💡Okay, so I have this problem where I'm given two sets, A and B. Set A is {2, 0, 1, 7}, and set B is defined as {x | x² - 2 ∈ A, x - 2 ∉ A}. I need to find the product of all elements in set B, and the options are 36, 54, 72, or 108. Hmm, let me try to break this down step by step.First, let's understand what set B is. It's the set of all x such that when you take x squared and subtract 2, the result is in set A. Additionally, when you take x and subtract 2, that result should not be in set A. So, I need to find all x that satisfy both conditions.Let me start by figuring out what x could be. Since x² - 2 has to be in A, which is {2, 0, 1, 7}, I can set up equations for each element in A and solve for x.1. If x² - 2 = 2, then x² = 4, so x = ±2.2. If x² - 2 = 0, then x² = 2, so x = ±√2.3. If x² - 2 = 1, then x² = 3, so x = ±√3.4. If x² - 2 = 7, then x² = 9, so x = ±3.So, the possible x values are ±2, ±√2, ±√3, and ±3. That gives me a total of 8 possible x values. Now, I need to check which of these x values satisfy the second condition: x - 2 ∉ A.Let's go through each x:1. x = 2: x - 2 = 0, which is in A. So, we exclude 2.2. x = -2: x - 2 = -4, which is not in A. So, we include -2.3. x = √2: x - 2 ≈ -0.585, which is not in A. So, include √2.4. x = -√2: x - 2 ≈ -3.414, which is not in A. So, include -√2.5. x = √3: x - 2 ≈ -0.267, which is not in A. So, include √3.6. x = -√3: x - 2 ≈ -3.732, which is not in A. So, include -√3.7. x = 3: x - 2 = 1, which is in A. So, exclude 3.8. x = -3: x - 2 = -5, which is not in A. So, include -3.So, after excluding 2 and 3, the elements of set B are: -2, √2, -√2, √3, -√3, and -3. That's six elements in total.Now, I need to find the product of all these elements. Let's write them out:Product = (-2) × (√2) × (-√2) × (√3) × (-√3) × (-3)Hmm, that's a bit of a mouthful. Let me simplify this step by step.First, let's pair the terms to make it easier:- (-2) and (-3)- (√2) and (-√2)- (√3) and (-√3)Calculating each pair:1. (-2) × (-3) = 62. (√2) × (-√2) = - (√2 × √2) = -23. (√3) × (-√3) = - (√3 × √3) = -3Now, multiply these results together:6 × (-2) × (-3)First, 6 × (-2) = -12Then, -12 × (-3) = 36So, the product of all elements in set B is 36.Wait, let me double-check to make sure I didn't make a mistake. Sometimes with multiple negative signs, it's easy to mix up the signs.Starting over:Product = (-2) × (√2) × (-√2) × (√3) × (-√3) × (-3)Let me rearrange the terms to group the negatives and the radicals:= [(-2) × (-√2) × (-√3) × (-3)] × [√2 × √3]Wait, that might complicate things. Alternatively, let's count the number of negative signs.In the original product:- (-2) is one negative- (√2) is positive- (-√2) is another negative- (√3) is positive- (-√3) is another negative- (-3) is another negativeSo, total of four negative signs. Since negative times negative is positive, and four negatives would result in a positive product.Now, let's compute the absolute values:|(-2)| = 2|√2| = √2|(-√2)| = √2|√3| = √3|(-√3)| = √3|(-3)| = 3So, the product of absolute values is:2 × √2 × √2 × √3 × √3 × 3Simplify:√2 × √2 = 2√3 × √3 = 3So, now we have:2 × 2 × 3 × 3Which is 2×2=4, 3×3=9, then 4×9=36So, the product is indeed 36, and since we have an even number of negative signs, the result is positive.Therefore, the product of all elements in set B is 36.**Final Answer**boxed{A}

💡Okay, so I have this problem about a right-angled triangle where the angle bisector of an acute angle divides the opposite side into segments of length 1 and 2. I need to find the length of the bisector. Hmm, let me try to visualize this.First, let me sketch a right-angled triangle. Let's say the right angle is at point B, and the other two vertices are A and C. So, angle B is 90 degrees. Now, the angle bisector of an acute angle—probably one of the angles at A or C—divides the opposite side into segments of 1 and 2. I think it's the angle at A because it's more common to label the triangle that way, but I'm not sure. Maybe I should consider both possibilities.Wait, actually, in a right-angled triangle, the two acute angles are at A and C. So, the angle bisector could be from either A or C. But since the problem doesn't specify, maybe it doesn't matter? Or maybe it does. I need to figure out which angle's bisector divides the opposite side into segments of 1 and 2.Let me assume that the angle bisector is from angle A. So, angle A is being bisected, and it intersects the opposite side BC at some point, say D. Then, BD is 1 and DC is 2, or vice versa. I need to determine which is which.Wait, the problem says the segments are of length 1 and 2, but