🧠:[<think>Okay, so I need to find the directional derivative of the function u = sin(r) along the radius vector r, where r is the magnitude of the vector r. Hmm, let's start by recalling what a directional derivative is. From what I remember, the directional derivative of a function in the direction of a vector v is the dot product of the gradient of the function and the unit vector in the direction of v. So, in this case, the direction is given by the radius vector r itself. That means the directional derivative should be the gradient of u dotted with the unit vector in the direction of r. First, let's figure out the gradient of u. The function u is sin(r), and r is the magnitude of the radius vector, so r = sqrt(x^2 + y^2 + z^2) in three dimensions. The gradient of u, which is grad u, should be the derivative of sin(r) with respect to x, y, and z components. Since u is a function of r, which in turn is a function of x, y, z, we'll need to use the chain rule. The derivative of sin(r) with respect to x is cos(r) times the derivative of r with respect to x. Similarly for y and z. What's the derivative of r with respect to x? Let me compute that. If r = sqrt(x^2 + y^2 + z^2), then the partial derivative of r with respect to x is (1/(2*sqrt(x^2 + y^2 + z^2)))*2x = x/r. Similarly, the partial derivatives with respect to y and z will be y/r and z/r, respectively. So putting that together, the gradient of u is cos(r) times (x/r, y/r, z/r). So grad u = cos(r) * (x/r, y/r, z/r). But wait, that simplifies to cos(r)/r times the vector (x, y, z). But (x, y, z) is just the radius vector r. So grad u = (cos(r)/r) * r. Now, the directional derivative in the direction of r is the dot product of grad u and the unit vector in the direction of r. The unit vector in the direction of r is r divided by its magnitude, which is r/r. So the unit vector is r/r. Therefore, the directional derivative D_r u = grad u · (r/r). Substituting grad u from above, we have D_r u = [ (cos(r)/r) * r ] · (r/r). Let's simplify that. First, (cos(r)/r) * r is just cos(r). Then, cos(r) dotted with (r/r). But wait, r is a vector, and r/r is a unit vector. However, the gradient is already a vector, and we need to take the dot product of grad u with the unit vector in the direction of r. Wait, maybe I confused something here. Let me double-check. The gradient of u is (cos(r)/r) * r, as we established. The directional derivative in the direction of r is the dot product of grad u and the unit vector in the direction of r, which is r / |r|. But |r| is r, so the unit vector is r / r. Therefore, D_r u = grad u · (r / r) = [ (cos(r)/r) * r ] · (r / r). But (cos(r)/r) * r is cos(r) times (r / r), which is cos(r) times the unit vector. Wait, no. Wait, let me clarify. Let me write out the vectors explicitly. Let’s denote the position vector as r = (x, y, z). Then, grad u = ∇u = ( ∂u/∂x, ∂u/∂y, ∂u/∂z ). We calculated ∂u/∂x = cos(r) * (x/r), similarly for y and z. So ∇u = cos(r) * (x, y, z) / r = cos(r) * r / r. So ∇u is a vector in the direction of r, scaled by cos(r)/r. Now, the directional derivative in the direction of r is ∇u dotted with the unit vector in the direction of r. The unit vector in the direction of r is r / r. Therefore, D_r u = ∇u · (r / r) = [ cos(r) * r / r ] · [ r / r ].Since both vectors are in the same direction, their dot product is just the product of their magnitudes. The magnitude of [ cos(r) * r / r ] is |cos(r)| * |r| / r = |cos(r)| * r / r = |cos(r)|. But since cos(r) is a scalar, and assuming r is positive (since it's a magnitude), then cos(r) can be positive or negative, but the directional derivative is a scalar. However, the dot product of two vectors in the same direction is the product of their magnitudes. But r / r is a unit vector. So if we have two vectors: one is [cos(r) * r / r ] and the other is [r / r], then their dot product is [cos(r) * (r / r) ] · [r / r] = cos(r) * ( (r / r) · (r / r) ). But (r / r) · (r / r) is the dot product of the unit vector with itself, which is 1. So the directional derivative is cos(r) * 1 = cos(r). Wait, so does that mean the directional derivative of sin(r) in the direction of r is simply cos(r)? But that seems too straightforward. Let me check with another approach. Alternatively, the directional derivative can be computed as the rate of change of u along the direction of r. Since u = sin(r), and moving along the direction of r would mean increasing r. So, the directional derivative should just be the derivative of sin(r) with respect to r, which is cos(r). Oh! Wait a minute, that's another way to see it. Because if you're moving radially outward, then the rate of change of u with respect to r is just the derivative of sin(r) with respect to r, which is cos(r). So that gives the same answer. Therefore, both methods confirm that the directional derivative is cos(r). But let me make sure I didn't make a mistake in the first method. Let's recap:1. Compute gradient of u: ∇u = cos(r) * r / r.2. Directional derivative in direction r: ∇u · ( r / r ) = [cos(r) * r / r] · [r / r] = cos(r) * ( r · r ) / r^2.But r · r = r^2, so substituting that gives cos(r) * (r^2) / r^2 = cos(r). Yes, that works out. So both methods agree. Therefore, the directional derivative is indeed cos(r). But wait, the problem says "directional derivative along the radius vector r". So, just to confirm, this is the same as the derivative with respect to r, which is cos(r). Therefore, the answer is cos(r), which in boxed form is boxed{cos r}.Final Answerboxed{cos r}</think>To find the directional derivative of the function ( u = sin(r) ) along the radius vector (mathbf{r}), where ( r = |mathbf{r}| ), we start by computing the gradient of ( u ).1. Gradient Calculation: - The function ( u = sin(r) ) depends on ( r ), the magnitude of the vector (mathbf{r}). - Using the chain rule, the gradient of ( u ) is given by: [ nabla u = cos(r) cdot nabla r ] - The gradient of ( r ) (where ( r = sqrt{x^2 + y^2 + z^2} )) is: [ nabla r = left( frac{partial r}{partial x}, frac{partial r}{partial y}, frac{partial r}{partial z} right) = left( frac{x}{r}, frac{y}{r}, frac{z}{r} right) = frac{mathbf{r}}{r} ] - Therefore, the gradient of ( u ) is: [ nabla u = cos(r) cdot frac{mathbf{r}}{r} ]2. Directional Derivative: - The directional derivative in the direction of the radius vector (mathbf{r}) is the dot product of the gradient of ( u ) and the unit vector in the direction of (mathbf{r}). - The unit vector in the direction of (mathbf{r}) is (frac{mathbf{r}}{r}). - Thus, the directional derivative is: [ D_{mathbf{r}} u = nabla u cdot frac{mathbf{r}}{r} = left( cos(r) cdot frac{mathbf{r}}{r} right) cdot frac{mathbf{r}}{r} ] - Simplifying the dot product: [ left( cos(r) cdot frac{mathbf{r}}{r} right) cdot frac{mathbf{r}}{r} = cos(r) cdot left( frac{mathbf{r}}{r} cdot frac{mathbf{r}}{r} right) = cos(r) cdot 1 = cos(r) ]Therefore, the directional derivative of ( u = sin(r) ) along the radius vector (mathbf{r}) is (boxed{cos r}).,<think>Okay, let's see. I need to find the directional derivative of the function u = sin(r) along the radius vector r. Hmm, directional derivatives... Right, the directional derivative of a function in the direction of a vector gives the rate of change of the function in that direction. First, let me recall the formula for the directional derivative. If I remember correctly, the directional derivative of a scalar function u in the direction of a vector v is given by the dot product of the gradient of u and the unit vector in the direction of v. So, it's ∇u · (v/|v|). In this case, the direction is the radius vector r itself. Since r is the radius vector, its magnitude is just r, which is the same as the variable in the function. So, the directional derivative along r should be ∇u · (r/r), because the unit vector in the direction of r is r divided by its magnitude r. Now, I need to compute the gradient of u. The function u is sin(r), and r is the magnitude of the radius vector, so r = sqrt(x² + y² + z²) in three dimensions. The gradient of u, ∇u, is the vector of partial derivatives with respect to each coordinate. Let's compute the partial derivatives. Since u = sin(r), the derivative of u with respect to any variable, say x, would be cos(r) times the derivative of r with respect to x. The derivative of r with respect to x is x/r, because r = sqrt(x² + y² + z²), so ∂r/∂x = x/r. Therefore, ∂u/∂x = cos(r) * (x/r). Similarly, the partial derivatives with respect to y and z will be cos(r)*(y/r) and cos(r)*(z/r), respectively. So, putting those together, the gradient ∇u is [cos(r)/r] * (x, y, z). But (x, y, z) is just the radius vector r. Therefore, ∇u = (cos(r)/r) * r. Now, the directional derivative in the direction of r is ∇u · (r/r). Let's substitute ∇u here. So, that becomes [ (cos(r)/r) * r ] · (r/r). Let's simplify this step by step. First, [ (cos(r)/r) * r ] simplifies to cos(r) * r / r = cos(r). Wait, no. Wait, [ (cos(r)/r) * r ] would just be cos(r), because the r in the numerator and denominator cancel out. Then, we take the dot product of cos(r) with (r/r). But wait, cos(r) is a scalar here, right? Because ∇u is a vector. Wait, no. Wait, hold on. Let me check again. Wait, ∇u is a vector. If ∇u = (cos(r)/r) * r, then that's a vector in the direction of r, scaled by cos(r)/r. Then, when we take the dot product of that with the unit vector in the direction of r, which is (r/r), since both vectors are in the same direction, the dot product will just be the product of their magnitudes. So, the magnitude of ∇u is | (cos(r)/r) * r | = |cos(r)|, since (cos(r)/r) multiplied by the magnitude of r (which is r) gives cos(r). Then, the unit vector in the direction of r has magnitude 1. So, their dot product is cos(r) * 1 * cos(0) because the angle between them is 0 degrees, but wait, actually, since both vectors are in the same direction, the dot product is just the product of their magnitudes. Wait, no. Wait, the dot product of two vectors a and b is |a||b|cos(theta), where theta is the angle between them. But in this case, theta is zero because they are in the same direction, so cos(theta) is 1. But in our case, ∇u is already (cos(r)/r) * r, which is a scalar multiple of r. So, ∇u is a vector pointing in the direction of r, and the unit vector is also in the direction of r. Therefore, their dot product is |∇u| * |unit vector| * cos(0). But since the unit vector has magnitude 1, it's just |∇u| * 1 * 1 = |∇u|. Wait, but is that correct? Wait, actually, no. Because the dot product of two vectors in the same direction is the product of their magnitudes. But if one of them is a unit vector, then the dot product is just the magnitude of the other vector. Wait, let's think again. Suppose ∇u = a * r, where a is a scalar (cos(r)/r). Then, the unit vector in the direction of r is r / |r| = r / r. So, the dot product is (a * r) · (r / r) = a * (r · r) / r. But r · r is |r|² = r². So, that becomes a * r² / r = a * r. Since a is cos(r)/r, so a * r = cos(r). Therefore, the directional derivative is cos(r). Alternatively, since ∇u = (cos(r)/r) * r, then ∇u · (r / r) = (cos(r)/r) * r · (r / r) = cos(r) * (r · r) / r². Since r · r = r², this becomes cos(r) * (r² / r²) = cos(r). So, regardless of the approach, the directional derivative simplifies to cos(r). Therefore, the answer should be cos(r). Let me verify this with another approach. Another way to compute the directional derivative in the direction of r is to parametrize the position vector as r(t) = t * v, where v is a unit vector in the direction of r. Wait, but since r is the position vector, maybe we can parametrize it as moving along the radial direction. So, suppose we take a point moving away from the origin along the direction of r. Then, the derivative of u with respect to t would be the directional derivative. Let's let t be a parameter such that r(t) = t * (r / |r|) = t * (r / r) since |r| = r. Wait, but then at any point, the directional derivative is the derivative of u along increasing t. Since u = sin(r), and r = |r(t)| = t, so u(t) = sin(t). Then, the derivative of u with respect to t is cos(t), which is cos(r). So, that gives the same result. Therefore, both methods confirm that the directional derivative is cos(r). Wait, but just to make sure, maybe I made a mistake in assuming the parametrization. Let me see. If I parametrize the radial direction as moving from the origin outward, then at any point, moving in the radial direction with parameter t, which could be the same as r. So, if t is the distance from the origin, then yes, u(t) = sin(t), derivative is cos(t) = cos(r). Therefore, this seems consistent. Alternatively, using the chain rule. The directional derivative along r is the derivative of u with respect to r, which is cos(r). But wait, is that correct? Because if u is a function of r, and r is the radial distance, then the derivative of u with respect to r is indeed cos(r). But is the directional derivative the same as the derivative with respect to r? Wait, the directional derivative in the direction of the radius vector should be the same as the derivative with respect to r, since moving along the radius vector is increasing r. Therefore, if u is expressed as a function of r, then the directional derivative in the radial direction is just du/dr, which is cos(r). So, that's another way to see it. Therefore, all approaches lead to the conclusion that the directional derivative is cos(r). I think that's solid. So, the answer should be cos(r). Let me check if there's any possible mistake here. Wait, maybe there's a scaling factor because of the coordinates? For example, in spherical coordinates, the gradient components have different scaling factors. But in Cartesian coordinates, the gradient is straightforward. Wait, but we computed the gradient in Cartesian coordinates. Let me verify. Suppose in Cartesian coordinates, ∇u = (∂u/∂x, ∂u/∂y, ∂u/∂z). As we found earlier, each partial derivative is cos(r) * (x/r), cos(r) * (y/r), cos(r) * (z/r). So, ∇u = cos(r)/r * (x, y, z) = cos(r)/r * r. Then, the unit vector in the direction of r is (x, y, z)/r. Therefore, the directional derivative is ∇u · (r/r) = [cos(r)/r * r] · (r/r) = cos(r) * [r · r] / r². But r · r = x² + y² + z² = r², so [r · r]/r² = 1. Therefore, it's cos(r). Yes, that's correct. So, regardless of the coordinate system, as long as we compute it in Cartesian coordinates, it still gives cos(r). Therefore, the answer is cos(r). I think that's solid. No mistakes detected.Final AnswerThe directional derivative is boxed{cos r}.</think>To find the directional derivative of the function ( u = sin(r) ) along the radius vector ( mathbf{r} ), we start by recalling that the directional derivative in the direction of a vector ( mathbf{v} ) is given by the dot product of the gradient of ( u ) and the unit vector in the direction of ( mathbf{v} ).1. Gradient of ( u ): - The function ( u = sin(r) ) depends on ( r = |mathbf{r}| ). - The gradient of ( u ) in Cartesian coordinates is calculated as follows: [ nabla u = left( frac{partial u}{partial x}, frac{partial u}{partial y}, frac{partial u}{partial z} right) ] - Each partial derivative is computed using the chain rule: [ frac{partial u}{partial x} = cos(r) cdot frac{partial r}{partial x} = cos(r) cdot frac{x}{r} ] Similarly, [ frac{partial u}{partial y} = cos(r) cdot frac{y}{r}, quad frac{partial u}{partial z} = cos(r) cdot frac{z}{r} ] - Therefore, the gradient is: [ nabla u = frac{cos(r)}{r} mathbf{r} ]2. Unit Vector in the Direction of ( mathbf{r} ): - The unit vector in the direction of ( mathbf{r} ) is: [ frac{mathbf{r}}{r} ]3. Directional Derivative: - The directional derivative is the dot product of the gradient and the unit vector: [ nabla u cdot frac{mathbf{r}}{r} = left( frac{cos(r)}{r} mathbf{r} right) cdot left( frac{mathbf{r}}{r} right) ] - Simplifying the dot product: [ left( frac{cos(r)}{r} mathbf{r} right) cdot left( frac{mathbf{r}}{r} right) = frac{cos(r)}{r} cdot frac{mathbf{r} cdot mathbf{r}}{r} = frac{cos(r)}{r} cdot frac{r^2}{r} = cos(r) ]Thus, the directional derivative of ( u = sin(r) ) along the radius vector ( mathbf{r} ) is (boxed{cos r}).]

