, maybe I can use vectors or coordinate geometry to find the relationship between the sides.Wait, let me try another approach. Let me denote the length of BC as 2x, so BD = DC = x.Let me denote the length of AP as h, the height from A to BC.Since AD makes a 60-degree angle with BC, and D is the midpoint, then in triangle APD, which is a right triangle at P, we have angle at D is 60 degrees.So, in triangle APD, angle at D is 60 degrees, angle at P is 90 degrees, so angle at A is 30 degrees.Therefore, in triangle APD, sides are in the ratio 1 : √3 : 2.So, PD is opposite 30 degrees, so PD = h / √3.AP is opposite 60 degrees, so AP = h.Wait, no, wait. In triangle APD, angle at D is 60 degrees, so side opposite is AP, which is h. Side opposite 30 degrees is PD, which is h / √3. Hypotenuse AD is 2 * (h / √3) = 2h / √3.So, PD = h / √3.But PD is the distance from P to D along BC, which is |x - y|, where y is BP.So, |x - y| = h / √3.Also, the length of AD is 2h / √3.But AD is also a median, so using the formula for the length of a median:AD² = (2AB² + 2AC² - BC²) / 4.But maybe that's too involved. Alternatively, since I have coordinates, let me express AD in terms of coordinates.Point A is at (p, h), since AP is the height h, and P is at (p, 0). So, point A is (p, h).Point D is at (x, 0).So, vector AD is from (p, h) to (x, 0), which is (x - p, -h).The length of AD is sqrt[(x - p)² + h²] = 2h / √3.So, (x - p)² + h² = (4h²) / 3.Therefore, (x - p)² = (4h² / 3) - h² = (4h² - 3h²) / 3 = h² / 3.So, (x - p)² = h² / 3, so |x - p| = h / √3, which matches our earlier result.So, PD = |x - p| = h / √3.Now, let's find cot B and cot C.In triangle ABP, which is right-angled at P, cot B = adjacent / opposite = BP / AP = y / h.Similarly, in triangle APC, right-angled at P, cot C = adjacent / opposite = PC / AP = (2x - y) / h.So, cot B = y / h, cot C = (2x - y) / h.Therefore, |cot B - cot C| = |(y / h) - ((2x - y)/h)| = |(y - 2x + y)/h| = |(2y - 2x)/h| = |2(y - x)| / h.But from earlier, |x - y| = h / √3, so |y - x| = h / √3.Therefore, |2(y - x)| / h = 2 * (h / √3) / h = 2 / √3.Wait, but the assistant's solution had |cot B - cot C| = 3, but I'm getting 2 / √3. So, there's a discrepancy here.Wait, maybe I made a mistake in the sign. Let me check.In the assistant's solution, they had cot B = -y / (x + y), and cot C = (2x + y) / (x + y). Then, |cot B - cot C| = | - (y + 2x + y) / (x + y) | = | - (2x + 2y) / (x + y) | = 2.Wait, no, that doesn't seem right. Wait, let me look back.Wait, the assistant's solution had:cot B = -y / (x + y),cot C = (2x + y) / (x + y),so |cot B - cot C| = | -y/(x+y) - (2x + y)/(x+y) | = | (-y - 2x - y) / (x + y) | = | (-2x - 2y) / (x + y) | = | -2(x + y) / (x + y) | = 2.Wait, but in the assistant's solution, they had 3, but according to this, it's 2. Hmm, perhaps the assistant made a mistake.Wait, but in my calculation, I got 2 / √3, which is approximately 1.1547, but the assistant's solution had 3.Wait, maybe I need to re-examine the problem.Wait, the problem states that median AD makes an angle of 60 degrees with BC. So, in my coordinate system, AD makes a 60-degree angle with BC, which is along the x-axis. So, the slope of AD is tan(60°) = √3, so the slope is √3.But in my coordinate system, point A is at (p, h), and point D is at (x, 0). So, the slope of AD is (0 - h) / (x - p) = -h / (x - p) = √3.So, -h / (x - p) = √3 => h = -√3 (x - p) => h = √3 (p - x).So, h = √3 (p - x).But PD is |x - p| = |p - x| = h / √3.So, PD = h / √3.Now, let's express y, which is BP = p.So, y = p.And PC = 2x - y = 2x - p.So, cot B = y / h = p / h.cot C = (2x - y) / h = (2x - p) / h.So, |cot B - cot C| = |(p / h) - ((2x - p)/h)| = |(p - 2x + p)/h| = |(2p - 2x)/h| = |2(p - x)| / h.But h = √3 (p - x), so |2(p - x)| / h = |2(p - x)| / (√3 (p - x)) ) = 2 / √3.So, |cot B - cot C| = 2 / √3.But the assistant's solution had 3. So, perhaps the assistant made a mistake in their calculation.Wait, let me check the assistant's solution again.They said:Let P be the foot of the altitude from A to BC. Let x = BD = CD, and let y = BP. Since angle ADP = 60 degrees, using the property that in a 30-60-90 triangle the ratio of the lengths of the sides opposite the 30°, 60°, and 90° angles are 1:√3:2, and the relation AP = PD = x + y, we can derive:AP = PD = (x + y)/2 * √3Wait, that seems incorrect. Because in a 30-60-90 triangle, the sides are 1 : √3 : 2, where 1 is opposite 30°, √3 opposite 60°, and 2 is the hypotenuse.In triangle APD, angle at D is 60°, angle at P is 90°, so angle at A is 30°. So, side opposite 30° is PD, side opposite 60° is AP, and hypotenuse is AD.So, PD = (AP) / √3, and AD = 2 * PD = 2 * (AP / √3) = 2AP / √3.But the assistant said AP = PD = (x + y)/2 * √3, which seems incorrect because AP is opposite 60°, so AP = PD * √3.Wait, PD is opposite 30°, so PD = AP / √3, so AP = PD * √3.So, the assistant's equation AP = PD = (x + y)/2 * √3 is incorrect. It should be AP = PD * √3.So, AP = PD * √3.But PD is |x - y|, so AP = |x - y| * √3.But in the assistant's solution, they set AP = PD = (x + y)/2 * √3, which is different.So, perhaps the assistant made a mistake in their setup.In my calculation, I got |cot B - cot C| = 2 / √3, which is approximately 1.1547, but the assistant's solution had 3.Wait, maybe I need to consider the direction of the angle. If the angle between AD and BC is 60 degrees, but depending on the position of P, the angle could be on the other side, making PD = x + y instead of |x - y|.Wait, in my coordinate system, if P is to the left of D, then PD = x - y, but if P is to the right of D, PD = y - x.But in the assistant's solution, they assumed AP = PD = (x + y)/2 * √3, which suggests that PD = x + y, which would only be possible if P is beyond D, making PD = x + y, but that would require y > x, which might not always be the case.Wait, but in reality, P is the foot of the altitude, so it can be anywhere along BC, depending on the triangle. So, if P is beyond D, then PD = y - x, but if P is before D, PD = x - y.But in the assistant's solution, they set AP = PD = (x + y)/2 * √3, which seems to assume that PD = x + y, which would only be possible if P is beyond D, making y > x.But in that case, PD = y - x, not x + y.Wait, perhaps the assistant made a mistake in the sign or the relationship between PD and AP.Alternatively, maybe the assistant considered PD as the length from P to D, regardless of direction, so PD = |x - y|, but then they set AP = PD * √3, which would be correct.But in their solution, they wrote AP = PD = (x + y)/2 * √3, which is incorrect because AP = PD * √3, not AP = PD.So, perhaps the assistant made a mistake in their calculation, leading to the incorrect result of 3.In my calculation, I got |cot B - cot C| = 2 / √3, which simplifies to 2√3 / 3.But let me check if that's correct.Wait, let's consider a specific case where x = 1, so BC = 2, and let's set y such that PD = h / √3.Let me choose x = 1, so BD = DC = 1, BC = 2.Let me choose y = BP = 1, so P is at D. Then, PD = 0, which would mean h = 0, which is not possible. So, y cannot be 1.Let me choose y = 0.5, so BP = 0.5, PC = 1.5.Then, PD = |1 - 0.5| = 0.5.So, h = PD * √3 = 0.5 * √3 ≈ 0.866.Now, cot B = y / h = 0.5 / 0.866 ≈ 0.577.cot C = PC / h = 1.5 / 0.866 ≈ 1.732.So, |cot B - cot C| ≈ |0.577 - 1.732| ≈ 1.155, which is approximately 2 / √3 ≈ 1.1547.So, that seems consistent.Alternatively, if I choose y = 1.5, then PD = |1 - 1.5| = 0.5, h = 0.5 * √3 ≈ 0.866.Then, cot B = y / h = 1.5 / 0.866 ≈ 1.732.cot C = PC / h = (2 - 1.5) / 0.866 ≈ 0.577.So, |cot B - cot C| ≈ |1.732 - 0.577| ≈ 1.155, same as before.So, in both cases, |cot B - cot C| ≈ 2 / √3.Therefore, my calculation seems correct, and the assistant's solution had an error.So, the correct answer should be 2 / √3, which can be rationalized as 2√3 / 3.But wait, the problem asks for |cot B - cot C|, so maybe I need to rationalize it.2 / √3 = (2√3) / 3.So, the value is 2√3 / 3.But wait, in the assistant's solution, they had 3, which is different.Wait, maybe I need to consider the direction of the angle. If the angle between AD and BC is 60 degrees, but depending on the position of A, the angle could be on the other side, making PD = x + y instead of |x - y|.Wait, but in my coordinate system, PD is |x - y|, because P is between B and C.Wait, unless P is beyond C, making PD = x + y, but that would require y > 2x, which is not possible because y is BP, and BC is 2x, so y cannot exceed 2x.Wait, no, if P is beyond C, then PC would be negative, but in reality, P is between B and C, so y is between 0 and 2x.Wait, but in that case, PD is |x - y|, which is always positive.So, I think my calculation is correct, and the assistant's solution had a mistake.Therefore, the correct value of |cot B - cot C| is 2 / √3, which is 2√3 / 3.But wait, let me check another approach.Let me use the formula for cotangent in terms of the sides of the triangle.In triangle ABC, cot B = (AB² + BC² - AC²) / (4 * area).Similarly, cot C = (AC² + BC² - AB²) / (4 * area).So, |cot B - cot C| = |(AB² + BC² - AC² - AC² - BC² + AB²) / (4 * area)| = |(2AB² - 2AC²) / (4 * area)| = |(AB² - AC²) / (2 * area)|.But I'm not sure if that helps directly.Alternatively, since I have the coordinates, I can compute AB and AC.From earlier, AB² = p² + 3(p - x)²,AC² = (2x - p)² + 3(p - x)².So, AB² - AC² = p² + 3(p - x)² - [(2x - p)² + 3(p - x)²] = p² - (4x² - 4xp + p²) = p² - 4x² + 4xp - p² = 4xp - 4x² = 4x(p - x).So, AB² - AC² = 4x(p - x).Now, the area of triangle ABC is (base * height)/2 = (2x * h)/2 = xh.So, |cot B - cot C| = |(AB² - AC²) / (2 * area)| = |4x(p - x) / (2xh)| = |2(p - x)/h|.But from earlier, h = √3 (p - x), so |2(p - x)/h| = |2(p - x)| / (√3 (p - x)) ) = 2 / √3.So, same result.Therefore, the correct answer is 2 / √3, which is 2√3 / 3.But the assistant's solution had 3, which is incorrect.So, I think the correct answer is 2√3 / 3.But wait, let me check if I made a mistake in the sign.In the assistant's solution, they had cot B = -y / (x + y), which suggests that they considered y as a negative length, but in reality, y is a positive length.Wait, in my coordinate system, y = BP = p, which is positive, so cot B = y / h = p / h.But in the assistant's solution, they had cot B = -y / (x + y), which would imply that they considered y as negative, but that doesn't make sense because lengths are positive.So, perhaps the assistant made a mistake in the sign, leading to an incorrect result.Therefore, I think the correct answer is 2√3 / 3.But wait, let me check another way.Let me consider the triangle with specific values.Let me set x = 1, so BC = 2.Let me set y = BP = 1 - t, so PC = 1 + t, where t is some value.Then, PD = |1 - (1 - t)| = |t| = t.So, PD = t.From triangle APD, which is 30-60-90, we have AP = PD * √3 = t√3.So, h = t√3.Now, cot B = BP / AP = (1 - t) / (t√3).cot C = PC / AP = (1 + t) / (t√3).So, |cot B - cot C| = |(1 - t)/(t√3) - (1 + t)/(t√3)| = |(1 - t - 1 - t)/(t√3)| = |(-2t)/(t√3)| = | -2 / √3 | = 2 / √3.So, same result.Therefore, regardless of the value of t, |cot B - cot C| is always 2 / √3.So, the correct answer is 2 / √3, which is 2√3 / 3.Therefore, the assistant's solution was incorrect, and the correct answer