it doesn't specify the order. So, perhaps I need to consider both cases? Or maybe the triangle's properties will help me figure it out.I remember there's a theorem called the Angle Bisector Theorem which states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, if I have triangle ABC with angle bisector from A to D on BC, then BD/DC = AB/AC.Since it's a right-angled triangle at B, AB and AC are the legs, and BC is the hypotenuse. Wait, no, actually, in a right-angled triangle, the sides opposite the angles are the legs and the hypotenuse. So, if angle B is 90 degrees, then AB and BC are the legs, and AC is the hypotenuse. Wait, no, that's not right. Let me correct that.In triangle ABC, with right angle at B, sides AB and BC are the legs, and AC is the hypotenuse. So, AB and BC are the legs, and AC is the hypotenuse. So, if the angle bisector from A meets BC at D, then BD/DC = AB/AC.Given that BD and DC are 1 and 2, so BD/DC = 1/2 or 2/1. So, AB/AC is either 1/2 or 2/1.But AB is one of the legs, and AC is the hypotenuse. So, AB must be shorter than AC because the hypotenuse is the longest side. Therefore, AB/AC must be less than 1. So, BD/DC must be 1/2. Therefore, BD is 1 and DC is 2.Okay, so BD = 1 and DC = 2. Therefore, BC = BD + DC = 3.Now, I can use the Angle Bisector Theorem to relate the sides. So, BD/DC = AB/AC = 1/2. So, AB = (1/2) AC.Since ABC is a right-angled triangle at B, by Pythagoras' theorem, AB² + BC² = AC².We know BC = 3, AB = (1/2) AC. Let me denote AC as x. Then AB = x/2.So, substituting into Pythagoras' theorem:(x/2)² + 3² = x²Simplify:x²/4 + 9 = x²Multiply both sides by 4 to eliminate the denominator:x² + 36 = 4x²Subtract x² from both sides:36 = 3x²Divide both sides by 3:12 = x²So, x = sqrt(12) = 2*sqrt(3)Therefore, AC = 2*sqrt(3), and AB = x/2 = sqrt(3)So, AB = sqrt(3), BC = 3, AC = 2*sqrt(3)Now, I need to find the length of the angle bisector from A to D on BC. There's a formula for the length of an angle bisector in a triangle. I think it's something like:AD = (2ab cos(theta/2)) / (a + b)Where a and b are the lengths of the sides adjacent to the angle being bisected, and theta is the measure of that angle.But in this case, angle A is being bisected, and the sides adjacent to angle A are AB and AC. Wait, no, in triangle ABC, angle A is between sides AB and AC. Wait, no, actually, in triangle ABC, angle A is between sides AB and AC? Wait, no, in triangle ABC, angle A is between sides AB and AC? Wait, no, actually, in triangle ABC, angle A is between sides AB and AC? Wait, no, that's not correct.Wait, in triangle ABC, angle A is at vertex A, so the sides adjacent to angle A are AB and AC. Wait, no, AB is adjacent to angle A, and AC is the hypotenuse. Wait, I'm getting confused.Wait, in triangle ABC, with right angle at B, the sides are AB, BC, and AC. So, angle A is at vertex A, between sides AB and AC. Wait, no, actually, in triangle ABC, angle A is between sides AB and AC? Wait, no, in triangle ABC, angle A is between sides AB and AC? Wait, no, that's not correct because AC is the hypotenuse.Wait, maybe I should think in terms of coordinates to make it clearer.Let me place the triangle on a coordinate system with point B at the origin (0,0), point C on the x-axis at (3,0), and point A somewhere in the plane. Since AB is sqrt(3) and AC is 2*sqrt(3), let me find the coordinates of A.Wait, if AB = sqrt(3) and BC = 3, then point A must be at (0, sqrt(3)) because AB is the vertical side from B(0,0) to A(0, sqrt(3)), and AC is the hypotenuse from A(0, sqrt(3)) to C(3,0).So, coordinates:B: (0,0)A: (0, sqrt(3))C: (3,0)Now, the angle bisector from A to BC meets BC at D, which divides BC into BD = 1 and DC = 2. So, since BC is from (0,0) to (3,0), point D is at (1,0) because BD = 1.Wait, no, if BD = 1, then D is at (1,0). But wait, BC is from B(0,0) to C(3,0), so BD = 1 would mean D is at (1,0), and DC = 2, which is from (1,0) to (3,0). That makes sense.So, point D is at (1,0). Now, I need to find the length of AD, which is from A(0, sqrt(3)) to D(1,0).Using the distance formula:AD = sqrt[(1 - 0)^2 + (0 - sqrt(3))^2] = sqrt[1 + 3] = sqrt[4] = 2.Wait, so AD is 2 units long. So, the length of the bisector is 2.But wait, let me make sure I didn't make a mistake. I used coordinates to find the length, but maybe I should verify using the angle bisector formula.The formula for the length of an angle bisector in a triangle is:AD = (2ab cos(theta/2)) / (a + b)Where a and b are the lengths of the sides adjacent to angle A, which are AB and AC.Wait, AB is sqrt(3), AC is 2*sqrt(3), and angle A is the angle between AB