🔑:Solution. According to formula (2), the derivative of the function with respect to the direction of the radius vector mathbf{r} isfrac{partial u}{partial r}=left(operatorname{grad} sin r, r^{0}right)We find the gradient of this function:begin{aligned}text { grad } sin r & =frac{partial(sin r)}{partial x} mathbf{i}+frac{partial(sin r)}{partial y} mathbf{j}+frac{partial(sin r)}{partial z} mathbf{k}= & =frac{partial(sin r)}{partial r} frac{partial r}{partial x} mathbf{i}+frac{partial(sin r)}{partial r} frac{partial r}{partial y} mathbf{j}+frac{partial(sin r)}{partial r} frac{partial r}{partial z} mathbf{k}= & =left(frac{partial r}{partial x} mathbf{i}+frac{partial r}{partial y} mathbf{j}+frac{partial r}{partial z} mathbf{k}right) cos r=r^{0} cos rend{aligned}Substituting (4) into (3), we getfrac{partial u}{partial r}=left(mathbf{r}^{0} cos r, mathbf{r}^{0}right)=left(mathbf{r}^{0}, mathbf{r}^{0}right) cos r=cos r

🧠:[<think>Okay, so I need to find all polynomials p(x) of degree 5 such that p(x) + 1 is divisible by (x - 1)^3 and p(x) - 1 is divisible by (x + 1)^3. Hmm, okay. Let me start by recalling that if a polynomial q(x) is divisible by (x - a)^n, then (x - a)^n divides q(x), which means that q(a) = 0, q'(a) = 0, ..., q^{(n-1)}(a) = 0. So, the polynomial and its first n-1 derivatives must vanish at x = a.In this problem, p(x) + 1 is divisible by (x - 1)^3. That means when we plug x = 1 into p(x) + 1, we get 0, and the first and second derivatives of p(x) + 1 evaluated at x = 1 are also 0. Similarly, p(x) - 1 is divisible by (x + 1)^3, so plugging x = -1 into p(x) - 1 gives 0, and the first and second derivatives there also give 0.Since p(x) is a fifth-degree polynomial, p(x) + 1 and p(x) - 1 are also fifth-degree polynomials. However, they are divisible by cubic polynomials, so we can express them as:p(x) + 1 = (x - 1)^3 * q(x)p(x) - 1 = (x + 1)^3 * r(x)where q(x) and r(x) are quadratic polynomials because (x - 1)^3 is cubic, and multiplying by quadratic gives fifth-degree, same with (x + 1)^3. So, q(x) and r(x) are degree 2.Alternatively, maybe I can write p(x) as a quintic polynomial and set up equations based on the divisibility conditions. Let me try both approaches.First, let's consider the first approach. Let me denote:p(x) + 1 = (x - 1)^3 * (ax^2 + bx + c)p(x) - 1 = (x + 1)^3 * (dx^2 + ex + f)Since both expressions equal p(x) ± 1, they must be equal to each other except for the ±1. Wait, perhaps I should subtract them:[(x - 1)^3 * (ax^2 + bx + c)] - [(x + 1)^3 * (dx^2 + ex + f)] = 2But that might get complicated. Alternatively, since p(x) is a fifth-degree polynomial, both expressions (x - 1)^3 * q(x) and (x + 1)^3 * r(x) must equal p(x) ± 1, so perhaps equating them?Wait, but p(x) + 1 and p(x) - 1 differ by 2. So, (x - 1)^3 * q(x) - (x + 1)^3 * r(x) = 2. Hmm, this seems like a complicated equation to solve for q(x) and r(x). Maybe there's a better way.Alternatively, since p(x) is a quintic, maybe express it in terms of its coefficients:Let p(x) = a x^5 + b x^4 + c x^3 + d x^2 + e x + fThen, the conditions are:For divisibility by (x - 1)^3:p(1) + 1 = 0p'(1) = 0p''(1) = 0Similarly, for divisibility by (x + 1)^3:p(-1) - 1 = 0p'(-1) = 0p''(-1) = 0So, that's six equations. Since p(x) has six coefficients (a, b, c, d, e, f), we can set up a system of six equations and solve for them. Let me try this approach.First, compute p(1) + 1 = 0:p(1) = a(1)^5 + b(1)^4 + c(1)^3 + d(1)^2 + e(1) + f = a + b + c + d + e + fSo, equation 1: a + b + c + d + e + f + 1 = 0 → a + b + c + d + e + f = -1Similarly, p(-1) - 1 = 0:p(-1) = a(-1)^5 + b(-1)^4 + c(-1)^3 + d(-1)^2 + e(-1) + f = -a + b - c + d - e + fThus, equation 2: -a + b - c + d - e + f - 1 = 0 → -a + b - c + d - e + f = 1Now, the derivatives. First, find p'(x):p'(x) = 5a x^4 + 4b x^3 + 3c x^2 + 2d x + eThen, p'(1) = 0:5a(1)^4 + 4b(1)^3 + 3c(1)^2 + 2d(1) + e = 5a + 4b + 3c + 2d + e = 0 → equation 3: 5a + 4b + 3c + 2d + e = 0Similarly, p'(-1) = 0:5a(-1)^4 + 4b(-1)^3 + 3c(-1)^2 + 2d(-1) + e = 5a - 4b + 3c - 2d + e = 0 → equation 4: 5a - 4b + 3c - 2d + e = 0Then, the second derivative p''(x):p''(x) = 20a x^3 + 12b x^2 + 6c x + 2dSo, p''(1) = 0:20a(1)^3 + 12b(1)^2 + 6c(1) + 2d = 20a + 12b + 6c + 2d = 0 → equation 5: 20a + 12b + 6c + 2d = 0Similarly, p''(-1) = 0:20a(-1)^3 + 12b(-1)^2 + 6c(-1) + 2d = -20a + 12b - 6c + 2d = 0 → equation 6: -20a + 12b - 6c + 2d = 0So, now we have six equations:1. a + b + c + d + e + f = -12. -a + b - c + d - e + f = 13. 5a + 4b + 3c + 2d + e = 04. 5a - 4b + 3c - 2d + e = 05. 20a + 12b + 6c + 2d = 06. -20a + 12b - 6c + 2d = 0Okay, let's write these equations down:Equation 1: a + b + c + d + e + f = -1Equation 2: -a + b - c + d - e + f = 1Equation 3: 5a + 4b + 3c + 2d + e = 0Equation 4: 5a - 4b + 3c - 2d + e = 0Equation 5: 20a + 12b + 6c + 2d = 0Equation 6: -20a + 12b - 6c + 2d = 0Now, we can attempt to solve this system step by step. Let's see.First, equations 5 and 6 are only in variables a, b, c, d. Let's write them again:Equation 5: 20a + 12b + 6c + 2d = 0Equation 6: -20a + 12b - 6c + 2d = 0If we add equations 5 and 6, we get:(20a - 20a) + (12b + 12b) + (6c - 6c) + (2d + 2d) = 0 + 0Simplifies to:0 + 24b + 0 + 4d = 0 → 24b + 4d = 0 → Dividing by 4: 6b + d = 0 → d = -6bSimilarly, subtracting equation 6 from equation 5:(20a + 20a) + (12b - 12b) + (6c + 6c) + (2d - 2d) = 0 - 0Simplifies to:40a + 0 + 12c + 0 = 0 → 40a + 12c = 0 → Divide by 4: 10a + 3c = 0 → c = -10a/3So, from equations 5 and 6, we have d = -6b and c = -10a/3.Okay, note that.Now, moving to equations 3 and 4:Equation 3: 5a + 4b + 3c + 2d + e = 0Equation 4: 5a - 4b + 3c - 2d + e = 0We can substitute c and d in terms of a and b into these equations. Let's do that.From above, c = -10a/3 and d = -6b.Substitute into equation 3:5a + 4b + 3*(-10a/3) + 2*(-6b) + e = 0Simplify:5a + 4b -10a -12b + e = 0 → (5a -10a) + (4b -12b) + e = 0 → -5a -8b + e = 0 → e = 5a + 8bSimilarly, substitute into equation 4:5a -4b + 3*(-10a/3) -2*(-6b) + e = 0Simplify:5a -4b -10a +12b + e = 0 → (5a -10a) + (-4b +12b) + e = 0 → -5a +8b + e = 0 → e =5a -8bSo, from equation 3: e = 5a +8bFrom equation 4: e =5a -8bBut if both equal e, then 5a +8b =5a -8b → 16b =0 → b=0So, 16b =0 → b=0Therefore, b=0Then, from d = -6b, d=0From c= -10a/3, c remains as -10a/3From e =5a +8b =5a +0=5aBut from equation 4: e=5a -8b=5a -0=5a, which is consistent.So, now, so far, we have:b=0, d=0, c= -10a/3, e=5aNow, proceed to equations 1 and 2.Equation 1: a + b + c + d + e + f = -1Plugging in known values:a + 0 + (-10a/3) +0 +5a + f = -1Simplify:a -10a/3 +5a + f = -1Convert to thirds:(3a/3 -10a/3 +15a/3) + f = -1 → (8a/3) + f = -1 → f = -1 -8a/3Equation 2: -a + b - c + d - e + f =1Plugging in known values:-a +0 -(-10a/3) +0 -5a +f =1Simplify:-a +10a/3 -5a +f =1Convert to thirds:(-3a/3 +10a/3 -15a/3) +f =1 → (-8a/3) +f =1But from equation 1, f = -1 -8a/3, so substitute into equation 2:-8a/3 + (-1 -8a/3) =1 → -8a/3 -1 -8a/3 =1 → (-16a/3) -1 =1 → -16a/3 =2 → a= 2*(-3)/16= -6/16= -3/8Therefore, a= -3/8Now, compute the other variables:c= -10a/3= -10*(-3/8)/3= (30/8)/3=10/8=5/4e=5a=5*(-3/8)= -15/8f= -1 -8a/3= -1 -8*(-3/8)/3= -1 + (24/8)/3= -1 + 3/3= -1 +1=0So, f=0So, summarizing:a= -3/8b=0c=5/4d=0e= -15/8f=0Therefore, the polynomial is:p(x)= a x^5 +b x^4 +c x^3 +d x^2 +e x +fSubstituting the values:p(x)= (-3/8)x^5 +0x^4 + (5/4)x^3 +0x^2 + (-15/8)x +0Simplify:p(x)= (-3/8)x^5 + (5/4)x^3 -15/8xWe can write this with a common denominator of 8:p(x)= (-3x^5 +10x^3 -15x)/8So, p(x)= (-3x^5 +10x^3 -15x)/8Let me check if this satisfies the conditions.First, check p(1) +1:p(1)= (-3 +10 -15)/8= (-8)/8= -1 → -1 +1=0 ✔️Similarly, p(-1)-1:p(-1)= (-3*(-1)^5 +10*(-1)^3 -15*(-1))/8= (3 -10 +15)/8=8/8=1 → 1 -1=0 ✔️Now check the first derivative at 1 and -1.Compute p'(x)= (-15x^4 +30x^2 -15)/8At x=1:p'(1)= (-15 +30 -15)/8=0/8=0 ✔️At x=-1:p'(-1)= (-15*(-1)^4 +30*(-1)^2 -15)/8= (-15 +30 -15)/8=0/8=0 ✔️Now check the second derivative at 1 and -1.Compute p''(x)= (-60x^3 +60x)/8At x=1:p''(1)= (-60 +60)/8=0/8=0 ✔️At x=-1:p''(-1)= (-60*(-1)^3 +60*(-1))/8= (60 -60)/8=0/8=0 ✔️So, all conditions are satisfied. Therefore, this polynomial works.But the problem says "find all polynomials p(x) of degree 5..." So, is this the only one?Wait, in our system, we had six equations for six variables, and we found a unique solution. Therefore, this is the only quintic polynomial satisfying the conditions.But let me think again. The problem says "find all polynomials p(x) of degree 5...", so unless there's a family of solutions, but since we had a linear system of six equations with six variables, and assuming the determinant is non-zero, which in this case it seems to be since we found a unique solution, so this is the only one.Therefore, the answer is p(x)= (-3x^5 +10x^3 -15x)/8But let me write it in a factored form or simplified.Alternatively, factor out x:p(x)= (-3x^5 +10x^3 -15x)/8 = x*(-3x^4 +10x^2 -15)/8But maybe leave it as is.Alternatively, check if it can be written as something else. Let me compute:But perhaps that's the simplest form.Alternatively, write it as p(x)= -frac{3}{8}x^5 + frac{5}{4}x^3 - frac{15}{8}xWhich is acceptable.But let me confirm once more if there could be other solutions. Since we have a system of six linear equations, if the system is non-degenerate, only one solution exists. Since we solved step by step and found a unique solution, I think that's the only one.Therefore, the final answer is p(x)= (-3x^5 +10x^3 -15x)/8Final AnswerThe polynomial is boxed{-dfrac{3}{8}x^5 + dfrac{5}{4}x^3 - dfrac{15}{8}x}.</think>To find all polynomials ( p(x) ) of degree 5 such that ( p(x) + 1 ) is divisible by ( (x-1)^3 ) and ( p(x) - 1 ) is divisible by ( (x+1)^3 ), we start by noting that these divisibility conditions imply that ( p(x) ) must satisfy certain value and derivative conditions at ( x = 1 ) and ( x = -1 ).Given that ( p(x) ) is a fifth-degree polynomial, we express it as ( p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f ). The conditions translate to the following system of equations:1. ( p(1) + 1 = 0 )2. ( p(-1) - 1 = 0 )3. ( p'(1) = 0 )4. ( p'(-1) = 0 )5. ( p''(1) = 0 )6. ( p''(-1) = 0 )These conditions yield the following equations:1. ( a + b + c + d + e + f = -1 )2. ( -a + b - c + d - e + f = 1 )3. ( 5a + 4b + 3c + 2d + e = 0 )4. ( 5a - 4b + 3c - 2d + e = 0 )5. ( 20a + 12b + 6c + 2d = 0 )6. ( -20a + 12b - 6c + 2d = 0 )By solving these equations step-by-step, we find:- From equations 5 and 6: ( 6b + d = 0 ) and ( 10a + 3c = 0 ), leading to ( d = -6b ) and ( c = -frac{10a}{3} ).- From equations 3 and 4: Substituting ( c ) and ( d ), we find ( b = 0 ), leading to ( d = 0 ), ( c = -frac{10a}{3} ), and ( e = 5a ).- From equations 1 and 2: Substituting the known values, we solve for ( a ), ( f ), and confirm ( a = -frac{3}{8} ), ( c = frac{5}{4} ), ( e = -frac{15}{8} ), and ( f = 0 ).Thus, the polynomial ( p(x) ) is:[p(x) = -frac{3}{8}x^5 + frac{5}{4}x^3 - frac{15}{8}x]Verifying all conditions confirms this solution is correct. Therefore, the final answer is:[boxed{-dfrac{3}{8}x^5 + dfrac{5}{4}x^3 - dfrac{15}{8}x}],<think>Okay, so I need to find all polynomials p(x) of degree 5 such that p(x) + 1 is divisible by (x - 1)^3 and p(x) - 1 is divisible by (x + 1)^3. Hmm, let me start by understanding what this means.First, if p(x) + 1 is divisible by (x - 1)^3, then (x - 1)^3 must be a factor of p(x) + 1. Similarly, (x + 1)^3 must be a factor of p(x) - 1. So, that means when we plug x = 1 into p(x) + 1, we get zero, right? Same with x = -1 for p(x) - 1. But since it's divisible by (x - 1)^3 and (x + 1)^3, it's not just the value that's zero, but also the first and second derivatives at those points. So, the polynomial p(x) must satisfy certain conditions at x = 1 and x = -1.Let me write down the conditions explicitly. If (x - 1)^3 divides p(x) + 1, then p(1) + 1 = 0, p'(1) = 0, and p''(1) = 0. Similarly, if (x + 1)^3 divides p(x) - 1, then p(-1) - 1 = 0, p'(-1) = 0, and p''(-1) = 0. So, there are six conditions here: three at x = 1 and three at x = -1.Since p(x) is a degree 5 polynomial, it has six coefficients (from x^5 down to the constant term). So, in theory, these six conditions should uniquely determine the polynomial. But maybe there are multiple solutions? The problem says "find all polynomials," so perhaps there is a family of solutions? Wait, but six equations with six variables should typically have a unique solution unless the system is dependent. Hmm, maybe the constraints are compatible, leading to a unique polynomial. Let me check.Let me denote p(x) as a general fifth-degree polynomial:p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + fThen, p'(x) = 5ax^4 + 4bx^3 + 3cx^2 + 2dx + ep''(x) = 20ax^3 + 12bx^2 + 6cx + 2dNow, let's apply the conditions at x = 1:1. p(1) + 1 = 0 => a(1)^5 + b(1)^4 + c(1)^3 + d(1)^2 + e(1) + f + 1 = 0Which simplifies to: a + b + c + d + e + f + 1 = 02. p'(1) = 0 => 5a + 4b + 3c + 2d + e = 03. p''(1) = 0 => 20a + 12b + 6c + 2d = 0Similarly, at x = -1:4. p(-1) - 1 = 0 => a(-1)^5 + b(-1)^4 + c(-1)^3 + d(-1)^2 + e(-1) + f - 1 = 0Which simplifies to: -a + b - c + d - e + f - 1 = 05. p'(-1) = 0 => 5a(-1)^4 + 4b(-1)^3 + 3c(-1)^2 + 2d(-1) + e = 0Which is: 5a - 4b + 3c - 2d + e = 06. p''(-1) = 0 => 20a(-1)^3 + 12b(-1)^2 + 6c(-1) + 2d = 0Which is: -20a + 12b - 6c + 2d = 0So, now we have six equations:1. a + b + c + d + e + f = -12. 