is 2√3 / 3.But wait, the problem asks for |cot B - cot C|, and in my calculation, it's 2 / √3, which is approximately 1.1547, but the assistant's solution had 3.Wait, perhaps I need to consider the absolute value differently.Wait, in my calculation, |cot B - cot C| = 2 / √3, but in the assistant's solution, they had | - (3x + 2y) / (x + y) | = 3.Wait, perhaps the assistant made a mistake in their algebra.In the assistant's solution, they had:cot B = -y / (x + y),cot C = (2x + y) / (x + y),so |cot B - cot C| = | -y/(x+y) - (2x + y)/(x+y) | = | (-y - 2x - y) / (x + y) | = | (-2x - 2y) / (x + y) | = | -2(x + y) / (x + y) | = 2.Wait, but in their solution, they had 3, but according to this, it's 2.So, perhaps the assistant made a mistake in their algebra.Wait, let me check their steps.They said:cot B = -y / (x + y),cot C = (2x + y) / (x + y),so |cot B - cot C| = | -y/(x+y) - (2x + y)/(x+y) | = | (-y - 2x - y) / (x + y) | = | (-2x - 2y) / (x + y) | = | -2(x + y) / (x + y) | = 2.But in their solution, they wrote:|cot B - cot C| = | - (3x + 2y) / (x + y) | = 3.Wait, that seems incorrect. They must have made a mistake in their algebra.So, in conclusion, the correct answer is 2 / √3, which is 2√3 / 3.But wait, let me check if the problem specifies the angle between AD and BC as 60 degrees, but depending on the orientation, it could be 60 degrees or 120 degrees.Wait, no, the problem says 60 degrees, so it's acute.Therefore, the correct answer is 2√3 / 3.But wait, in the assistant's solution, they had 3, which is different.Wait, perhaps I need to consider that in their solution, they had AP = PD = (x + y)/2 * √3, which is incorrect, leading to their wrong result.So, in conclusion, the correct value of |cot B - cot C| is 2√3 / 3.But wait, let me check the problem again.Wait, the problem says "median AD makes an angle of 60 degrees with side BC".So, in my coordinate system, that's the angle between AD and BC is 60 degrees, which is consistent with the slope of AD being √3.Therefore, my calculation is correct, and the answer is 2√3 / 3.But wait, let me check if the problem is in degrees or radians, but it's in degrees, so 60 degrees is correct.Therefore, I think the correct answer is 2√3 / 3.But wait, in the assistant's solution, they had 3, which is different.Wait, perhaps I made a mistake in the relationship between PD and AP.Wait, in triangle APD, which is 30-60-90, PD is opposite 30 degrees, so PD = AP / √3.Therefore, AP = PD * √3.But in my calculation, I had AP = h = √3 (p - x), and PD = |x - p| = h / √3.So, that's correct.Therefore, I think my answer is correct, and the assistant's solution had an error.Therefore, the value of |cot B - cot C| is 2√3 / 3.

💡Alright, so I've got this problem here: "x > 2" is a (　　) condition for "(x-1)^2 > 1". The options are A, B, C, D, which correspond to different types of conditions. I need to figure out whether "x > 2" is a sufficient, necessary, both, or neither condition for "(x-1)^2 > 1".First, let me recall what sufficient and necessary conditions mean. A sufficient condition means that if it's met, then the statement is true, but it doesn't necessarily mean it's the only way for the statement to be true. On the other hand, a necessary condition means that for the statement to be true, the condition must be met, but meeting the condition doesn't guarantee the statement is true.So, in this case, I need to see if "x > 2" is sufficient, necessary, both, or neither for "(x-1)^2 > 1".Let me start by solving the inequality "(x-1)^2 > 1". To do that, I can take the square root of both sides, but I have to remember that when dealing with inequalities involving squares, I have to consider both the positive and negative roots.So, if "(x-1)^2 > 1", then either (x - 1) > 1 or (x - 1) < -1. Let me write that down:1. (x - 1) > 1 Adding 1 to both sides gives: x > 22. (x - 1) < -1 Adding 1 to both sides gives: x < 0So, the solution to the inequality "(x-1)^2 > 1" is x > 2 or x < 0. That means that for "(x-1)^2 > 1" to be true, x must either be greater than 2 or less than 0.Now, the condition given is "x > 2". So, if x is greater than 2, then certainly "(x-1)^2 > 1" is true because x > 2 is one of the solutions we found. Therefore, "x > 2" is a sufficient condition for "(x-1)^2 > 1".But is it a necessary condition? Well, a necessary condition would mean that "(x-1)^2 > 1" can only be true if "x > 2" is true. However, from our earlier solution, we saw that x can also be less than 0 for the inequality to hold. So, there are values of x (specifically, x < 0) where "(x-1)^2 > 1" is true, but "x > 2" is not. Therefore, "x > 2" is not a necessary condition because the inequality can be true without "x > 2" being true.So, putting it all together, "x > 2" is a sufficient condition because it guarantees that "(x-1)^2 > 1" is true, but it's not necessary because there are other values of x that also make the inequality true.Let me double-check my reasoning. If I take x = 3, which is greater than 2, then (3 - 1)^2 = 4 > 1, which is true. If I take x = -1, which is less than 0, then (-1 - 1)^2 = 4 > 1, which is also true. So, both x > 2 and x < 0 satisfy the inequality, but only x > 2 is given as the condition. Therefore, "x > 2" is sufficient because it ensures the inequality holds, but it's not necessary because the inequality can hold without x being greater than 2.I think that makes sense. So, the answer should be that "x > 2" is a sufficient but not necessary condition for "(x-1)^2 > 1".

💡Alright, so I've got this problem here with two vectors, a and b. They're defined in terms of cosine and sine functions with different arguments. Vector a is (cos(3x/2), sin(3x/2)) and vector b is (cos(x/2), -sin(x/2)). The variable x is in the interval [0, π/2]. The problem has two parts: first, to find the dot product of a and b, and the magnitude of the sum of a and b. Second, to find the value of λ such that the minimum value of the function f(x) = a·b - 2λ|a + b| is -3/2.Okay, let's start with part 1. The dot product of two vectors is calculated by multiplying their corresponding components and then adding those products together. So, for vectors a and b, the dot product a·b should be:a·b = cos(3x/2) * cos(x/2) + sin(3x/2) * (-sin(x/2)).Wait, that's right. So, it's cos(3x/2)cos(x/2) minus sin(3x/2)sin(x/2). Hmm, that looks familiar. Isn't that the formula for cosine of a sum? Yes, cos(A + B) = cos A cos B - sin A sin B. So, if I let A = 3x/2 and B = x/2, then A + B = 2x. So, the dot product a·b is cos(2x). That simplifies things a lot.Alright, so a·b = cos(2x). Got that.Now, the next part is to find the magnitude of a + b. So, first, let's find the vector a + b. Vector a is (cos(3x/2), sin(3x/2)) and vector b is (cos(x/2), -sin(x/2)). So, adding them component-wise:a + b = (cos(3x/2) + cos(x/2), sin(3x/2) - sin(x/2)).To find the magnitude |a + b|, we need to compute the square root of the sum of the squares of the components. So:|a + b| = sqrt[(cos(3x/2) + cos(x/2))^2 + (sin(3x/2) - sin(x/2))^2].Hmm, that looks a bit complicated. Maybe we can simplify it using some trigonometric identities. Let's expand the squares:First component squared: [cos(3x/2) + cos(x/2)]^2 = cos²(3x/2) + 2cos(3x/2)cos(x/2) + cos²(x/2).Second component squared: [sin(3x/2) - sin(x/2)]^2 = sin²(3x/2) - 2sin(3x/2)sin(x/2) + sin²(x/2).Adding these together:cos²(3x/2) + sin²(3x/2) + cos²(x/2) + sin²(x/2) + 2cos(3x/2)cos(x/2) - 2sin(3x/2)sin(x/2).Now, cos²θ + sin²θ = 1 for any θ, so this simplifies to:1 + 1 + 2[cos(3x/2)cos(x/2) - sin(3x/2)sin(x/2)].Which is 2 + 2[cos(3x/2 + x/2)] because cos(A + B) = cos A cos B - sin A sin B. So, 3x/2 + x/2 = 2x. Therefore, this becomes:2 + 2cos(2x).So, |a + b| = sqrt[2 + 2cos(2x)].We can factor out a 2 from inside the square root:sqrt[2(1 + cos(2x))].And 1 + cos(2x) is equal to 2cos²x, so substituting that in:sqrt[2 * 2cos²x] = sqrt[4cos²x] = 2|cosx|.Since x is in [0, π/2], cosx is non-negative, so |cosx| = cosx. Therefore, |a + b| = 2cosx.Alright, so part 1 is done. The dot product a·b is cos(2x), and the magnitude |a + b| is 2cosx.Moving on to part 2. We have the function f(x) = a·b - 2λ|a + b|. From part 1, we know that a·b = cos(2x) and |a + b| = 2cosx. So, substituting these into f(x):f(x) = cos(2x) - 2λ * 2cosx = cos(2x) - 4λcosx.We need to find the value of λ such that the minimum value of f(x) is -3/2.First, let's express f(x) in terms of cosx. We know that cos(2x) can be written as 2cos²x - 1. So:f(x) = 2cos²x - 1 - 4λcosx.Let me rewrite that:f(x) = 2cos²x - 4λcosx - 1.This is a quadratic function in terms of cosx. Let's let t = cosx. Since x is in [0, π/2], t ranges from 0 to 1. So, f(x) can be expressed as:f(t) = 2t² - 4λt - 1.We need to find the minimum value of this quadratic function on the interval t ∈ [0, 1], and set that minimum equal to -3/2, then solve for λ.Quadratic functions have their extrema either at the vertex or at the endpoints of the interval. The vertex of a quadratic function f(t) = at² + bt + c is at t = -b/(2a). In our case, a = 2, b = -4λ, so the vertex is at t = -(-4λ)/(2*2) = 4λ/4 = λ.So, the vertex is at t = λ. Now, we need to consider where this vertex lies relative to our interval [0, 1].Case 1: If λ < 0. Then the vertex is at t = λ < 0, which is outside our interval. So, the minimum would occur at the left endpoint t = 0.Compute f(0) = 2*(0)^2 - 4λ*(0) - 1 = -1.But the problem states that the minimum value is -3/2, which is less than -1. So, this case is not possible because the minimum cannot be less than -1 if the vertex is outside the interval on the left.Case 2: If 0 ≤ λ ≤ 1. Then the vertex t = λ is within our interval [0, 1]. So, the minimum occurs at t = λ.Compute f(λ) = 2λ² - 4λ*λ - 1 = 2λ² - 4λ² - 1 = -2λ² - 1.We are told that the minimum value is -3/2, so:-2λ² - 1 = -3/2.Let's solve for λ:-2λ² - 1 = -3/2Add 1 to both sides:-2λ² = -3/2 + 1 = -1/2Multiply both sides by -1:2λ² = 1/2Divide both sides by 2:λ² = 1/4Take square roots:λ = ±1/2But since we are in the case where 0 ≤ λ ≤ 1, λ must be positive. So, λ = 1/2.Case 3: If λ > 1. Then the vertex is at t = λ > 1, outside our interval on the right. So, the minimum occurs at the right endpoint t = 1.Compute f(1) = 2*(1)^2 - 4λ*(1) - 1 = 2 - 4λ - 1 = 1 - 4λ.Set this equal to -3/2:1 - 4λ = -3/2Subtract 1 from both sides:-4λ = -5/2Divide both sides by -4:λ = (-5/2)/(-4) = 5/8.But in this case, we assumed λ > 1, but 5/8 is less than 1, which contradicts our assumption. Therefore, this case is not possible.So, the only valid solution is from Case 2, where λ = 1/2.Let me double-check to make sure I didn't make any mistakes.Starting with f(t) = 2t² - 4λt - 1. The vertex is at t = λ. If λ is within [0,1], then f(λ) = -2λ² - 1. Setting this equal to -3/2 gives λ = 1/2, which is indeed within [0,1]. So, that seems correct.If λ were outside [0,1], the minimum would be at the endpoints, but in those cases, we either get a minimum of -1 or 1 - 4λ. Since 1 - 4λ would require λ = 5/8, which is less than 1, it doesn't fit the case where λ > 1. So, the only valid solution is λ = 1/2.Alright, I think that makes sense. So, the value of λ is 1/2.**Final Answer**The value of λ is boxed{dfrac{1}{2}}.