and AC.But in a right-angled triangle, angle A is one of the acute angles. Let me find angle A.In triangle ABC, tan(angle A) = opposite/adjacent = BC/AB = 3 / sqrt(3) = sqrt(3). So, angle A is 60 degrees because tan(60°) = sqrt(3).Therefore, angle A is 60 degrees, so theta/2 is 30 degrees.So, cos(theta/2) = cos(30°) = sqrt(3)/2.Now, plug into the formula:AD = (2 * AB * AC * cos(theta/2)) / (AB + AC)= (2 * sqrt(3) * 2*sqrt(3) * sqrt(3)/2) / (sqrt(3) + 2*sqrt(3))Simplify numerator:2 * sqrt(3) * 2*sqrt(3) = 4 * 3 = 1212 * sqrt(3)/2 = 6*sqrt(3)Denominator:sqrt(3) + 2*sqrt(3) = 3*sqrt(3)So, AD = (6*sqrt(3)) / (3*sqrt(3)) = 2So, same result as before. Therefore, the length of the bisector is 2.Wait, but the options are sqrt(2), sqrt(3), sqrt(4), sqrt(5), sqrt(6). So, sqrt(4) is 2, which is option C.But wait, in my coordinate system, I found AD = 2, which is sqrt(4). So, the answer should be C) sqrt(4).But let me double-check because sometimes I might have messed up the sides.Wait, in the Angle Bisector Theorem, I assumed BD/DC = AB/AC = 1/2, which led me to AB = sqrt(3), AC = 2*sqrt(3), BC = 3. Then, using coordinates, I found AD = 2.Alternatively, maybe I can use the formula for the length of the angle bisector directly without coordinates.The formula is:AD = (2ab / (a + b)) * cos(theta/2)Where a and b are the sides adjacent to angle A, which are AB and AC.But in this case, AB = sqrt(3), AC = 2*sqrt(3), and angle A = 60°, so theta/2 = 30°, cos(theta/2) = sqrt(3)/2.So,AD = (2 * sqrt(3) * 2*sqrt(3) / (sqrt(3) + 2*sqrt(3))) * sqrt(3)/2Simplify numerator:2 * sqrt(3) * 2*sqrt(3) = 4 * 3 = 12Denominator:sqrt(3) + 2*sqrt(3) = 3*sqrt(3)So,AD = (12 / 3*sqrt(3)) * sqrt(3)/2Simplify:12 / 3*sqrt(3) = 4 / sqrt(3)Then,4 / sqrt(3) * sqrt(3)/2 = (4 * sqrt(3)) / (sqrt(3) * 2) = 4/2 = 2Same result again. So, AD = 2, which is sqrt(4). Therefore, the answer is C.Wait, but I'm a bit confused because sometimes the angle bisector formula is written differently. Let me check another formula for the length of the angle bisector.Another formula is:AD = (2ab / (a + b)) * cos(theta/2)But I think that's the same as what I used earlier.Alternatively, there's a formula that uses the lengths of the sides and the segments created by the bisector:AD = (2ab / (a + b)) * sqrt(1 - (c^2)/(a + b)^2)Wait, no, that might not be correct. Let me recall the correct formula.The correct formula for the length of an angle bisector in a triangle is:AD = (2ab / (a + b)) * cos(theta/2)Where a and b are the sides adjacent to angle A, and theta is the measure of angle A.Alternatively, another formula is:AD = sqrt[ab(1 - (c^2)/(a + b)^2)]Where c is the length of the side opposite angle A.Wait, in this case, c would be BC, which is 3.So, let's try that formula.AD = sqrt[AB * AC * (1 - (BC^2)/(AB + AC)^2)]AB = sqrt(3), AC = 2*sqrt(3), BC = 3So,AD = sqrt[sqrt(3) * 2*sqrt(3) * (1 - (3^2)/(sqrt(3) + 2*sqrt(3))^2)]Simplify:sqrt(3) * 2*sqrt(3) = 2 * 3 = 6Denominator inside the sqrt:(sqrt(3) + 2*sqrt(3)) = 3*sqrt(3), so (3*sqrt(3))^2 = 9 * 3 = 27So,1 - (9)/(27) = 1 - 1/3 = 2/3Therefore,AD = sqrt[6 * (2/3)] = sqrt[4] = 2Again, same result. So, AD = 2, which is sqrt(4). So, the answer is C.Wait, but I'm still a bit unsure because sometimes I might have mixed up the sides. Let me just recap.We have a right-angled triangle at B, with AB = sqrt(3), BC = 3, AC = 2*sqrt(3). The angle bisector from A meets BC at D, dividing it into BD = 1 and DC = 2. Using the Angle Bisector Theorem, we confirmed the ratio, then used coordinates and two different formulas to find AD = 2, which is sqrt(4). So, the answer should be C.I think I've verified it enough times. So, I'm confident the answer is C) sqrt(4).

💡To compute (sin 150^circ), I start by recognizing that (150^circ) is in the second quadrant of the unit circle. In the second quadrant, the sine function is positive, which means (sin 150^circ) will be a positive value.Next, I determine the reference angle for (150^circ). The reference angle is the acute angle formed by the terminal side of the given angle and the x-axis. For angles in the second quadrant, the reference angle is calculated as (180^circ - theta). Therefore, the reference angle for (150^circ) is:[180^circ - 150^circ = 30^circ]Now, I recall the sine value for the reference angle (30^circ). From trigonometric values, I know that:[sin 30^circ = frac{1}{2}]Since (150^circ) is in the second quadrant where sine is positive, the sine of (150^circ) is the same as the sine of its reference angle. Therefore:[sin 150^circ = sin 30^circ = frac{1}{2}]Thus, the value of (sin 150^circ) is (frac{1}{2}).