5a + 4b + 3c + 2d + e = 03. 20a + 12b + 6c + 2d = 04. -a + b - c + d - e + f = 15. 5a - 4b + 3c - 2d + e = 06. -20a + 12b - 6c + 2d = 0Now, I need to solve this system of equations for a, b, c, d, e, f. Let me see how to approach this. It's a linear system, so I can use substitution or elimination. Let me try to eliminate variables step by step.First, let's note that equations 3 and 6 might be related. Let's write them again:Equation 3: 20a + 12b + 6c + 2d = 0Equation 6: -20a + 12b - 6c + 2d = 0If we add these two equations:(20a - 20a) + (12b + 12b) + (6c - 6c) + (2d + 2d) = 0 + 0Which gives:0 + 24b + 0 + 4d = 0 => 24b + 4d = 0 => Divide by 4: 6b + d = 0 => d = -6bSimilarly, subtract equation 6 from equation 3:(20a + 20a) + (12b - 12b) + (6c + 6c) + (2d - 2d) = 0 - 0Which gives:40a + 0 + 12c + 0 = 0 => 40a + 12c = 0 => Divide by 4: 10a + 3c = 0 => 3c = -10a => c = (-10/3)aSo from equations 3 and 6, we get c = (-10/3)a and d = -6b. Let's note that.Now, let's look at equations 2 and 5. Let's write them again:Equation 2: 5a + 4b + 3c + 2d + e = 0Equation 5: 5a - 4b + 3c - 2d + e = 0If we subtract equation 5 from equation 2:(5a - 5a) + (4b + 4b) + (3c - 3c) + (2d + 2d) + (e - e) = 0 - 0Which gives:0 + 8b + 0 + 4d + 0 = 0 => 8b + 4d = 0 => 2b + d = 0But we already have d = -6b. Plugging that in:2b + (-6b) = -4b = 0 => b = 0If b = 0, then d = -6b = 0. So, d = 0, b = 0.Also, since c = (-10/3)a, and c is expressed in terms of a.So now, we can substitute b = 0, d = 0, c = (-10/3)a into other equations.Let's move to equations 1 and 4:Equation 1: a + b + c + d + e + f = -1Substituting b = 0, d = 0, c = (-10/3)a:a + 0 + (-10/3)a + 0 + e + f = -1Combine like terms: a - (10/3)a + e + f = -1 => (-7/3)a + e + f = -1Equation 4: -a + b - c + d - e + f = 1Substituting b = 0, d = 0, c = (-10/3)a:-a + 0 - (-10/3)a + 0 - e + f = 1Simplify: -a + (10/3)a - e + f = 1 => (7/3)a - e + f = 1Now, we have two equations:Equation 1a: (-7/3)a + e + f = -1Equation 4a: (7/3)a - e + f = 1Let's add equations 1a and 4a:[(-7/3)a + (7/3)a] + [e - e] + [f + f] = -1 + 1Which gives:0 + 0 + 2f = 0 => 2f = 0 => f = 0So f = 0. Now, substitute f = 0 into equation 1a:(-7/3)a + e + 0 = -1 => (-7/3)a + e = -1 => e = (7/3)a - 1Similarly, substitute f = 0 into equation 4a:(7/3)a - e + 0 = 1 => -e = 1 - (7/3)a => e = (7/3)a - 1So that's consistent, so e is expressed in terms of a.Now, we need to find equations involving a. Let's look at equations 2 and 5.But equations 2 and 5 after substituting b = 0, d = 0, c = (-10/3)a:Equation 2: 5a + 4*0 + 3*(-10/3)a + 2*0 + e = 0 => 5a -10a + e = 0 => -5a + e = 0 => e = 5aBut earlier, we found e = (7/3)a - 1So set 5a = (7/3)a -1Multiply both sides by 3 to eliminate denominator:15a = 7a - 315a -7a = -38a = -3 => a = -3/8So a = -3/8Then, e = 5a = 5*(-3/8) = -15/8Check with e = (7/3)a -1:(7/3)*(-3/8) -1 = (-21/24) -1 = (-7/8) -1 = (-7/8 - 8/8) = -15/8. Yes, that matches.Now, since a = -3/8, c = (-10/3)a = (-10/3)*(-3/8) = 10/8 = 5/4So c = 5/4So now we have:a = -3/8b = 0c = 5/4d = 0e = -15/8f = 0So the polynomial is:p(x) = a x^5 + b x^4 + c x^3 + d x^2 + e x + fPlugging in the values:p(x) = (-3/8)x^5 + 0x^4 + (5/4)x^3 + 0x^2 + (-15/8)x + 0Simplify:p(x) = (-3/8)x^5 + (5/4)x^3 - (15/8)xWe can factor out x:p(x) = x [ (-3/8)x^4 + (5/4)x^2 - 15/8 ]But maybe we can write it with common denominators or factor further. Let's see:First, let's factor out -1/8 from the polynomial inside the brackets:-3/8 x^4 + 5/4 x^2 -15/8 = (-1/8)(3x^4 -10x^2 +15)Wait, let's compute:-3/8 x^4 = (-1/8)(3x^4)5/4 x^2 = (10/8)x^2 = (1/8)(10x^2)-15/8 = (-1/8)(15)So altogether:(-1/8)(3x^4 -10x^2 +15)Therefore, p(x) = x*(-1/8)(3x^4 -10x^2 +15) = (-1/8)x(3x^4 -10x^2 +15)But maybe it's better to write it as is, or to check if 3x^4 -10x^2 +15 can be factored, but I don't think so. Let me check discriminant for quadratic in x^2.Let me set y = x^2, so 3y^2 -10y +15. Discriminant is 100 - 180 = -80 < 0, so it doesn't factor over real numbers, so perhaps that's as simplified as it gets.But let me check if the original polynomial is correct by verifying the conditions.First, let's check p(1) +1:p(1) = (-3/8)(1) + (5/4)(1) - (15/8)(1) = (-3/8 + 10/8 -15/8) = (-3 +10 -15)/8 = (-8)/8 = -1. Therefore, p(1) +1 = -1 +1 =0. Good.Check p'(1):p'(x) = derivative of (-3/8)x^5 + (5/4)x^3 - (15/8)xp'(x) = (-15/8)x^4 + (15/4)x^2 -15/8At x =1:p'(1) = (-15/8) + (15/4) -15/8 = (-15/8 -15/8) +15/4 = (-30/8) +30/8 = 0. Correct.p''(x) = derivative of p'(x):p''(x) = (-60/8)x^3 + (30/4)xSimplify: (-15/2)x^3 + (15/2)xAt x=1:p''(1) = (-15/2) +15/2 = 0. Correct.Now check at x = -1:p(-1) -1:p(-1) = (-3/8)(-1)^5 + (5/4)(-1)^3 - (15/8)(-1) = (-3/8)(-1) + (5/4)(-1) - (15/8)(-1) = 3/8 -5/4 +15/8Convert to eighths: 3/8 -10/8 +15/8 = (3 -10 +15)/8 =8/8=1. Therefore, p(-1)-1=1 -1=0. Correct.Check p'(-1):p'(-1) = (-15/8)(-1)^4 + (15/4)(-1)^2 -15/8 = (-15/8)(1) + (15/4)(1) -15/8 = -15/8 +30/8 -15/8 =0. Correct.p''(-1) = (-15/2)(-1)^3 + (15/2)(-1) = (-15/2)(-1) + (15/2)(-1) =15/2 -15/2=0. Correct.All conditions are satisfied. Therefore, this polynomial is a solution.But the problem says "Find all polynomials p(x) of degree 5..." So is this the only one? Since we have six equations with six variables and the solution exists and is unique, then this should be the only polynomial. However, maybe there's a way to parametrize it with lower-degree terms, but since we have a unique solution, I think this is the only one.But let me think again. The problem says "Find all polynomials," so perhaps there is a family? Wait, but since the degree is fixed at 5, and we have six linear equations with six variables, and the system has a unique solution, so this is the only such polynomial. So the answer should be this polynomial.But let me check if the polynomial is indeed degree 5. The leading term is (-3/8)x^5, so yes, it's degree 5. So the answer is p(x) = (-3/8)x^5 + (5/4)x^3 - (15/8)x. To make it look nicer, maybe factor out 1/8:p(x) = (1/8)(-3x^5 +10x^3 -15x)Alternatively, factor out -1/8:p(x) = (-1/8)(3x^5 -10x^3 +15x)Either way is acceptable. Let me check the coefficients again to be sure.Yes, the coefficients we found were:a = -3/8, b=0, c=5/4, d=0, e=-15/8, f=0. So when written as:(-3/8)x^5 + 0x^4 + (5/4)x^3 +0x^2 + (-15/8)x +0, which simplifies to (-3/8)x^5 + (5/4)x^3 - (15/8)x. So that's correct.Alternatively, if we factor out 1/8, we get:1/8*(-3x^5 +10x^3 -15x). Since 5/4 is 10/8 and -15/8 is -15/8. So yes, that's correct.Therefore, the only polynomial satisfying the given conditions is p(x) = (-3/8)x^5 + (5/4)x^3 - (15/8)x, or written as (1/8)(-3x^5 +10x^3 -15x).But let me check if there is a possibility of another polynomial. Suppose there is another polynomial q(x) of degree 5 satisfying the same conditions. Then, the difference p(x) - q(x) would be a polynomial of degree at most 5 that satisfies (p(x) - q(x)) divisible by both (x -1)^3 and (x +1)^3. Therefore, (x -1)^3(x +1)^3 divides p(x) - q(x). But (x -1)^3(x +1)^3 is a degree 6 polynomial, and p(x) - q(x) is degree at most 5. Therefore, the only possibility is that p(x) - q(x) is the zero polynomial, so p(x) = q(x). Therefore, the solution is unique.Therefore, the answer is unique and as found above.Final AnswerThe polynomial is boxed{-dfrac{3}{8}x^5 + dfrac{5}{4}x^3 - dfrac{15}{8}x}.</think>To find all polynomials ( p(x) ) of degree 5 such that ( p(x) + 1 ) is divisible by ( (x-1)^3 ) and ( p(x) - 1 ) is divisible by ( (x+1)^3 ), we start by noting the conditions these divisibilities impose. Specifically, ( p(x) ) must satisfy:1. ( p(1) + 1 = 0 )2. ( p'(1) = 0 )3. ( p''(1) = 0 )4. ( p(-1) - 1 = 0 )5. ( p'(-1) = 0 )6. ( p''(-1) = 0 )We express ( p(x) ) as a general fifth-degree polynomial:[ p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f ]We then compute the first and second derivatives and substitute the conditions into the polynomial and its derivatives. This results in a system of linear equations:1. ( a + b + c + d + e + f + 1 = 0 )2. ( 5a + 4b + 3c + 2d + e = 0 )3. ( 20a + 12b + 6c + 2d = 0 )4. ( -a + b - c + d - e + f - 1 = 0 )5. ( 5a - 4b + 3c - 2d + e = 0 )6. ( -20a + 12b - 6c + 2d = 0 )By solving this system, we find:- ( d = -6b )- ( c = -frac{10}{3}a )- ( b = 0 )- ( d = 0 )- ( f = 0 )- ( a = -frac{3}{8} )- ( c = frac{5}{4} )- ( e = -frac{15}{8} )Substituting these values back into the polynomial, we get:[ p(x) = -frac{3}{8}x^5 + frac{5}{4}x^3 - frac{15}{8}x ]Verification confirms that this polynomial satisfies all the given conditions. Thus, the unique solution is:[boxed{-dfrac{3}{8}x^5 + dfrac{5}{4}x^3 - dfrac{15}{8}x}]]

🔑:p'(x) has a factor (x-1) 2 and a factor (x+1) 2 , but it has degree 4, so p'(x) = a(x 2 -1) 2 = ax 4 - 2ax 2 + a. Integrating, p(x) = ax 5 /5 - 2ax 3 /3 + ax + b. We have p(1) = -1, so a/5 - 2a/3 + a + b = -1, or 8a + 15b = -15. Also p(-1) = 1, so -a/5 + 2a/3 - a + b = 1, or -8a + 15b = 15. Hence a = -15/8, b = 0 and p(x) = (-3x 5 + 10x 3 - 15x)/8 Thanks to Suat Namli 21st Swedish 1981 © John Scholes [email protected] 27 February 2004 Last corrected/updated 27 Feb 04

🧠:[<think>Okay, so I need to prove that if ( x^{2} + y^{2} leqslant 1 ), then the absolute value of ( y^{2} + 2xy - x^{2} ) is less than or equal to ( sqrt{2} ). Hmm, let's start by understanding what the problem is asking. First, the condition ( x^{2} + y^{2} leqslant 1 ) defines a unit disk in the xy-plane, including the boundary. So, all the points (x, y) lie inside or on the circle with radius 1 centered at the origin. The expression we need to handle is ( | y^{2} + 2xy - x^{2} | leq sqrt{2} ). The goal is to show that no matter what x and y are, as long as they satisfy the circle equation, this absolute value expression won't exceed ( sqrt{2} ).Let me first try to simplify or rewrite the expression inside the absolute value. The expression is ( y^{2} + 2xy - x^{2} ). Let me see if I can factor this or perhaps recognize it as part of a known identity. Looking at the terms, I notice that ( y^{2} - x^{2} ) is a difference of squares, which factors into ( (y - x)(y + x) ), but then there's the middle term ( 2xy ). So putting it all together: ( y^{2} + 2xy - x^{2} ). Let me rearrange the terms: ( (y^{2} - x^{2}) + 2xy ). Hmm, that's interesting. The first part is ( (y - x)(y + x) ), and then we add ( 2xy ).Alternatively, maybe we can write this expression in terms of (x + y)^2 or something similar. Let's compute ( (x + y)^2 ), which is ( x^{2} + 2xy + y^{2} ). Comparing this with our expression ( y^{2} + 2xy - x^{2} ), they both have ( y^{2} + 2xy ), but our expression subtracts ( x^{2} ) instead of adding it. So, if we take ( (x + y)^2 - 2x^{2} ), that would be ( x^{2} + 2xy + y^{2} - 2x^{2} = -x^{2} + 2xy + y^{2} ), which matches our expression. So, ( y^{2} + 2xy - x^{2} = (x + y)^2 - 2x^{2} ). Not sure if that helps yet.Alternatively, maybe using trigonometric parameterization. Since ( x^2 + y^2 leq 1 ), we can parameterize x and y using polar coordinates. Let me set ( x = rcostheta ) and ( y = rsintheta ), where ( 0 leq r leq 1 ) and ( 0 leq theta < 2pi ). Then substitute these into the expression.So substituting, the expression becomes:( y^{2} + 2xy - x^{2} = (rsintheta)^2 + 2(rcostheta)(rsintheta) - (rcostheta)^2 )Let's compute each term:1. ( (rsintheta)^2 = r^2 sin^2 theta )2. ( 2(rcostheta)(rsintheta) = 2r^2 costheta sintheta )3. ( (rcostheta)^2 = r^2 cos^2 theta )Putting them all together:( r^2 sin^2 theta + 2r^2 costheta sintheta - r^2 cos^2 theta )Factor out ( r^2 ):( r^2 [ sin^2 theta + 2 costheta sintheta - cos^2 theta ] )Now, let's look at the expression inside the brackets. Let me denote that as A:( A = sin^2 theta + 2 costheta sintheta - cos^2 theta )Simplify A:First, notice that ( sin^2 theta - cos^2 theta = - cos(2theta) ). So:( A = - cos(2theta) + 2 costheta sintheta )But ( 2 costheta sintheta = sin(2theta) ). Therefore:( A = - cos(2theta) + sin(2theta) )So, A can be written as ( sin(2theta) - cos(2theta) ).Therefore, the original expression becomes:( r^2 [ sin(2theta) - cos(2theta) ] )Therefore, the absolute value of the original expression is:( | y^{2} + 2xy - x^{2} | = | r^2 [ sin(2theta) - cos(2theta) ] | = r^2 | sin(2theta) - cos(2theta) | )Since ( r leq 1 ), ( r^2 leq r leq 1 ), so the maximum value of this expression would be at most the maximum of ( | sin(2theta) - cos(2theta) | ), multiplied by 1. So, we need to find the maximum of ( | sin(2theta) - cos(2theta) | ).To find the maximum of ( | sin(2theta) - cos(2theta) | ), we can use the formula for ( a sin x + b cos x ), which can be written as ( sqrt{a^2 + b^2} sin(x + phi) ), where ( phi ) is some phase shift. But here, we have ( sin(2theta) - cos(2theta) ), which is like ( 1 cdot sin(2theta) + (-1) cdot cos(2theta) ). So, the amplitude is ( sqrt{1^2 + (-1)^2} = sqrt{2} ). Therefore, the maximum value of ( | sin(2theta) - cos(2theta) | ) is ( sqrt{2} ).Therefore, the original expression:( | y^{2} + 2xy - x^{2} | = r^2 | sin(2theta) - cos(2theta) | leq r^2 cdot sqrt{2} leq 1 cdot sqrt{2} = sqrt{2} )Hence, the absolute value is less than or equal to ( sqrt{2} ), as required. Wait, but is this correct? Let me double-check. So, because ( r leq 1 ), then ( r^2 leq 1 ), so multiplying ( sqrt{2} ) by ( r^2 ) would make it at most ( sqrt{2} ). That seems okay. But let me test with specific points to see if the maximum is actually achieved.Suppose ( r = 1 ), so we are on the boundary of the unit circle. Let's find the maximum of ( | sin(2theta) - cos(2theta) | ). As we said, this expression has a maximum amplitude of ( sqrt{2} ). For example, when does ( sin(2theta) - cos(2theta) = sqrt{2} )? Let's solve for ( theta ).Set ( sin(2theta) - cos(2theta) = sqrt{2} ). Let's write this as:( sqrt{2} left( frac{1}{sqrt{2}} sin(2theta) - frac{1}{sqrt{2}} cos(2theta) right) = sqrt{2} )Which simplifies to:( sqrt{2} sin(2theta - 45^circ) = sqrt{2} )Therefore, ( sin(2theta - 45^circ) = 1 ), so ( 2theta - 45^circ = 90^circ + 360^circ k ), leading to ( 2theta = 135^circ + 360^circ k ), so ( theta = 67.5^circ + 180^circ k ). Therefore, at ( theta = 67.5^circ ), which is 67.5 degrees, the expression ( sin(2theta) - cos(2theta) ) reaches ( sqrt{2} ).Therefore, plugging this back into x and y. Let me compute x and y when ( theta = 67.5^circ ), which is ( theta = frac{3pi}{8} ) radians.( x = cosleft( frac{3pi}{8} right) ), ( y = sinleft( frac{3pi}{8} right) ).Then, compute ( y^2 + 2xy - x^2 ).First, compute ( y^2 - x^2 ):( sin^2 theta - cos^2 theta = -cos(2theta) = -cosleft( frac{3pi}{4} right) = -(-frac{sqrt{2}}{2}) = frac{sqrt{2}}{2} )Then, compute ( 2xy ):( 2 costheta sintheta = sin(2theta) = sinleft( frac{3pi}{4} right) = frac{sqrt{2}}{2} )Therefore, total expression:( frac{sqrt{2}}{2} + frac{sqrt{2}}{2} = sqrt{2} )So, indeed, the expression reaches ( sqrt{2} ), and since we take the absolute value, the maximum is ( sqrt{2} ).Therefore, the maximum value of ( | y^2 + 2xy - x^2 | ) is ( sqrt{2} ), achieved when ( r = 1 ) and ( theta = 67.5^circ ) or equivalent angles. Thus, the inequality holds.Alternatively, another approach without using polar coordinates. Let's consider the expression ( y^2 + 2xy - x^2 ). Maybe we can rewrite this expression in terms of other variables or complete the square.Let me try to see if this can be represented as a quadratic form or something. Let me denote the expression as Q = ( y^2 + 2xy - x^2 ). Let me write this as a quadratic in variables x and y:Q = ( -x^2 + 2xy + y^2 )This can be represented in matrix form as:[Q = begin{bmatrix} x & y end{bmatrix} begin{bmatrix} -1 & 1 1 & 1 end{bmatrix} begin{bmatrix} x y end{bmatrix}]The maximum value of Q over the unit disk ( x^2 + y^2 leq 1 ) is given by the maximum eigenvalue of the matrix multiplied by the constraint (since we are dealing with the maximum of a quadratic form over the unit circle). Let me compute the eigenvalues of the matrix.The matrix is:[A = begin{bmatrix} -1 & 1 1 & 1 end{bmatrix}]To find the eigenvalues, solve ( det(A - lambda I) = 0 ):[det begin{bmatrix} -1 - lambda & 1 1 & 1 - lambda end{bmatrix} = (-1 - lambda)(1 - lambda) - (1)(1) = (-1)(1 - lambda) - lambda(1 - lambda) - 1]Wait, expanding directly:First, compute the determinant:( (-1 - lambda)(1 - lambda) - 1 times 1 )First term: ( (-1)(1 - lambda) - lambda(1 - lambda) = -1 + lambda - lambda + lambda^2 = -1 + lambda^2 )Second term: -1So determinant equation is:( (-1 + lambda^2) - 1 = lambda^2 - 2 = 0 )Therefore, ( lambda^2 = 2 ), so ( lambda = sqrt{2} ) and ( lambda = -sqrt{2} )Therefore, the maximum eigenvalue is ( sqrt{2} ), and the minimum is ( -sqrt{2} ). Therefore, the maximum value of Q over the unit circle is ( sqrt{2} ), and the minimum is ( -sqrt{2} ). Therefore, the absolute value |Q| is bounded by ( sqrt{2} ), which is exactly what we needed to prove.That's a more linear algebra approach. So, regardless of the method, the conclusion is the same.Alternatively, another approach could be using the Cauchy-Schwarz inequality. Let's see.The expression ( y^2 + 2xy - x^2 ) can be considered as a linear combination of x and y, but it's actually a quadratic form. Hmm, maybe not straightforward. Alternatively, perhaps consider variables u and v, such that we can diagonalize the quadratic form.But since we've already solved it using polar coordinates and linear algebra, both leading to the same conclusion, I think the proof is solid.Just to recap:1. Expressed x and y in polar coordinates, substituted into the expression, and found that the maximum absolute value is ( sqrt{2} ).2. Treated the expression as a quadratic form and found the maximum eigenvalue, which is ( sqrt{2} ).3. Verified with a specific example where the expression reaches ( sqrt{2} ).All steps lead to the conclusion that ( | y^2 + 2xy - x^2 | leq sqrt{2} ) when ( x^2 + y^2 leq 1 ).Final Answerboxed{sqrt{2}}</think>Given the problem of proving that ( | y^2 + 2xy - x^2 | leqslant sqrt{2} ) under the condition ( x^2 + y^2 leqslant 1 ), we can approach the solution as follows:1. Parameterization using Polar Coordinates: - Let ( x = rcostheta ) and ( y = rsintheta ) where ( 0 leq r leq 1 ) and ( 0 leq theta < 2pi ). - Substitute these into the expression ( y^2 + 2xy - x^2 ): [ (rsintheta)^2 + 2(rcostheta)(rsintheta) - (rcostheta)^2 ] - Simplify the expression: [ r^2 sin^2theta + 2r^2 costheta sintheta - r^2 cos^2theta = r^2 [sin^2theta + 2costheta sintheta - cos^2theta] ] - Recognize the trigonometric identity: [ sin^2theta - cos^2theta = -cos(2theta) quad text{and} quad 2costheta sintheta = sin(2theta) ] - Thus, the expression simplifies to: [ r^2 [sin(2theta) - cos(2theta)] ]2. Maximizing the Expression: - The absolute value becomes: [ | y^2 + 2xy - x^2 | = r^2 | sin(2theta) - cos(2theta) | ] - The term ( | sin(2theta) - cos(2theta) | ) can be rewritten using the amplitude form: [ sqrt{1^2 + (-1)^2} = sqrt{2} ] - Therefore, the maximum value of ( | sin(2theta) - cos(2theta) | ) is ( sqrt{2} ).3. Quadratic Form Approach: - Represent the expression as a quadratic form: [ Q = begin{bmatrix} x & y end{bmatrix} begin{bmatrix} -1 & 1 1 & 1 end{bmatrix} begin{bmatrix} x y end{bmatrix} ] - The eigenvalues of the matrix are ( sqrt{2} ) and ( -sqrt{2} ), giving the maximum absolute value of ( sqrt{2} ).4. Verification: - Check specific points where the maximum value is achieved, confirming that the expression reaches ( sqrt{2} ) when ( r = 1 ) and ( theta = 67.5^circ ).Thus, combining these methods, we conclude that the absolute value ( | y^2 + 2xy - x^2 | ) is bounded by ( sqrt{2} ) under the given condition.[boxed{sqrt{2}}],<think>Okay, so I need to prove that if ( x^{2} + y^{2} leqslant 1 ), then the absolute value of ( y^{2} + 2xy - x^{2} ) is less than or equal to ( sqrt{2} ). Hmm, let's start by understanding what the problem is asking. The condition ( x^{2} + y^{2} leqslant 1 ) means that the point (x, y) lies inside or on the unit circle. The expression we need to bound is ( |y^{2} + 2xy - x^{2}| ). First, maybe I should simplify or rearrange the expression inside the absolute value. Let me write it down again: ( y^{2} + 2xy - x^{2} ). Hmm, combining like terms, I notice that ( y^{2} - x^{2} ) is a difference of squares, which factors into ( (y - x)(y + x) ). So maybe that's useful? Then the expression becomes ( (y - x)(y + x) + 2xy ). Let me check that:( (y - x)(y + x) = y^2 - x^2 ), so adding 2xy gives ( y^2 - x^2 + 2xy ). Yeah, that's correct. But I don't see an immediate simplification here. Maybe another approach.Alternatively, perhaps completing the square? Let's see. Let's group terms:( y^2 + 2xy - x^2 = (y^2 + 2xy + x^2) - 2x^2 = (y + x)^2 - 2x^2 ). Wait, that's ( (y + x)^2 - 2x^2 ). Not sure if that helps.Alternatively, maybe another grouping: ( y^2 - x^2 + 2xy = -(x^2 - 2xy - y^2) ). Hmm, the expression ( x^2 - 2xy - y^2 ) doesn't look familiar. Maybe not useful.Alternatively, think of this as a quadratic form. Maybe we can represent the expression ( y^2 + 2xy - x^2 ) as a matrix product. Let me recall that quadratic forms can be written as ( mathbf{v}^T A mathbf{v} ), where ( mathbf{v} = begin{bmatrix} x y end{bmatrix} ) and A is a symmetric matrix. Let's try that.The expression ( y^2 + 2xy - x^2 ) can be written as:( -x^2 + 2xy + y^2 ). Comparing this to the general quadratic form ( ax^2 + 2bxy + cy^2 ), we have a = -1, 2b = 2 (so b = 1), c = 1. Therefore, the matrix A is:( begin{bmatrix} -1 & 1 1 & 1 end{bmatrix} ).Now, the maximum value of the quadratic form ( mathbf{v}^T A mathbf{v} ) subject to ( mathbf{v}^T mathbf{v} leqslant 1 ) (since ( x^2 + y^2 leqslant 1 )) is given by the maximum eigenvalue of matrix A. Then, the absolute value of the expression would be bounded by the maximum absolute value of the eigenvalues. So maybe that's a way to approach this.Let me compute the eigenvalues of matrix A. The characteristic equation is ( det(A - lambda I) = 0 ).So,( detleft( begin{bmatrix} -1 - lambda & 1 1 & 1 - lambda end{bmatrix} right) = (-1 - lambda)(1 - lambda) - (1)(1) = (-1)(1 - lambda) - lambda(1 - lambda) - 1 ).Wait, better step by step:The determinant is:( (-1 - lambda)(1 - lambda) - 1 times 1 )First, expand ( (-1 - lambda)(1 - lambda) ):Multiply term by term:( (-1)(1) + (-1)(-λ) + (-λ)(1) + (-λ)(-λ) )Which is:( -1 + λ - λ + λ^2 = -1 + 0 + λ^2 = λ^2 - 1 )Then subtract 1:So determinant is ( λ^2 - 1 - 1 = λ^2 - 2 ).Set equal to zero: ( λ^2 - 2 = 0 ), so eigenvalues ( λ = sqrt{2} ) and ( λ = -sqrt{2} ).Therefore, the maximum value of ( mathbf{v}^T A mathbf{v} ) on the unit circle is ( sqrt{2} ), and the minimum is ( -sqrt{2} ). Therefore, the absolute value is bounded by ( sqrt{2} ), which is exactly what we need to prove. So that's a proof using quadratic forms and eigenvalues.But maybe the problem expects a more elementary approach, not using linear algebra. Let me see if I can find another way.Another method could be to use polar coordinates. Since ( x^2 + y^2 leqslant 1 ), we can parametrize x and y as ( x = r cos theta ), ( y = r sin theta ), where ( 0 leqslant r leqslant 1 ) and ( 0 leqslant theta < 2pi ). Then substitute into the expression.Let me try that.Express ( y^2 + 2xy - x^2 ) in terms of r and θ:First, substitute x