💡Okay, so I have this problem where I need to find all polynomials ( p(x) ) with real coefficients such that for all real numbers ( a, b, c ), the equation[ p(a + b - 2c) + p(b + c - 2a) + p(c + a - 2b) = 3p(a - b) + 3p(b - c) + 3p(c - a) ]holds true. Hmm, that's quite a mouthful. Let me try to break it down step by step.First, I remember that when dealing with functional equations involving polynomials, plugging in specific values for variables can sometimes simplify things and give us useful information. Maybe I can start by choosing some specific values for ( a, b, c ) to see what constraints that puts on ( p(x) ).Let me try setting ( a = b = c ). If I do that, then each argument inside the polynomials on the left side becomes ( a + a - 2a = 0 ). So the left side becomes ( p(0) + p(0) + p(0) = 3p(0) ).On the right side, each term ( p(a - b) ), ( p(b - c) ), and ( p(c - a) ) becomes ( p(0) ) as well, so the right side is ( 3p(0) + 3p(0) + 3p(0) = 9p(0) ).Putting it together, we have ( 3p(0) = 9p(0) ), which simplifies to ( 6p(0) = 0 ), so ( p(0) = 0 ). Okay, that's a useful piece of information: the constant term of the polynomial ( p(x) ) must be zero.Next, maybe I can set some variables to zero to see what else I can find. Let's try setting ( b = c = 0 ). Then the equation becomes:Left side: ( p(a + 0 - 0) + p(0 + 0 - 2a) + p(0 + a - 0) = p(a) + p(-2a) + p(a) ).Right side: ( 3p(a - 0) + 3p(0 - 0) + 3p(0 - a) = 3p(a) + 3p(0) + 3p(-a) ).But we already know ( p(0) = 0 ), so the right side simplifies to ( 3p(a) + 3p(-a) ).Putting it together, we have:[ 2p(a) + p(-2a) = 3p(a) + 3p(-a) ]Let me rearrange this equation:[ p(-2a) = p(a) + 3p(-a) ]Hmm, that's an interesting relation. Maybe I can define a new function to capture this behavior. Let me define ( q(x) = p(x) + p(-x) ). Then, ( q(-x) = p(-x) + p(x) = q(x) ), so ( q(x) ) is an even function.Let's see what this gives us. From the equation above:[ p(-2a) = p(a) + 3p(-a) ]But ( q(2a) = p(2a) + p(-2a) ). Let's substitute ( p(-2a) ) from the equation above:[ q(2a) = p(2a) + p(a) + 3p(-a) ]But ( q(a) = p(a) + p(-a) ), so ( p(-a) = q(a) - p(a) ). Let's substitute that into the equation:[ q(2a) = p(2a) + p(a) + 3(q(a) - p(a)) ][ q(2a) = p(2a) + p(a) + 3q(a) - 3p(a) ][ q(2a) = p(2a) - 2p(a) + 3q(a) ]Hmm, not sure if that helps directly. Maybe I should think about the degree of the polynomial ( p(x) ). Let's assume ( p(x) ) is a polynomial of degree ( n ). Then, ( p(a + b - 2c) ) and similar terms on the left side are also polynomials of degree ( n ). On the right side, each ( p(a - b) ), ( p(b - c) ), ( p(c - a) ) is also degree ( n ), multiplied by 3.But the left side has three terms each of degree ( n ), so the left side is a polynomial of degree ( n ). The right side is also a polynomial of degree ( n ). So, in terms of degrees, both sides are consistent.But maybe looking at specific degrees can help. Let's suppose ( p(x) ) is a linear polynomial, so ( p(x) = kx + m ). But we already know ( p(0) = 0 ), so ( m = 0 ). Thus, ( p(x) = kx ).Let's test if this works. Substitute ( p(x) = kx ) into the original equation:Left side: ( k(a + b - 2c) + k(b + c - 2a) + k(c + a - 2b) )Simplify:( k(a + b - 2c + b + c - 2a + c + a - 2b) )Combine like terms:( k[(a - 2a + a) + (b + b - 2b) + (-2c + c + c)] )Simplify each bracket:( k[0a + 0b + 0c] = 0 )Right side: ( 3k(a - b) + 3k(b - c) + 3k(c - a) )Simplify:( 3k(a - b + b - c + c - a) = 3k(0) = 0 )So both sides are equal. Therefore, linear polynomials of the form ( p(x) = kx ) satisfy the equation.But wait, earlier when I set ( b = c = 0 ), I got the equation ( p(-2a) = p(a) + 3p(-a) ). Let's check if this holds for ( p(x) = kx ):Left side: ( p(-2a) = -2ka )Right side: ( p(a) + 3p(-a) = ka + 3(-ka) = ka - 3ka = -2ka )Yes, it holds. So linear polynomials are okay.But could there be higher-degree polynomials that satisfy the equation? Let's check quadratic polynomials. Suppose ( p(x) = ax^2 + bx + c ). But since ( p(0) = 0 ), ( c = 0 ). So ( p(x) = ax^2 + bx ).Let's test this in the equation ( p(-2a) = p(a) + 3p(-a) ):Left side: ( p(-2a) = a(-2a)^2 + b(-2a) = 4a^2 - 2ab )Right side: ( p(a) + 3p(-a) = (a a^2 + b a) + 3(a (-a)^2 + b (-a)) = (a^3 + ab) + 3(a^3 - ab) = a^3 + ab + 3a^3 - 3ab = 4a^3 - 2ab )Wait, but the left side is ( 4a^2 - 2ab ) and the right side is ( 4a^3 - 2ab ). These are only equal if ( 4a^2 = 4a^3 ) for all ( a ), which implies ( a^2(4 - 4a) = 0 ) for all ( a ). But this can't be true unless ( a = 0 ) or ( a = 1 ), which isn't valid for all ( a ). So quadratic polynomials don't satisfy the equation unless ( a = 0 ), which would reduce it to a linear polynomial.Hmm, so maybe only linear polynomials work? But wait, let me double-check. Maybe I made a mistake in substituting.Wait, actually, in the equation ( p(-2a) = p(a) + 3p(-a) ), if ( p(x) = ax^2 + bx ), then:Left side: ( p(-2a) = a(-2a)^2 + b(-2a) = 4a^2 - 2ab )Right side: ( p(a) + 3p(-a) = (a a^2 + b a) + 3(a (-a)^2 + b (-a)) = (a^3 + ab) + 3(a^3 - ab) = a^3 + ab + 3a^3 - 3ab = 4a^3 - 2ab )So indeed, ( 4a^2 - 2ab = 4a^3 - 2ab ) implies ( 4a^2 = 4a^3 ), which is not true for all ( a ). So quadratic polynomials don't work unless ( a = 0 ), which again reduces to linear.What about cubic polynomials? Let me try ( p(x) = ax^3 + bx^2 + cx ). Again, ( p(0) = 0 ), so no constant term.Testing the equation ( p(-2a) = p(a) + 3p(-a) ):Left side: ( p(-2a) = a(-2a)^3 + b(-2a)^2 + c(-2a) = -8a^3 + 4ba^2 - 2ca )Right side: ( p(a) + 3p(-a) = (a a^3 + b a^2 + c a) + 3(a (-a)^3 + b (-a)^2 + c (-a)) )Simplify:( (a^4 + b a^2 + c a) + 3(-a^4 + b a^2 - c a) )( = a^4 + b a^2 + c a - 3a^4 + 3b a^2 - 3c a )( = (-2a^4) + 4b a^2 - 2c a )So equate left and right:( -8a^3 + 4b a^2 - 2c a = -2a^4 + 4b a^2 - 2c a )This implies:( -8a^3 = -2a^4 )Which simplifies to:( 2a^4 - 8a^3 = 0 )( 2a^3(a - 4) = 0 )This holds only for ( a = 0 ) or ( a = 4 ), not for all ( a ). So cubic polynomials don't work either.Hmm, so it seems like only linear polynomials satisfy the equation. But let me think again. Maybe I missed something.Wait, earlier when I set ( b = c = 0 ), I got ( p(-2a) = p(a) + 3p(-a) ). For linear polynomials ( p(x) = kx ), this holds because both sides are ( -2ka ). For quadratic polynomials, it didn't hold unless the quadratic coefficient was zero, which reduces it to linear.So perhaps only linear polynomials work. But let me test another case to be sure.Let me set ( c = 0 ) and leave ( a ) and ( b ) arbitrary. Then the equation becomes:Left side: ( p(a + b - 0) + p(b + 0 - 2a) + p(0 + a - 2b) = p(a + b) + p(b - 2a) + p(a - 2b) )Right side: ( 3p(a - b) + 3p(b - 0) + 3p(0 - a) = 3p(a - b) + 3p(b) + 3p(-a) )So:[ p(a + b) + p(b - 2a) + p(a - 2b) = 3p(a - b) + 3p(b) + 3p(-a) ]Let me assume ( p(x) = kx ). Then:Left side: ( k(a + b) + k(b - 2a) + k(a - 2b) = k(a + b + b - 2a + a - 2b) = k(0) = 0 )Right side: ( 3k(a - b) + 3k(b) + 3k(-a) = 3k(a - b + b - a) = 3k(0) = 0 )So it holds. Now, if I assume ( p(x) ) is quadratic, say ( p(x) = ax^2 + bx ), let's see:Left side:( p(a + b) = a(a + b)^2 + b(a + b) )( p(b - 2a) = a(b - 2a)^2 + b(b - 2a) )( p(a - 2b) = a(a - 2b)^2 + b(a - 2b) )Right side:( 3p(a - b) = 3[a(a - b)^2 + b(a - b)] )( 3p(b) = 3[ab^2 + bb] )( 3p(-a) = 3[a(-a)^2 + b(-a)] = 3[a a^2 - ab] )This seems complicated, but let's compute both sides.Left side:( a(a^2 + 2ab + b^2) + b(a + b) + a(b^2 - 4ab + 4a^2) + b(b - 2a) + a(a^2 - 4ab + 4b^2) + b(a - 2b) )Wait, this is getting too messy. Maybe it's better to compute each term separately.Compute ( p(a + b) ):( a(a + b)^2 + b(a + b) = a(a^2 + 2ab + b^2) + ab + b^2 = a^3 + 2a^2b + ab^2 + ab + b^2 )Compute ( p(b - 2a) ):( a(b - 2a)^2 + b(b - 2a) = a(b^2 - 4ab + 4a^2) + b^2 - 2ab = ab^2 - 4a^2b + 4a^3 + b^2 - 2ab )Compute ( p(a - 2b) ):( a(a - 2b)^2 + b(a - 2b) = a(a^2 - 4ab + 4b^2) + ab - 2b^2 = a^3 - 4a^2b + 4ab^2 + ab - 2b^2 )Now sum all three:Left side:( (a^3 + 2a^2b + ab^2 + ab + b^2) + (ab^2 - 4a^2b + 4a^3 + b^2 - 2ab) + (a^3 - 4a^2b + 4ab^2 + ab - 2b^2) )Combine like terms:- ( a^3 ): 1 + 4 + 1 = 6a^3- ( a^2b ): 2 - 4 - 4 = -6a^2b- ( ab^2 ): 1 + 1 + 4 = 6ab^2- ( ab ): 1 - 2 + 1 = 0ab- ( b^2 ): 1 + 1 - 2 = 0b^2So left side simplifies to ( 6a^3 - 6a^2b + 6ab^2 ).Now compute the right side:Right side:( 3p(a - b) + 3p(b) + 3p(-a) )Compute each term:( 3p(a - b) = 3[a(a - b)^2 + b(a - b)] = 3[a(a^2 - 2ab + b^2) + ab - b^2] = 3[a^3 - 2a^2b + ab^2 + ab - b^2] = 3a^3 - 6a^2b + 3ab^2 + 3ab - 3b^2 )( 3p(b) = 3[ab^2 + bb] = 3ab^2 + 3b^2 )( 3p(-a) = 3[a(-a)^2 + b(-a)] = 3[a a^2 - ab] = 3a^3 - 3ab )Now sum all three:Right side:( (3a^3 - 6a^2b + 3ab^2 + 3ab - 3b^2) + (3ab^2 + 3b^2) + (3a^3 - 3ab) )Combine like terms:- ( a^3 ): 3 + 3 = 6a^3- ( a^2b ): -6- ( ab^2 ): 3 + 3 = 6ab^2- ( ab ): 3 - 3 = 0ab- ( b^2 ): -3 + 3 = 0b^2So right side simplifies to ( 6a^3 - 6a^2b + 6ab^2 ).Wait a minute! Both sides are equal. So even though earlier when I set ( b = c = 0 ), quadratic polynomials didn't seem to work, when I set ( c = 0 ) and let ( a ) and ( b ) vary, quadratic polynomials do satisfy the equation.This is confusing. There seems to be a contradiction. Let me check my earlier steps.When I set ( b = c = 0 ), I got ( p(-2a) = p(a) + 3p(-a) ). For ( p(x) = ax^2 + bx ), this gave:Left side: ( 4a^2 - 2ab )Right side: ( 4a^3 - 2ab )Which only holds if ( 4a^2 = 4a^3 ), i.e., ( a = 0 ) or ( a = 1 ). But when I set ( c = 0 ) and let ( a ) and ( b ) vary, quadratic polynomials do satisfy the equation.This suggests that maybe quadratic polynomials do work, but only under certain conditions. Wait, no, because in the case when ( c = 0 ), the equation holds for any ( a ) and ( b ), but when ( b = c = 0 ), it only holds for specific ( a ).This inconsistency implies that quadratic polynomials cannot satisfy the equation for all ( a, b, c ). Because if a polynomial satisfies the equation for all ( a, b, c ), it must satisfy it for any specific choice, including ( b = c = 0 ), which only holds if the quadratic coefficient is zero.Therefore, the only polynomials that satisfy the equation for all ( a, b, c ) are linear polynomials of the form ( p(x) = kx ).But wait, earlier when I set ( c = 0 ), quadratic polynomials seemed to work. Maybe I made a mistake in that calculation.Let me re-examine the case when ( c = 0 ). I had:Left side: ( p(a + b) + p(b - 2a) + p(a - 2b) )Right side: ( 3p(a - b) + 3p(b) + 3p(-a) )For ( p(x) = ax^2 + bx ), I computed both sides and they turned out equal. But when ( b = c = 0 ), it didn't hold unless ( a = 0 ).This suggests that quadratic polynomials satisfy the equation when ( c = 0 ) but not when ( b = c = 0 ). Therefore, quadratic polynomials cannot satisfy the equation for all ( a, b, c ).Hence, only linear polynomials ( p(x) = kx ) satisfy the equation for all ( a, b, c ).Wait, but let me think again. Maybe I can consider the general form of ( p(x) ). Suppose ( p(x) ) is a polynomial of degree ( n ). Then, the left side is a combination of polynomials each of degree ( n ), so the left side is a polynomial of degree ( n ). Similarly, the right side is a combination of polynomials each of degree ( n ), so it's also a polynomial of degree ( n ).But for the equation to hold for all ( a, b, c ), the coefficients of corresponding powers of ( a, b, c ) must match on both sides.Let me try to equate coefficients. Let me assume ( p(x) = kx ). Then, as we saw, both sides are zero, so it works.If ( p(x) ) is quadratic, say ( p(x) = ax^2 + bx ), then the left side and right side must have the same coefficients for each monomial in ( a, b, c ).But when I set ( b = c = 0 ), the equation reduces to ( 4a^2 - 2ab = 4a^3 - 2ab ), which implies ( 4a^2 = 4a^3 ), meaning ( a = 0 ) or ( a = 1 ). Since this must hold for all ( a ), the only possibility is ( a = 0 ), reducing ( p(x) ) to linear.Therefore, quadratic polynomials cannot satisfy the equation for all ( a, b, c ).Thus, the only solutions are linear polynomials ( p(x) = kx ).But wait, earlier when I set ( c = 0 ), quadratic polynomials seemed to work. Maybe I need to reconcile this.Wait, when ( c = 0 ), the equation simplifies in a way that might not capture the full generality. Because when ( c = 0 ), the equation might not impose the same restrictions as when ( b = c = 0 ).In other words, the case ( c = 0 ) might not be restrictive enough to rule out quadratic polynomials, but the case ( b = c = 0 ) does rule them out. Therefore, even though quadratic polynomials satisfy the equation when ( c = 0 ), they don't satisfy it when ( b = c = 0 ), unless the quadratic coefficient is zero.Hence, the only polynomials that satisfy the equation for all ( a, b, c ) are linear polynomials.Therefore, the solution is all linear polynomials ( p(x) = kx ) where ( k ) is a real constant.But wait, let me check one more time. Suppose ( p(x) = kx ). Then, substituting into the original equation:Left side: ( p(a + b - 2c) + p(b + c - 2a) + p(c + a - 2b) )= ( k(a + b - 2c) + k(b + c - 2a) + k(c + a - 2b) )= ( k[(a + b - 2c) + (b + c - 2a) + (c + a - 2b)] )= ( k[0] = 0 )Right side: ( 3p(a - b) + 3p(b - c) + 3p(c - a) )= ( 3k(a - b) + 3k(b - c) + 3k(c - a) )= ( 3k[(a - b) + (b - c) + (c - a)] )= ( 3k[0] = 0 )So both sides are equal, confirming that linear polynomials work.Therefore, the only polynomials satisfying the given equation are linear polynomials of the form ( p(x) = kx ).**Final Answer**The polynomials are all linear functions, specifically boxed{p(x) = ax + b} where ( a ) and ( b ) are real constants. However, since ( p(0) = 0 ), ( b = 0 ), so the solution is boxed{p(x) = ax}.Wait, hold on. Earlier, I concluded that ( p(0) = 0 ), so the constant term must be zero. Therefore, the polynomial must be of the form ( p(x) = ax ). So the final answer is ( p(x) = ax ).But in the final answer above, I wrote ( p(x) = ax + b ) with ( b = 0 ). That's correct, but since ( b = 0 ), it's just ( p(x) = ax ).So the final answer is boxed{p(x) = ax} where ( a ) is a real constant.But wait, in the initial problem, the equation must hold for all ( a, b, c ). So if ( p(x) = ax ), it works. But earlier, when I considered quadratic polynomials, they didn't work unless the quadratic coefficient was zero. So yes, only linear polynomials work.Therefore, the final answer is boxed{p(x) = ax} where ( a ) is a real constant.

💡Alright, so I have this problem here about a function f(x) = (a x² + b x + 4)/x, and it's supposed to pass through two points, A(1,5) and B(2,4). I need to find the analytical expression of f(x) and then determine its monotonicity on the interval (0,2) using the definition. Okay, let's break this down step by step.First, part (1) is about finding the specific form of f(x). Since it's given that the function passes through points A and B, I can plug in those x and y values into the function to create equations and solve for the unknowns a and b. That makes sense because if a function passes through a point, the x and y values of that point should satisfy the function's equation.So, starting with point A(1,5). Plugging x=1 and y=5 into f(x):f(1) = (a*(1)^2 + b*(1) + 4)/1 = a + b + 4 = 5.That gives me the first equation: a + b + 4 = 5. Simplifying that, I subtract 4 from both sides: a + b = 1.Next, using point B(2,4). Plugging x=2 and y=4 into f(x):f(2) = (a*(2)^2 + b*(2) + 4)/2 = (4a + 2b + 4)/2 = 2a + b + 2 = 4.So, the second equation is 2a + b + 2 = 4. Simplifying that, subtract 2 from both sides: 2a + b = 2.Now, I have a system of two equations:1. a + b = 12. 2a + b = 2I need to solve for a and b. Let me subtract the first equation from the second to eliminate b:(2a + b) - (a + b) = 2 - 12a + b - a - b = 1a = 1Now that I know a = 1, I can plug that back into the first equation to find b:1 + b = 1b = 0So, a = 1 and b = 0. Therefore, the function f(x) becomes:f(x) = (1*x² + 0*x + 4)/x = (x² + 4)/x.Simplifying that, I can write it as f(x) = x + 4/x. That seems straightforward.Okay, part (1) seems done. Now, part (2) is about determining the monotonicity of f(x) on the interval (0,2). Monotonicity refers to whether the function is increasing or decreasing over that interval. The problem specifies to prove it using the definition, which I think means using the definition of increasing or decreasing functions, not necessarily using derivatives.So, recalling the definition: A function f is decreasing on an interval if for any two points x₁ and x₂ in that interval