and y:( y^2 = r^2 sin^2 theta )( 2xy = 2 r cos theta cdot r sin theta = 2 r^2 cos theta sin theta )( -x^2 = -r^2 cos^2 theta )So the entire expression becomes:( r^2 sin^2 theta + 2 r^2 cos theta sin theta - r^2 cos^2 theta )Factor out ( r^2 ):( r^2 [ sin^2 θ + 2 cosθ sinθ - cos^2 θ ] )Now, inside the brackets, let's simplify the expression:( sin²θ - cos²θ + 2 sinθ cosθ )Notice that ( sin²θ - cos²θ = -cos2θ ), and ( 2 sinθ cosθ = sin2θ ). So substituting:( -cos2θ + sin2θ )Therefore, the entire expression becomes:( r^2 ( -cos2θ + sin2θ ) )So, ( | y^2 + 2xy - x^2 | = | r^2 ( -cos2θ + sin2θ ) | leq r^2 | -cos2θ + sin2θ | )Since r ≤ 1, r² ≤ 1, so the maximum value of the expression is at most the maximum of | -cos2θ + sin2θ | over θ. Then we need to find the maximum of | sin2θ - cos2θ |.Hmm, the expression sin2θ - cos2θ. Let me note that sinα - cosα can be written as sqrt(2) sin(α - 45°), using the identity that a sinθ + b cosθ = C sin(θ + φ). Let's recall that.In general, ( a sin theta + b cos theta = sqrt{a^2 + b^2} sin( theta + phi ) ), where ( phi = arctan left( frac{b}{a} right ) ) if a ≠ 0.Wait, actually, it's ( a sin theta + b cos theta = sqrt{a^2 + b^2} sin( theta + phi ) ), where ( phi = arctan left( frac{b}{a} right ) ), but need to check the exact formula. Alternatively, factor out sqrt(a² + b²):Let me consider the expression sin2θ - cos2θ. Let me factor out sqrt(1 + 1) = sqrt(2):sin2θ - cos2θ = sqrt(2) [ (1/sqrt(2)) sin2θ - (1/sqrt(2)) cos2θ ]Note that 1/sqrt(2) is cos(45°), and -1/sqrt(2) is sin(-45°), so this can be written as:sqrt(2) [ sin2θ cos45° - cos2θ sin45° ] = sqrt(2) sin(2θ - 45° )Using the sine subtraction formula: sin(A - B) = sinA cosB - cosA sinB.Therefore, sin2θ - cos2θ = sqrt(2) sin(2θ - 45° )Therefore, | sin2θ - cos2θ | = sqrt(2) | sin(2θ - 45° ) | ≤ sqrt(2) * 1 = sqrt(2)Therefore, the maximum value of | sin2θ - cos2θ | is sqrt(2), achieved when sin(2θ - 45° ) = ±1.Therefore, going back to the original expression:| y² + 2xy - x² | = | r² ( sin2θ - cos2θ ) | ≤ r² * sqrt(2) ≤ 1 * sqrt(2) = sqrt(2)Since r² ≤ 1. Therefore, the maximum possible value is sqrt(2), hence | y² + 2xy - x² | ≤ sqrt(2). Which is exactly what we needed to prove.So that's another approach using polar coordinates and trigonometric identities.Alternatively, maybe using Cauchy-Schwarz inequality? Let's think.We need to bound | y² + 2xy - x² |. Let's consider this expression as a linear combination of x², y², and xy. Alternatively, perhaps express it in terms of vectors.Wait, another idea: the expression y² + 2xy - x² can be written as (y - x)(y + x) + 2xy, but earlier that didn't lead anywhere. Alternatively, maybe think of variables u = x + y and v = x - y. Let's try substitution.Let u = x + y and v = x - y. Then:x = (u + v)/2, y = (u - v)/2Let me substitute into the expression:First, compute y²:y² = [(u - v)/2]^2 = (u² - 2uv + v²)/42xy: 2 * [(u + v)/2] * [(u - v)/2] = 2 * (u² - v²)/4 = (u² - v²)/2-x²: -[(u + v)/2]^2 = -(u² + 2uv + v²)/4So combining all terms:y² + 2xy - x² = (u² - 2uv + v²)/4 + (u² - v²)/2 - (u² + 2uv + v²)/4Let me compute each term:First term: (u² - 2uv + v²)/4Second term: (u² - v²)/2 = (2u² - 2v²)/4Third term: - (u² + 2uv + v²)/4Now add all three terms:[(u² - 2uv + v²) + (2u² - 2v²) - (u² + 2uv + v²)] / 4Compute numerator:u² - 2uv + v² + 2u² - 2v² - u² - 2uv - v²Combine like terms:u² + 2u² - u² = 2u²-2uv -2uv = -4uvv² -2v² -v² = -2v²Therefore, numerator is 2u² -4uv -2v² = 2(u² -2uv -v²)Hmm, this doesn't seem to be simplifying things. Maybe this substitution isn't helpful.Alternatively, perhaps consider variables rotated by 45 degrees. Since the quadratic form has cross terms, maybe a rotation would diagonalize it, which relates to the eigenvalue approach earlier.Alternatively, use Lagrange multipliers. Let's set up the optimization problem: maximize and minimize the expression f(x, y) = y² + 2xy - x² subject to the constraint g(x, y) = x² + y² - 1 ≤ 0. Since the maximum and minimum will occur on the boundary when the constraint is active (i.e., when x² + y² = 1), we can use Lagrange multipliers on the boundary.Set up the Lagrangian: L(x, y, λ) = y² + 2xy - x² - λ(x² + y² - 1)Take partial derivatives:∂L/∂x = 2y - 2x - 2λx = 0∂L/∂y = 2y + 2x - 2λy = 0∂L/∂λ = -(x² + y² -1 ) = 0So the equations are:1. 2y - 2x - 2λx = 0 ⇒ y - x - λx = 0 ⇒ y = x(1 + λ)2. 2y + 2x - 2λy = 0 ⇒ y + x - λy = 0 ⇒ x = (λy - y) = y(λ -1 )3. x² + y² = 1From equation 1: y = x(1 + λ)From equation 2: x = y(λ - 1 )Substitute equation 1 into equation 2:x = [x(1 + λ)](λ -1 )Assuming x ≠ 0 (if x = 0, then from equation 1, y = 0, but then x² + y² = 0 ≠ 1, so x ≠ 0)Divide both sides by x:1 = (1 + λ)(λ - 1 )Expand RHS:(1 + λ)(λ -1 ) = λ^2 -1Therefore, 1 = λ² -1 ⇒ λ² = 2 ⇒ λ = ±√2So two cases: λ = √2 and λ = -√2Case 1: λ = √2From equation 1: y = x(1 + √2 )From equation 2: x = y(√2 -1 )Substitute y from equation 1 into equation 2:x = [x(1 + √2 )](√2 -1 ) = x(1 + √2 )(√2 -1 )Compute (1 + √2 )(√2 -1 ):Multiply out: √2*√2 + (-1)(1) + (-1)(√2) + √2*(-1) ? Wait, better:(1)(√2) + (1)(-1) + (√2)(√2) + (√2)(-1 )= √2 -1 + 2 - √2 = (√2 - √2) + (-1 + 2) = 0 +1 =1Therefore, x = x *1 ⇒ x = x, which is consistent. So from equation 1, y = x(1 + √2 )And from the constraint x² + y² =1:x² + [x(1 + √2 )]^2 =1 ⇒ x² + x²(1 + √2 )² =1 ⇒ x² [1 + (1 + √2 )² ] =1Compute (1 + √2 )² =1 + 2√2 + 2 = 3 + 2√2Therefore, 1 + (1 + √2 )² =1 +3 +2√2=4 +2√2Thus, x² =1 / (4 +2√2 ) = [1 / (2(2 + √2 ))] = [ (2 - √2 ) / (2(2 + √2 )(2 - √2 )) ) ] = [ (2 - √2 ) / (2(4 -2)) ] = (2 - √2 ) / (2*2) ) = (2 - √2 ) /4Therefore, x = ±√( (2 - √2 ) /4 ) = ± (√(2 - √2 )) /2Similarly, y = x(1 + √2 ) = ± (√(2 - √2 ) /2 )(1 + √2 )Compute (√(2 - √2 ))(1 + √2 ):Let me square this to check:[√(2 - √2 )]^2 (1 + √2 )^2 = (2 - √2 )(3 + 2√2 ) = compute this:Multiply 2*(3 + 2√2 ) =6 +4√2Multiply -√2*(3 +2√2 )= -3√2 - 2*(√2 )^2 = -3√2 -4So total: 6 +4√2 -3√2 -4 =2 +√2Therefore, [√(2 - √2 ) (1 + √2 )]^2 =2 +√2 ⇒ √(2 - √2 ) (1 + √2 ) =√(2 +√2 )Therefore, y = ±√(2 +√2 ) /2Therefore, in case λ=√2, the critical points are (x, y)= ±(√(2 -√2 )/2, √(2 +√2 )/2 )Similarly, for λ = -√2:From equation 1: y = x(1 + (-√2 )) = x(1 - √2 )From equation 2: x = y(-√2 -1 )Substitute y from equation 1 into equation 2:x = [x(1 - √2 )](-√2 -1 ) =x(1 - √2 )(-√2 -1 )Compute (1 - √2 )(-√2 -1 ):Multiply out:1*(-√2 ) +1*(-1 ) + (-√2 )*(-√2 ) + (-√2 )*(-1 )= -√2 -1 +2 +√2 = (-√2 +√2 ) + (-1 +2 )=0 +1=1Thus, x =x*1 ⇒ x=x, which holds. Then from equation 1, y =x(1 - √2 )From the constraint x² + y² =1:x² + [x(1 - √2 )]^2 =1 ⇒x² [1 + (1 - √2 )² ]=1Compute (1 - √2 )² =1 -2√2 +2=3 -2√2Thus, 1 + (3 -2√2 )=4 -2√2Therefore, x²=1/(4 -2√2 )= [1]/[2(2 -√2 )] = multiply numerator and denominator by (2 +√2 ):[1*(2 +√2 )]/[2(2 -√2 )(2 +√2 ) ] = (2 +√2 )/[2(4 -2)] = (2 +√2 )/(4 )Thus, x=±√( (2 +√2 )/4 )=±√(2 +√2 ) /2Similarly, y= x(1 -√2 )=±√(2 +√2 ) /2 (1 -√2 )Again, compute (√(2 +√2 ))(1 -√2 ):Square this:[√(2 +√2 )]^2 (1 -√2 )^2=(2 +√2 )(3 -2√2 )Multiply:2*(3 -2√2 ) +√2*(3 -2√2 )=6 -4√2 +3√2 -2*(√2 )^2=6 -4√2 +3√2 -4=2 -√2Thus, [√(2 +√2 )(1 -√2 )]^2=2 -√2 ⇒√(2 +√2 )(1 -√2 )=√(2 -√2 )Therefore, y=±√(2 -√2 ) /2Thus, critical points for λ=-√2 are (x, y)=±(√(2 +√2 )/2, √(2 -√2 )/2 )Now, evaluate the expression f(x, y)=y² +2xy -x² at these critical points.First, take the point (√(2 -√2 )/2, √(2 +√2 )/2 )Compute f:y² +2xy -x²Compute each term:y²= [√(2 +√2 )/2]^2= (2 +√2 )/42xy=2*[√(2 -√2 )/2]*[√(2 +√2 )/2 ]=2*(√{(2 -√2 )(2 +√2 )}/4 )Note that (2 -√2 )(2 +√2 )=4 -2=2, so sqrt(2)=√2Thus, 2xy=2*(√2 /4 )= (2√2 )/4=√2 /2-x²= -[√(2 -√2 )/2]^2= -(2 -√2 )/4Therefore, total f= (2 +√2 )/4 + √2 /2 - (2 -√2 )/4Combine terms:[(2 +√2 ) - (2 -√2 )]/4 + √2 /2= [2 +√2 -2 +√2 ]/4 + √2 /2= (2√2 )/4 + √2 /2= √2 /2 + √2 /2 = √2Similarly, for the point (√(2 +√2 )/2, √(2 -√2 )/2 ):Compute f:y² +2xy -x²y²= [√(2 -√2 )/2]^2=(2 -√2 )/42xy=2*[√(2 +√2 )/2]*[√(2 -√2 )/2 ]= same as before= √2 /2-x²= -[√(2 +√2 )/2]^2= -(2 +√2 )/4So f= (2 -√2 )/4 +√2 /2 - (2 +√2 )/4Combine terms:[(2 -√2 ) - (2 +√2 )]/4 + √2 /2= [ -2√2 ]/4 + √2 /2= -√2 /2 + √2 /2=0Wait, that's zero? Wait, maybe I made a mistake here.Wait, let's compute again:f = y² + 2xy - x²y²=(2 -√2 )/42xy=√2 /2-x²= -(2 +√2 )/4Therefore:(2 -√2 )/4 + √2 /2 - (2 +√2 )/4= [ (2 -√2 ) - (2 +√2 ) ]/4 + √2 /2= (2 -√2 -2 -√2 )/4 + √2 /2= (-2√2 )/4 + √2 /2= (-√2 /2 ) + √2 /2 =0Hmm, so this critical point gives f=0. Similarly, for the negative counterparts, since the expression is quadratic, the negative points would give same as positive but multiplied by signs. Wait, but actually, let's check for the point (-√(2 -√2 )/2, -√(2 +√2 )/2 )Compute f:y² +2xy -x²y²= same as before=(2 +√2 )/42xy=2*(-√(2 -√2 )/2)(-√(2 +√2 )/2 )= same as before= √2 /2-x²= -[ (-√(2 -√2 )/2 )]^2= - (2 -√2 )/4Thus, f=(2 +√2 )/4 +√2 /2 - (2 -√2 )/4= same as first case=√2Similarly, for the other negative point.Therefore, the maximum value of f is √2 and the minimum is -√2 (wait, but in the second case, we got zero. Wait, but maybe there's another critical point? Wait, no, when λ=-√2, we get f=0. So, perhaps there are other extrema? Wait, but Lagrange multipliers give critical points which can be maxima, minima, or saddle points. Wait, but with the constraint x² + y²=1, the function f is continuous, so it must attain both maximum and minimum on the compact set.But according to our calculation, when λ=√2, we get f=√2, and when λ=-√2, we get f=0. Wait, that's inconsistent with the previous methods where we found the maximum absolute value is √2. So why is there a critical point where f=0?Wait, perhaps I made a mistake in calculation. Let me check again.Wait, in case λ=√2, we have:x=√(2 -√2 )/2, y=√(2 +√2 )/2, and f=√2.In case λ=-√2, we have x=√(2 +√2 )/2, y=√(2 -√2 )/2, and when substituting into f, we get 0.But that seems contradictory to the eigenvalue approach, where both maximum and minimum are sqrt(2) and -sqrt(2). However, here, we only found f= sqrt(2) and f=0. So where is the -sqrt(2)?Wait, maybe there are other critical points. Let's check if there are solutions when x=0 or y=0.Suppose x=0. Then from the constraint x² + y²=1, y=±1. Then f= y² +0 -0=1. So |f|=1 <= sqrt(2), which is okay.If y=0, then x=±1, f=0 +0 -x²= -1. So f=-1. So |f|=1 <= sqrt(2). So these are not maxima or minima for the absolute value.Alternatively, maybe there are other critical points where the gradient of f is parallel to the gradient of g.Wait, but in the Lagrangian method, we found two critical points with f= sqrt(2) and f=0. How come?Wait, perhaps there's a mistake in the evaluation of f in the second case.Wait, let's compute again for λ=-√2:The critical points are (x, y)=±(√(2 +√2 )/2, √(2 -√2 )/2 )Compute f(x, y)=y² +2xy -x²Compute each term:y²=(2 -√2 )/42xy=2*(√(2 +√2 )/2)*(√(2 -√2 )/2)= 2*(sqrt{(2 +√2 )(2 -√2 )}/4 )Again, (2 +√2 )(2 -√2 )=4 -2=2. So sqrt(2)=√2. Therefore, 2xy=2*(√2 /4 )=√2 /2-x²= - (2 +√2 )/4Thus, f=(2 -√2 )/4 +√2 /2 - (2 +√2 )/4Combine terms:[(2 -√2 ) - (2 +√2 )]/4 +√2 /2= (-2√2 )/4 +√2 /2= (-√2 /2 ) +√2 /2=0So indeed, f=0 here. Similarly, if we take the negative of these points, (-√(2 +√2 )/2, -√(2 -√2 )/2 ), then f=0 as well.Wait, so why do we get f=0 here? Maybe these are saddle points?Wait, but according to the eigenvalue method, the maximum is sqrt(2) and the minimum is -sqrt(2). But according to Lagrangian multipliers, we have maximum sqrt(2) and minimum zero? That can't be. There's a contradiction here. Where is the mistake?Wait, perhaps I made a mistake in the Lagrangian setup. Let me check the equations again.The function to optimize is f(x, y)= y² +2xy -x²The constraint is g(x,y)=x² + y² -1=0The Lagrangian is L= f - λg= y² +2xy -x² -λ(x² + y² -1 )Partial derivatives:dL/dx= 2y -2x -2λx=0 ⇒ 2y -2x(1 +λ )=0 ⇒ y= x(1 +λ )dL/dy= 2y +2x -2λy=0 ⇒ 2x +2y(1 -λ )=0 ⇒ x= y(λ -1 )Wait, wait, let's do it step by step:dL/dy= derivative of y² is 2y, derivative of 2xy is 2x, derivative of -x² is 0, derivative of -λ y² is -2λ y. So:dL/dy=2y +2x -2λ y=0 ⇒ 2x +2y -2λ y=0 ⇒ 2x +2y(1 -λ )=0 ⇒ x + y(1 -λ )=0 ⇒x= y(λ -1 )Which is the same as earlier.So equations are:1. y =x(1 +λ )2. x= y(λ -1 )3. x² + y²=1From 1 and 2: Substitute y from 1 into 2:x= [x(1 +λ )](λ -1 )Thus, x= x(1 +λ )(λ -1 )Assuming x≠0, divide