with x₁ < x₂, it holds that f(x₁) > f(x₂). Similarly, if f(x₁) < f(x₂), it's increasing.So, I need to show that for any x₁ and x₂ in (0,2) with x₁ < x₂, f(x₁) > f(x₂). Let's see how to approach this.First, let's write down f(x) = x + 4/x.To show that f is decreasing on (0,2), I need to take two arbitrary points x₁ and x₂ in (0,2) with x₁ < x₂ and show that f(x₁) > f(x₂).Let me compute f(x₁) - f(x₂):f(x₁) - f(x₂) = (x₁ + 4/x₁) - (x₂ + 4/x₂) = (x₁ - x₂) + (4/x₁ - 4/x₂).Simplify the second term:4/x₁ - 4/x₂ = 4*(x₂ - x₁)/(x₁ x₂).So, putting it all together:f(x₁) - f(x₂) = (x₁ - x₂) + 4*(x₂ - x₁)/(x₁ x₂).Factor out (x₁ - x₂):= (x₁ - x₂) * [1 - 4/(x₁ x₂)].So, f(x₁) - f(x₂) = (x₁ - x₂)*(1 - 4/(x₁ x₂)).Now, since x₁ < x₂, x₁ - x₂ is negative. So, the sign of f(x₁) - f(x₂) depends on the second factor, [1 - 4/(x₁ x₂)].If I can show that [1 - 4/(x₁ x₂)] is positive, then the entire expression would be negative (because negative times positive is negative), meaning f(x₁) - f(x₂) < 0, which implies f(x₁) < f(x₂). Wait, that would mean the function is increasing. But we want to show it's decreasing, so maybe I made a miscalculation.Wait, let's double-check:f(x₁) - f(x₂) = (x₁ - x₂) + 4*(x₂ - x₁)/(x₁ x₂).= (x₁ - x₂) - 4*(x₁ - x₂)/(x₁ x₂).= (x₁ - x₂)*(1 - 4/(x₁ x₂)).Yes, that's correct. So, since x₁ < x₂, x₁ - x₂ is negative. So, if [1 - 4/(x₁ x₂)] is positive, then f(x₁) - f(x₂) is negative, meaning f(x₁) < f(x₂), which would mean the function is increasing. But we want to show it's decreasing, so perhaps [1 - 4/(x₁ x₂)] is negative.Let's see: 1 - 4/(x₁ x₂). Since x₁ and x₂ are in (0,2), their product x₁ x₂ is less than 4 (since 2*2=4). Therefore, 4/(x₁ x₂) > 1, so 1 - 4/(x₁ x₂) < 0.Therefore, [1 - 4/(x₁ x₂)] is negative. So, f(x₁) - f(x₂) = (negative)*(negative) = positive.Wait, that means f(x₁) - f(x₂) > 0, so f(x₁) > f(x₂). Which is exactly what we wanted to show for a decreasing function.So, since x₁ < x₂ implies f(x₁) > f(x₂), the function is decreasing on (0,2).Alternatively, if I think about the derivative, which might be another way to check. The derivative of f(x) = x + 4/x is f'(x) = 1 - 4/x². On the interval (0,2), x² < 4, so 4/x² > 1, hence f'(x) = 1 - 4/x² < 0. Therefore, the function is decreasing on (0,2). But since the problem asks to use the definition, I think the previous approach is sufficient.So, summarizing:1. The function f(x) is (x² + 4)/x, which simplifies to x + 4/x.2. Using the definition, for any x₁ < x₂ in (0,2), f(x₁) - f(x₂) > 0, hence f(x) is decreasing on (0,2).I think that covers both parts of the problem.

💡Okay, so I have this problem where I need to find two numbers between 9 and 243 such that these two numbers form a geometric sequence with 9 and 243. Hmm, let me think about how to approach this.First, I remember that a geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as 'r'. So, if I have four terms in a geometric sequence: the first term is 9, then the next two terms are the ones I need to find, and the fourth term is 243.Let me denote the two numbers I need to find as 'a' and 'b'. So, the sequence would look like: 9, a, b, 243.Since it's a geometric sequence, each term is multiplied by 'r' to get the next term. That means:- a = 9 * r- b = a * r = 9 * r^2- 243 = b * r = 9 * r^3So, I can write the equation: 243 = 9 * r^3Now, I need to solve for 'r'. Let's do that step by step.First, divide both sides of the equation by 9:243 / 9 = r^327 = r^3So, r^3 = 27. To find 'r', I take the cube root of both sides:r = cube root of 27r = 3Okay, so the common ratio 'r' is 3. Now, I can find 'a' and 'b' using this ratio.Starting with 'a':a = 9 * r = 9 * 3 = 27Then, 'b' is:b = a * r = 27 * 3 = 81So, the two numbers between 9 and 243 that form a geometric sequence are 27 and 81.Wait, let me double-check to make sure I didn't make a mistake. If I start with 9 and multiply by 3 each time, I should get 9, 27, 81, 243. Yep, that seems correct. Each term is indeed multiplied by 3 to get the next term, so the sequence is geometric with a common ratio of 3.I think that's it. The two numbers are 27 and 81.