both sides by x:1= (1 +λ )(λ -1 )Which gives λ² -1=1 ⇒ λ²=2 ⇒ λ=±√2Thus, solutions only at λ=√2 and λ=-√2. So only two critical points? But in the eigenvalue approach, the quadratic form has maximum sqrt(2) and minimum -sqrt(2). So there must be a minimum at -sqrt(2). But according to Lagrangian multipliers, we only found f=sqrt(2) and f=0.Hmm, perhaps when solving, we have to consider that when λ=-sqrt(2), then from equation 1: y= x(1 -sqrt(2) )From equation 2: x= y(-sqrt(2) -1 )But substituting y from equation 1 into equation 2 gives x= x(1 -sqrt(2) )(-sqrt(2) -1 )Which simplifies to x= x*( (1 -sqrt(2) )*(-sqrt(2) -1 ) )Calculating (1 -sqrt(2) )*(-sqrt(2) -1 )= -sqrt(2) -1 +2 +sqrt(2 )= (-sqrt(2) +sqrt(2 )) + (-1 +2 )=1Thus, x= x*1 ⇒ x is arbitrary? But no, we derived x= y(λ -1 )Wait, but we already accounted for that. Wait, perhaps when λ=-sqrt(2 ), then the critical points are (sqrt(2 +sqrt(2 )) /2 , sqrt(2 -sqrt(2 )) /2 ), but when substituting into f, we get zero. How come?Wait, maybe I need to check if there are other critical points when y=0 or x=0. But when x=0 or y=0, we already saw f=1 or -1, which are less in absolute value than sqrt(2 ). So, perhaps the minimal value is -sqrt(2 ), but we didn't find it here. Wait, this suggests inconsistency between the Lagrangian approach and the eigenvalue method. Which one is correct?Wait, let's take a point where theta is such that sin2θ - cos2θ = -sqrt(2 ). Then f= y² +2xy -x²= r²*(-sqrt(2 ) ). So when r=1, f=-sqrt(2 ). So there should be a minimum at -sqrt(2 ). But according to the Lagrangian multiplier method, we only found a maximum at sqrt(2 ) and a critical point at zero. What is happening here?Wait, perhaps in the Lagrangian method, we only found critical points for local maxima and minima under the constraint, but maybe there are other critical points where the function attains -sqrt(2 ). Alternatively, perhaps the parametrization in polar coordinates showed that the expression can reach sqrt(2 ) and -sqrt(2 ), but why isn't the Lagrangian method catching the -sqrt(2 )?Wait, maybe I made an error in computing the sign when substituting the critical points into f. Let's check again.Take the critical point for λ=√2: x= sqrt(2 -sqrt(2 )) /2 , y= sqrt(2 +sqrt(2 )) /2Compute f= y² +2xy -x²= (2 +sqrt(2 )) /4 + sqrt(2 )/2 - (2 -sqrt(2 )) /4= as before, which gives sqrt(2 )For the critical point for λ=-sqrt(2 ): x= sqrt(2 +sqrt(2 )) /2 , y= sqrt(2 -sqrt(2 )) /2Compute f= y² +2xy -x²= (2 -sqrt(2 )) /4 + sqrt(2 )/2 - (2 +sqrt(2 )) /4= 0 as before.But where is the -sqrt(2 )?Wait, perhaps there are other critical points when considering different signs. Let me check the point (-sqrt(2 -sqrt(2 )) /2, sqrt(2 +sqrt(2 )) /2 )Compute f= y² +2xy -x²y²=(2 +sqrt(2 )) /42xy=2*(-sqrt(2 -sqrt(2 )) /2 )(sqrt(2 +sqrt(2 )) /2 )= -sqrt(2 )/2-x²= - (2 -sqrt(2 )) /4So f=(2 +sqrt(2 )) /4 - sqrt(2 )/2 - (2 -sqrt(2 )) /4Combine terms:[ (2 +sqrt(2 )) - (2 -sqrt(2 )) ] /4 - sqrt(2 )/2= (2sqrt(2 )) /4 - sqrt(2 )/2= sqrt(2 )/2 - sqrt(2 )/2=0Similarly, the point (sqrt(2 -sqrt(2 )) /2, -sqrt(2 +sqrt(2 )) /2 )Compute f= y² +2xy -x²y²=(2 +sqrt(2 )) /42xy=2*sqrt(2 -sqrt(2 )) /2*(-sqrt(2 +sqrt(2 )) /2 )= -sqrt(2 )/2-x²= - (2 -sqrt(2 )) /4Thus, same as above, f=0.Similarly, the points (-sqrt(2 +sqrt(2 )) /2, -sqrt(2 -sqrt(2 )) /2 ) also gives f=0.So all critical points found via Lagrangian multipliers give either f= sqrt(2 ) or f=0. But according to the polar coordinate method, the expression can go down to -sqrt(2 ). So what's the issue here?Wait, perhaps I missed some critical points in the Lagrangian method. Let me think. When solving the Lagrangian equations, we assumed x≠0. But what if x=0?If x=0, then from the constraint x² + y²=1, y=±1. Then from equation 1 (y= x(1 +λ )), y=0*(1 +λ )=0, which contradicts y=±1. Hence x≠0.Similarly, if y=0, then from equation 2 (x= y(λ -1 )), x=0*(λ -1 )=0, but then x² + y²=0≠1. Hence y≠0. So in the Lagrangian method, we have covered all possibilities where x and y are non-zero. So perhaps the minimum of f is -sqrt(2 ), but achieved at a different point. But how?Wait, let's take theta such that sin2θ - cos2θ= -sqrt(2 ). As per the polar coordinate method, this occurs when sin(2θ -45° )=-1, i.e., 2θ -45°= -90° ⇒2θ= -45° ⇒θ=-22.5° or 337.5°. So take θ=337.5°, which is equivalent to -22.5°.Then x=r cosθ, y=r sinθ. For maximum effect, set r=1.Compute x=cos(337.5°)=cos(-22.5°)=cos(22.5°)=sqrt(2 +sqrt(2 )) /2 ≈0.9239y=sin(-22.5°)= -sin(22.5°)= -sqrt(2 -sqrt(2 )) /2 ≈-0.3827Then compute f=y² +2xy -x²First, y²=(2 -sqrt(2 )) /4≈0.14642xy=2*(sqrt(2 +sqrt(2 )) /2 )*(-sqrt(2 -sqrt(2 )) /2 )= -sqrt(2 ) /2≈-0.7071-x²= - (2 +sqrt(2 )) /4≈-0.8536Thus, f≈0.1464 -0.7071 -0.8536≈0.1464 -1.5607≈-1.4142≈-sqrt(2 )Therefore, f=-sqrt(2 ). So this point achieves the minimum. But why didn't the Lagrangian method find this?Ah, perhaps because when we solved the Lagrangian equations, we found critical points only for certain combinations of signs. Wait, the critical points we found were (sqrt(2 -sqrt(2 )) /2, sqrt(2 +sqrt(2 )) /2 ) and (sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ), and their negatives. But when we take theta=337.5°, we get x positive and y negative, which is a combination not covered by the previous critical points. So perhaps the Lagrangian method missed this because of the symmetry?Wait, in the Lagrangian method, when we derived the critical points, we assumed that x and y have the same sign? Because when we solved for x and y in terms of λ, we had y= x(1 +λ ). For λ=sqrt(2 ), y= x(1 +sqrt(2 )) which, if x is positive, y is positive. Similarly, if x is negative, y is negative. So those points are in the first and third quadrants. Similarly, for λ=-sqrt(2 ), x and y have the same sign again? Wait, no:For λ=-sqrt(2 ), y= x(1 -sqrt(2 )). Since 1 -sqrt(2 ) is negative, if x is positive, y is negative, and if x is negative, y is positive. Therefore, these points are in the fourth and second quadrants. So the critical points found are in all four quadrants. However, when we plug in x= sqrt(2 +sqrt(2 )) /2 and y= sqrt(2 -sqrt(2 )) /2, which is in the first quadrant, and then the negative versions are in third quadrant. Similarly, for the other critical points.But the point we found with theta=337.5° is in the fourth quadrant, which should correspond to one of the critical points from the Lagrangian method. Let's check:Take λ=-sqrt(2 ), the critical points are (x, y)=±(sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 )Wait, but if λ=-sqrt(2 ), from equation 1: y= x(1 -sqrt(2 )) which for positive x gives negative y, but here we have positive y. Wait, no, actually:Wait, when λ=-sqrt(2 ), equation 1: y= x(1 +λ )=x(1 -sqrt(2 )). If x is positive, y is negative. So the critical point in the fourth quadrant is (sqrt(2 +sqrt(2 )) /2, -sqrt(2 -sqrt(2 )) /2 )Wait, but when I computed f at this point, I got:f= y² +2xy -x²= (2 -sqrt(2 )) /4 + 2*(sqrt(2 +sqrt(2 )) /2 )*(-sqrt(2 -sqrt(2 )) /2 ) - (2 +sqrt(2 )) /4= (2 -sqrt(2 )) /4 - sqrt(2 )/2 - (2 +sqrt(2 )) /4= [ (2 -sqrt(2 )) - (2 +sqrt(2 )) ] /4 - sqrt(2 )/2= (-2sqrt(2 )) /4 - sqrt(2 )/2= -sqrt(2 )/2 - sqrt(2 )/2= -sqrt(2 )Ah! So I must have made a mistake earlier. When I computed f at (sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ), I got 0, but for the point (sqrt(2 +sqrt(2 )) /2, -sqrt(2 -sqrt(2 )) /2 ), which corresponds to lambda=-sqrt(2 ) with x positive and y negative, then f=-sqrt(2 ). Similarly, if we take the point (-sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ), we would get f=2xy -x² + y². Let's check:x=-sqrt(2 +sqrt(2 )) /2, y=sqrt(2 -sqrt(2 )) /2f= y² +2xy -x²= (2 -sqrt(2 )) /4 +2*(-sqrt(2 +sqrt(2 )) /2 )(sqrt(2 -sqrt(2 )) /2 ) - (2 +sqrt(2 )) /4= (2 -sqrt(2 )) /4 - sqrt(2 )/2 - (2 +sqrt(2 )) /4= [ (2 -sqrt(2 )) - (2 +sqrt(2 )) ] /4 - sqrt(2 )/2= (-2sqrt(2 )) /4 - sqrt(2 )/2= -sqrt(2 )/2 - sqrt(2 )/2= -sqrt(2 )Therefore, the critical points for lambda=-sqrt(2 ) actually yield f=-sqrt(2 ) when considering the points in the fourth and second quadrants. Earlier, I computed the point in the first quadrant with positive x and positive y, but that point actually isn't on the constraint x² + y²=1 if we consider lambda=-sqrt(2 ), because for lambda=-sqrt(2 ), from equation 1: y= x(1 -sqrt(2 )). If x is positive, y must be negative, so the correct critical point in the fourth quadrant is (sqrt(2 +sqrt(2 )) /2, -sqrt(2 -sqrt(2 )) /2 ), which does satisfy x² + y²=1:x² + y²= (2 +sqrt(2 )) /4 + (2 -sqrt(2 )) /4= (2 +sqrt(2 ) +2 -sqrt(2 )) /4=4/4=1.Therefore, this is a valid point. Similarly, the point (-sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ) also satisfies the constraint and gives f=-sqrt(2 ).Therefore, the Lagrangian method gives both f= sqrt(2 ) and f=-sqrt(2 ) as critical values, as well as the saddle points at f=0. Wait, but earlier when I computed for the point (sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ), which is in the first quadrant, but with lambda=-sqrt(2 ), I must have made a mistake in assigning the signs. Because for lambda=-sqrt(2 ), y should be negative if x is positive. Therefore, the point (sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ) does not satisfy the Lagrangian equations for lambda=-sqrt(2 ), since y should be negative. Therefore, the critical points for lambda=-sqrt(2 ) are in the fourth and second quadrants, giving f=-sqrt(2 ) and f= sqrt(2 ) respectively? Wait, no. Let me check.Wait, if lambda=-sqrt(2 ), equation 1: y= x(1 -sqrt(2 )). If x is positive, y is negative. If x is negative, y is positive. So the critical points are (sqrt(2 +sqrt(2 )) /2, -sqrt(2 -sqrt(2 )) /2 ) and (-sqrt(2 +sqrt(2 )) /2, sqrt(2 -sqrt(2 )) /2 ). Evaluating f at these points gives -sqrt(2 ). Therefore, these are the minima. Similarly, the critical points for lambda= sqrt(2 ) are in the first and third quadrants, giving f= sqrt(2 ).Therefore, there was a mistake earlier when evaluating the point in the first quadrant with lambda=-sqrt(2 ). Such a point does not exist. The critical points for lambda=-sqrt(2 ) are in the fourth and second quadrants, giving f=-sqrt(2 ). Hence, both maximum and minimum are ±sqrt(2 ), which matches the eigenvalue method and the polar coordinate method.Therefore, the Lagrangian multipliers method does confirm that the maximum and minimum of f are sqrt(2 ) and -sqrt(2 ), hence proving that |f| <= sqrt(2 ).Thus, the proof is complete using multiple methods: quadratic forms (eigenvalues), polar coordinates with trigonometric identities, and Lagrange multipliers. All methods confirm that the maximum absolute value is sqrt(2 ), hence the inequality holds.Final Answerboxed{sqrt{2}}</think>Given the problem of proving that ( mid y^{2} + 2xy - x^{2} mid leqslant sqrt{2} ) under the condition ( x^{2} + y^{2} leqslant 1 ), we can approach the solution using multiple methods: Method 1: Quadratic Forms and EigenvaluesThe expression ( y^{2} + 2xy - x^{2} ) can be represented as a quadratic form with matrix ( A = begin{bmatrix} -1 & 1 1 & 1 end{bmatrix} ). The eigenvalues of this matrix are found by solving the characteristic equation ( det(A - lambda I) = 0 ), which gives eigenvalues ( pm sqrt{2} ). The maximum absolute value of the quadratic form on the unit circle is the maximum absolute eigenvalue, which is ( sqrt{2} ). Method 2: Polar CoordinatesUsing polar coordinates ( x = r cos theta ) and ( y = r sin theta ) with ( r leqslant 1 ), the expression transforms to ( r^2 (sin 2theta - cos 2theta) ). This simplifies to ( r^2 sqrt{2} sin(2theta - 45^circ) ). The maximum absolute value of ( sin ) function is 1, hence the expression is bounded by ( sqrt{2} ). Method 3: Lagrange MultipliersMaximizing and minimizing ( f(x, y) = y^2 + 2xy - x^2 ) subject to ( x^2 + y^2 = 1 ), we find critical points using Lagrange multipliers. Solving the system of equations derived from the Lagrangian, we find eigenvalues ( pm sqrt{2} ). The critical points yield maximum and minimum values of ( pm sqrt{2} ). ConclusionAll methods confirm that the maximum absolute value of the expression ( mid y^{2} + 2xy - x^{2} mid ) is bounded by ( sqrt{2} ).[boxed{sqrt{2}}]]

🔑:begin{array}{c}text { Proof: Let } x=lambda cos alpha, y=lambda sin alpha, text { then } lambda^{2}=x^{2}+y^{2} leqslant 1, text { and } left|y^{2}+2 x y-x^{2}right|=mid lambda^{2}left(cos ^{2} alpharight. +2 sin alpha cos alpha-sin ^{2} alpha mid =lambda^{2} mid cos 2 alpha+sin 2 alpha mid leqslant lambda^{2} sqrt{2} leqslant sqrt{2} .end{array}

🧠:[<think>Okay, so I need to find the value of m such that points A, B, and C are collinear, given the vectors AB and BC in terms of vectors a and b, which are non-collinear. Hmm, let's start by recalling what collinearity means in terms of vectors. If three points are collinear, then the vectors connecting them must be scalar multiples of each other. That is, the vector from A to B and the vector from B to C should be proportional. Wait, but AB is from A to B, and BC is from B to C. So if A, B, C are collinear, then the vector AC should be a scalar multiple of AB or BC. But maybe more straightforwardly, since AB + BC = AC, but if the points are collinear, then AB and BC must be scalar multiples of each other. Let me check that.Suppose points A, B, C are collinear. Then, the direction from A to B to C is along the same line. So the vector AB and vector BC must be parallel. Since vectors AB and BC are in the same line, one must be a scalar multiple of the other. So, AB = k * BC for some scalar k. That seems right. So if I can set up that equation, maybe I can solve for m.But let me write that out. If AB = k * BC, then:2a + m b = k*(3a - b)Expanding the right side: 3k a - k bSo, equating components:For the a component: 2 = 3kFor the b component: m = -kSo from the first equation, k = 2/3. Then m = -k = -2/3. So m would be -2/3. Is that right?Wait, but let me make sure. Alternatively, since points A, B, C are collinear, the vector AB and vector AC must be collinear. But AC = AB + BC = (2a + m b) + (3a - b) = 5a + (m -1)b. So if AB and AC are collinear, then there exists a scalar t such that AC = t AB. So:5a + (m -1)b = t*(2a + m b)Which gives:5 = 2tandm -1 = t mFrom the first equation, t = 5/2. Then substituting into the second equation:m -1 = (5/2)mMultiply both sides by 2:2m - 2 = 5mSubtract 2m:-2 = 3mSo m = -2/3. Same answer. So either way, whether I take AB = k BC or AC = t AB, I end up with m = -2/3. So that must be the answer. Let me check again.Alternatively, maybe using coordinates. Since a and b are non-collinear, they form a basis for the plane. So we can represent them as vectors in some coordinate system. Let me assign coordinates to vectors a and b. Let’s assume that vector a is (1,0) and vector b is (0,1). Since they are non-collinear, this is a valid basis. Then AB = 2a + m b = (2, m). BC = 3a - b = (3, -1). If points A, B, C are collinear, then the vector AB and BC must be scalar multiples. Because starting from point A, moving by AB gets you to B, then moving by BC gets you to C. So the direction from A to B is AB, and from B to C is BC. For all three points to lie on a straight line, the direction from A to B and from B to C must be the same, hence AB and BC must be scalar multiples.Therefore, (2, m) = k*(3, -1). So 2 = 3k and m = -k. Therefore, k = 2/3, m = -2/3. Same result.Alternatively, if I think in terms of vectors in the plane. If points A, B, C are collinear, then the vectors AB and BC must be linearly dependent. Since vectors AB and BC are in the plane, their linear dependence implies that the determinant of the matrix formed by them is zero. Let's compute the determinant. If AB and BC are vectors in components, then their determinant should be zero.But since they are expressed in terms of a and b, which are basis vectors, their components are (2, m) and (3, -1). The determinant is (2)(-1) - (m)(3) = -2 -3m. For collinearity, determinant must be zero:-2 -3m = 0 => -3m = 2 => m = -2/3.Yep, same answer. So all methods lead to m = -2/3. So that must be the correct value. So the answer is boxed{-dfrac{2}{3}}.Final AnswerThe value of ( m ) is boxed{-dfrac{2}{3}}.</think>Given that (overrightarrow{a}) and (overrightarrow{b}) are two non-collinear vectors in a plane, (overrightarrow{AB} = 2overrightarrow{a} + moverrightarrow{b}), (overrightarrow{BC} = 3overrightarrow{a} - overrightarrow{b}), and points (A), (B), (C) are collinear, we need to find the value of (m).Since points (A), (B), and (C) are collinear, the vectors (overrightarrow{AB}) and (overrightarrow{BC}) must be scalar multiples of each other. Therefore, we can write:[overrightarrow{AB} = k overrightarrow{BC}]for some scalar (k). Substituting the given vectors, we get:[2overrightarrow{a} + moverrightarrow{b} = k(3overrightarrow{a} - overrightarrow{b})]Expanding the right-hand side, we have:[2overrightarrow{a} + moverrightarrow{b} = 3koverrightarrow{a} - koverrightarrow{b}]Equating the coefficients of (overrightarrow{a}) and (overrightarrow{b}) on both sides, we get the system of equations:[2 = 3k quad text{and} quad m = -k]Solving the first equation for (k), we find:[k = frac{2}{3}]Substituting this value into the second equation, we get:[m = -frac{2}{3}]Thus, the value of (m) is (boxed{-dfrac{2}{3}}).,<think>Okay, let me try to figure out this problem. So, we have two non-collinear vectors, a and b. That means they can't be scalar multiples of each other, right? So, they form a basis for the plane. The problem gives us vectors AB and BC in terms of a and b, and says that points A, B, C are collinear. We need to find the value of m.First, let me recall that if three points are collinear, the vectors connecting them must be scalar multiples of each other. So, AB and BC should be parallel if A, B, C are on the same line. Wait, but AB is from A to B, and BC is from B to C. So, if A, B, C are collinear, then the vector from A to C should be a linear combination of AB and BC, but since they are collinear, AC should be a scalar multiple of AB or BC. Hmm, maybe a better approach is to think in terms of vectors and see if BC is a scalar multiple of AB?Let me think. If points A, B, C are collinear, then vector AB and vector AC should be parallel. But vector AC is AB + BC. Wait, but maybe we can use the concept that vector BC is a scalar multiple of vector AB. Because if they are collinear, moving from B to C should be in the same direction as from A to B or opposite. But actually, AB is from A to B, BC is from B to C. So, if they are collinear, then BC must be a scalar multiple of AB. Because the direction from B to C must be along the same line as AB.So, let me write that down. If A, B, C are collinear, then vector BC = k * vector AB, where k is some scalar.Given that, vector BC is 3a - b, and vector AB is 2a + m b. So, setting 3a - b = k*(2a + m b). Since a and b are non-collinear, their coefficients must be equal on both sides. Therefore:3 = 2k (coefficients of a)-1 = k*m (coefficients of b)So, solving the first equation: 3 = 2k => k = 3/2.Then plug into the second equation: -1 = (3/2)*m => m = (-1)/(3/2) = -2/3.Wait, that seems straightforward, but let me check if my reasoning is correct.Alternative approach: If points A, B, C are collinear, then the vectors AB and BC must be linearly dependent. Since vectors AB and BC are in the plane with basis a and b, they are linearly dependent if and only if the determinant of their coefficients is zero.So, write AB = 2a + m b, BC = 3a - b. The determinant of the matrix formed by their coefficients is |2 3| |m -1|Which is (2)(-1) - (3)(m) = -2 - 3m. For vectors to be linearly dependent, determinant must be zero. So:-2 -3m = 0 => -3m = 2 => m = -2/3.Same answer! So, that confirms it. Therefore, m is -2/3.But let me think again. Suppose I use the concept of coordinates. Let me assign coordinates to points A, B, C. Let me set point A as the origin. Then vector AB = 2a + m b, so point B has coordinates (2, m) in the basis of a and b. Then vector BC is 3a - b, so to get from B to C, we add 3a - b. Therefore, point C has coordinates (2 + 3, m - 1) = (5, m - 1) in the basis.But since points A, B, C are collinear, the slope between A and B must be the same as the slope between B and C. However, since a and b are non-collinear, they form a coordinate system. The slope in terms of the basis vectors would be calculated as follows.The slope from A(0,0) to B(2, m) is (m - 0)/(2 - 0) = m/2.The slope from B(2, m) to C(5, m - 1) is [(m - 1) - m]/[5 - 2] = (-1)/3 = -1/3.For collinearity, these slopes must be equal. Therefore:m/2 = -1/3 => m = -2/3.Again, same result. So, that's three different methods leading to the same answer. So, I think the value of m is -2/3.But let me just step back and make sure I didn't miss anything. The problem states that a and b are non-collinear vectors, so they form a basis. So, all vectors can be expressed uniquely in terms of a and b. The key is that if points are collinear, the vectors AB and BC must be scalar multiples. Alternatively, the vectors AB and AC must be scalar multiples. Either way, the determinant approach seems solid. The coordinate approach is another way to see it, which also gives the same answer.Alternatively, using parametric equations. If points A, B, C are collinear, then there's a parameter t such that for t=0 we have point A, t=1 point B, and t=2 point C? Wait, not necessarily. Let me think. If we parametrize the line as A + t*(AB). Then, at t=0, it's A; at t=1, it's B. Then, point C would be at some t = k. But vector BC is from B to C, so that's equivalent to vector AB multiplied by (k - 1). So, BC = (k - 1)*AB. Then, BC = (k - 1)*AB. Which brings us back to the first approach where BC = scalar * AB, which gives the same equations. So, solving for k gives k -1 = scalar. So, regardless, it all comes down to BC being a scalar multiple of AB.Therefore, all methods converge to m = -2/3. So, I think that's the answer.Final AnswerThe value of ( m ) is boxed{-dfrac{2}{3}}.</think>Given that (overrightarrow{a}) and (overrightarrow{b}) are two non-collinear vectors in a plane, (overrightarrow{AB} = 2overrightarrow{a} + moverrightarrow{b}), (overrightarrow{BC} = 3overrightarrow{a} - overrightarrow{b}), and points (A), (B), (C) are collinear, we need to find the value of (m).Since points (A), (B), and (C) are collinear, the vectors (overrightarrow{AB}) and (overrightarrow{BC}) must be linearly dependent. This means there exists a scalar (k) such that (overrightarrow{BC} = k overrightarrow{AB}).Given:[overrightarrow{BC} = 3overrightarrow{a} - overrightarrow{b}][overrightarrow{AB} = 2overrightarrow{a} + moverrightarrow{b}]Setting (overrightarrow{BC} = k overrightarrow{AB}), we get:[3overrightarrow{a} - overrightarrow{b} = k(2overrightarrow{a} + moverrightarrow{b})]Equating the coefficients of (overrightarrow{a}) and (overrightarrow{b}) since they are non-collinear:[3 = 2k][-1 = km]Solving the first equation for (k):[k = frac{3}{2}]Substituting (k = frac{3}{2}) into the second equation:[-1 = left(frac{3}{2}right)m][m = -frac{2}{3}]Thus, the value of (m) is (boxed{-dfrac{2}{3}}).]

🔑:To solve for the value of the real number m given that overrightarrow{AB}=2overrightarrow{a}+moverrightarrow{b}, overrightarrow{BC}=3overrightarrow{a}-overrightarrow{b}, and points A, B, C are collinear, we use the property that for collinear vectors, one vector can be expressed as a scalar multiple of the other.Given that overrightarrow{AB} = 2overrightarrow{a} + moverrightarrow{b} and overrightarrow{BC} = 3overrightarrow{a} - overrightarrow{b}, and acknowledging the fact that A, B, C are collinear, it implies that:overrightarrow{AB} = lambda overrightarrow{BC}Substituting the given expressions:2overrightarrow{a} + moverrightarrow{b} = lambda(3overrightarrow{a} - overrightarrow{b})This equation needs to hold true for the vectors overrightarrow{a} and overrightarrow{b}, which are non-collinear and thus linearly independent. This implies we can equate the coefficients of overrightarrow{a} and overrightarrow{b} on both sides of the equation:For overrightarrow{a}:2 = 3lambdaFor overrightarrow{b}:m = -lambdaSolving the first equation for lambda gives:lambda = frac{2}{3}Substituting lambda into the second equation gives:m = -frac{2}{3}Therefore, the value of the real number m is boxed{-frac{2}{3}}.

🧠:[<think>Okay, so I need to find the ratio (C² - D²)/(A² - B²) given these four quadratic trinomials f1, f2, f3, f4. The differences of their roots are A, B, C, D respectively, and |A| ≠ |B|. Hmm. Let me start by recalling that for a quadratic equation ax² + bx + c = 0, the roots are given by [-b ± sqrt(b² - 4ac)]/(2a). The difference between the roots would then be [sqrt(b² - 4ac)/a] because the numerator would subtract the two roots: [(-b + sqrt(D))/2a] - [(-b - sqrt(D))/2a] = (2 sqrt(D))/2a = sqrt(D)/a. Wait, but sqrt(b² - 4ac) is the discriminant, right? So the difference between the roots is sqrt(Discriminant)/a, since a is the coefficient of x². So in general, for quadratic ax² + bx + c, difference of roots is sqrt(b² - 4ac)/a. So the absolute difference is sqrt(b² - 4ac)/|a|. But since the problem mentions the difference, not the absolute difference, but maybe they take the positive value? Or perhaps they mean the absolute difference. The problem statement says "the differences of their roots be A, B, C, D". So I need to check if it's absolute difference or just the difference. But since it's called a difference, and squared later, maybe it's the absolute value. But in the ratio, they square them, so maybe it doesn't matter. Let me check.Wait, the problem says |A| ≠ |B|. So perhaps A and B are the actual differences, but their absolute values are different. But since we are squaring them in the ratio, maybe the signs don't matter. Hmm, but maybe A and B are considered as positive numbers? Wait, the roots can be real or complex. But if the roots are complex, the difference would be a complex number. But the problem probably assumes that all roots are real, so the discriminant is non-negative. Therefore, A, B, C, D are real numbers, and since discriminant is non-negative, the difference is sqrt(Discriminant)/a, which is a positive real number divided by a, so depending on the sign of a, the difference could be positive or negative. But since the problem mentions |A| ≠ |B|, maybe they are considering the absolute values. But then again, A, B, C, D are differences, but in the ratio, we have C² - D² over A² - B². Since squares eliminate signs, perhaps it doesn't matter if we take absolute differences or not. Wait, but the problem says "the differences of their roots be A, B, C, D", so perhaps they are actual differences, not absolute. However, in real roots, the difference is sqrt(Discriminant)/a. But depending on the leading coefficient a, the difference could be positive or negative. Wait, sqrt(Discriminant) is positive, and a is the coefficient of x². So if a is positive, then the difference is positive, and if a is negative, the difference is negative? Wait, but let's compute it.Given quadratic equation ax² + bx + c = 0, the roots are [-b + sqrt(b² - 4ac)]/(2a) and [-b - sqrt(b² - 4ac)]/(2a). So subtracting them: [(-b + sqrt(D))/2a] - [(-b - sqrt(D))/2a] = [2 sqrt(D)]/(2a) = sqrt(D)/a. So sqrt(D) is the square root of the discriminant, which is a non-negative real number, so sqrt(D) is non-negative. Therefore, the difference between the roots is sqrt(D)/a. So if a is positive, the difference is positive, and if a is negative, the difference is negative. However, the problem states "differences of their roots be A, B, C, D", so A, B, C, D could be positive or negative, depending on the leading coefficients. But since in the ratio we have C² - D² over A² - B², which would be the same as (|C|² - |D|²)/(|A|² - |B|²) because squaring removes the sign. Therefore, maybe we can just consider the absolute values of A, B, C, D for this ratio.But let's confirm. Let me compute A, B, C, D as per each quadratic:For f1(x) = x² + 2x + a. The difference of roots is sqrt( (2)^2 - 4*1*a ) / 1 = sqrt(4 - 4a). So A = sqrt(4 - 4a).Wait, but according to the formula, it's sqrt(Discriminant)/a, but here a is the coefficient of x², which is 1. So A = sqrt(4 - 4a)/1 = sqrt(4 - 4a) = 2*sqrt(1 - a). Similarly, for f2(x) = x² + b x -1. Discriminant is b² - 4*1*(-1) = b² +4. So difference B = sqrt(b² +4)/1 = sqrt(b² +4).For f3(x) = 2x² + (6 - b)x + 3a + 1. Here, coefficient of x² is 2, so the difference C is sqrt( (6 - b)^2 - 4*2*(3a + 1) ) / 2. Let me compute discriminant: (6 - b)^2 - 8*(3a + 1). Then C = sqrt( (6 - b)^2 - 24a - 8 ) / 2.Similarly, f4(x) = 2x² + (3b - 2)x - a -3. Discriminant is (3b - 2)^2 - 4*2*(-a -3). So discriminant = (9b² -12b +4) + 8(a +3) = 9b² -12b +4 +8a +24 = 9b² -12b +8a +28. Then difference D = sqrt(9b² -12b +8a +28)/2.Therefore, C² would be [( (6 - b)^2 -24a -8 )]/4, since (sqrt(...)/2)^2 = (...)/4. Similarly, D² is (9b² -12b +8a +28)/4.Therefore, C² - D² = [ (6 - b)^2 -24a -8 -9b² +12b -8a -28 ] /4.Similarly, A² is (4 -4a) = 4(1 -a), and B² is (b² +4). Therefore, A² - B² = 4(1 -a) - (b² +4) = 4 -4a -b² -4 = -4a -b².So, now compute C² - D²:First expand (6 - b)^2 = 36 -12b +b². Then:C² numerator: 36 -12b +b² -24a -8 -9b² +12b -8a -28Simplify term by term:36 -8 -28 = 0-12b +12b = 0b² -9b² = -8b²-24a -8a = -32aSo numerator is -8b² -32a = -8(b² +4a)Therefore, C² - D² = (-8b² -32a)/4 = (-8)(b² +4a)/4 = -2(b² +4a)So C² - D² = -2(b² +4a)And A² - B² = -4a -b² = -(4a +b²)Therefore, the ratio (C² - D²)/(A² - B²) is [-2(b² +4a)]/[-(4a +b²)] = [ -2(b² +4a) ] / [ - (b² +4a) ] = ( -2 / -1 ) * (b² +4a)/(b² +4a) ) = 2 *1 = 2.Wait, that's interesting. So the ratio is 2. But let me check my steps again because that seems straightforward, but maybe I made an error.First, let's recompute C² - D²:C² is [ (6 - b)^2 - 24a -8 ] /4D² is [9b² -12b +8a +28]/4So C² - D² = [ (6 - b)^2 -24a -8 -9b² +12b -8a -28 ] /4Expand (6 - b)^2: 36 -12b +b²So substituting:36 -12b +b² -24a -8 -9b² +12b -8a -28Combine like terms:36 -8 -28 = 0-12b +12b = 0b² -9b² = -8b²-24a -8a = -32aSo numerator is -8b² -32a = -8(b² +4a)Therefore, C² - D² = (-8(b² +4a))/4 = -2(b² +4a)A² is 4(1 -a) = 4 -4aB² is b² +4So A² - B² = 4 -4a -b² -4 = -4a -b² = -(4a +b²)Thus, (C² - D²)/(A² - B²) = (-2(b² +4a))/( - (4a +b²) ) = ( -2 / -1 ) * ( (b² +4a)/(b² +4a) ) = 2*1 = 2.So the ratio is 2.But the problem states |A| ≠ |B|. So maybe we need to check if there are conditions where this ratio is valid. But according to our calculation, the ratio is always 2, regardless of a and b, as long as the discriminants are non-negative (so roots are real). However, the problem mentions |A| ≠ |B|. Since |A| is |sqrt(4 -4a)| = 2 sqrt(1 -a), and |B| is sqrt(b² +4). The condition |A| ≠ |B| implies 2 sqrt(1 -a) ≠ sqrt(b² +4). So unless this equality holds, the ratio is 2. But since in the problem statement it's given that |A| ≠ |B|, so our calculation is valid. Thus, the ratio is 2. So the answer is 2, so the boxed answer is boxed{2}.Wait, let me confirm once again.Yes, step-by-step:1. For each quadratic, compute the difference of roots as sqrt(discriminant)/leading coefficient.2. Square those differences for C², D², A², B².3. Compute C² - D² and A² - B².4. Find their ratio.Through computation, variables cancel out, leading to a ratio of 2. The key was that when we expanded and simplified, the terms involving a and b canceled in a way that the ratio became constant. The condition |A| ≠ |B| is likely to ensure that the denominator A² - B² isn't zero, because if |A| = |B|, then A² = B², making the denominator zero, which would be undefined. Since |A| ≠ |B|, the denominator is non-zero, so the ratio is valid. Hence, the ratio is 2.Final Answerboxed{2}</think>Given quadratic trinomials ( f_1(x) = x^2 + 2x + a ), ( f_2(x) = x^2 + bx - 1 ), ( f_3(x) = 2x^2 + (6 - b)x + 3a + 1 ), and ( f_4(x) = 2x^2 + (3b - 2)x - a - 3 ). The differences of their roots are ( A, B, C, ) and ( D ) respectively. We need to find the ratio (frac{C^2 - D^2}{A^2 - B^2}).1. Calculate the differences of the roots: - For ( f_1(x) ), the difference ( A ) is ( sqrt{4 - 4a} ). - For ( f_2(x) ), the difference ( B ) is ( sqrt{b^2 + 4} ). - For ( f_3(x) ), the difference ( C ) is ( frac{sqrt{(6 - b)^2 - 8(3a + 1)}}{2} ). - For ( f_4(x) ), the difference ( D ) is ( frac{sqrt{(3b - 2)^2 - 8(-a - 3)}}{2} ).2. Square the differences: - ( A^2 = 4(1 - a) ) - ( B^2 = b^2 + 4 ) - ( C^2 = frac{(6 - b)^2 - 24a - 8}{4} ) - ( D^2 = frac{9b^2 - 12b + 8a + 28}{4} )3. Compute ( C^2 - D^2 ) and ( A^2 - B^2 ): - Expand and simplify ( C^2 - D^2 ): [ C^2 - D^2 = frac{(6 - b)^2 - 24a - 8 - (9b^2 - 12b + 8a + 28)}{4} = frac{-8b^2 - 32a}{4} = -2(b^2 + 4a) ] - Compute ( A^2 - B^2 ): [ A^2 - B^2 = 4(1 - a) - (b^2 + 4) = -4a - b^2 = -(4a + b^2) ]4. Find the ratio: [ frac{C^2 - D^2}{A^2 - B^2} = frac{-2(b^2 + 4a)}{-(4a + b^2)} = frac{2(b^2 + 4a)}{4a + b^2} = 2 ]Thus, the ratio is (boxed{2}).,<think>Alright, so I need to find the ratio (C² - D²)/(A² - B²) given these four quadratic trinomials f₁, f₂, f₃, f₄. The differences of their roots are A, B, C, D respectively, and |A| ≠ |B|. Hmm, okay. Let me start by recalling that for a quadratic equation ax² + bx + c = 0, the difference of the roots can be found using the formula |√(b² - 4ac)/a|. Wait, is that right? Let me think.The roots of a quadratic equation are given by [-b ± √(b² - 4ac)]/(2a). So the difference between the roots would be [(-b + √(b² - 4ac)) - (-b - √(b² - 4ac))]/(2a) = [2√(b² - 4ac)]/(2a) = √(b² - 4ac)/a. But since the difference is a magnitude, it should be the absolute value? The problem mentions A, B, C, D as differences, so maybe they are considered positive. But the problem also says |A| ≠ |B|, so perhaps A and B can be positive or negative, but their absolute values aren't equal. But the ratio (C² - D²)/(A² - B²) would not depend on the signs, since squares are involved. Anyway, maybe I can just compute the differences as √(b² - 4ac)/a without worrying about the sign, because when I square them, the sign won't matter. Let's see.So first, let's compute A, B, C, D. For each quadratic, compute the discriminant, take the square root, divide by the leading coefficient, and then square them for C² and D², and subtract, same for A² and B².Starting with f₁(x) = x² + 2x + a. The discriminant here is (2)² - 4(1)(a) = 4 - 4a. Then the difference of roots A is √(4 - 4a)/1 = √(4 - 4a). So A² = 4 - 4a.Similarly, f₂(x) = x² + b x - 1. Discriminant is b² - 4(1)(-1) = b² + 4. Then difference of roots B is √(b² + 4)/1, so B² = b² + 4.For f₃(x) = 2x² + (6 - b)x + 3a + 1. Let's compute discriminant: [(6 - b)]² - 4*2*(3a + 1). Then difference of roots C is √(discriminant)/2. So C² would be discriminant/(2²) = discriminant/4. Therefore, C² - D² would involve similar terms. Wait, but D is from f₄(x). Let's compute D as well.f₄(x) = 2x² + (3b - 2)x - a - 3. Its discriminant is (3b - 2)² - 4*2*(-a - 3). Then difference of roots D is √(discriminant)/2, so D² is discriminant/4.So let's compute each discriminant step by step.Starting with f₁: discriminant is 4 - 4a, so A² = 4 - 4a.f₂: discriminant is b² + 4, so B² = b² + 4.f₃: discriminant is (6 - b)² - 8*(3a + 1). Let's compute that:(6 - b)² = 36 - 12b + b²8*(3a + 1) = 24a + 8Thus discriminant for f₃ is 36 - 12b + b² - 24a - 8 = b² - 12b - 24a + 28Therefore, C² = (b² - 12b - 24a + 28)/4Similarly, for f₄: discriminant is (3b - 2)² - 8*(-a - 3)Compute (3b - 2)² = 9b² - 12b + 48*(-a - 3) = -8a -24Thus discriminant is 9b² -12b +4 - (-8a -24) = 9b² -12b +4 +8a +24 =9b² -12b +8a +28Therefore, D² = (9b² -12b +8a +28)/4Now, compute C² - D²:[(b² -12b -24a +28)/4] - [(9b² -12b +8a +28)/4] =[ (b² -12b -24a +28) - (9b² -12b +8a +28) ] /4 =(b² -12b -24a +28 -9b² +12b -8a -28)/4 =(-8b² -32a)/4 =(-8b² -32a)/4 = -2b² -8aSimilarly, compute A² - B²:(4 -4a) - (b² +4) = 4 -4a -b² -4 = -b² -4aSo the ratio (C² - D²)/(A² - B²) is (-2b² -8a)/(-b² -4a) = [ -2(b² +4a) ] / [ - (b² +4a) ] = 2Wait, that simplifies to 2? Let me check again.Yes, in the numerator, factor out -2: -2(b² + 4a). In the denominator, factor out -1: -1*(b² +4a). So dividing them gives (-2)/( -1 ) = 2. The (b² +4a) cancels out. Therefore, the ratio is 2. But wait, the problem states |A| ≠ |B|. Does that condition affect anything?Let me check if in the denominator A² - B² is zero. If A² = B², then denominator would be zero, but the problem states |A| ≠ |B|, so A² - B² ≠ 0. Therefore, the ratio is valid. So the answer is 2. Therefore, the ratio is 2, so boxed answer is boxed{2}Wait, but let me just confirm once again. Let's retrace the steps.First, compute A², B², C², D².A² = discriminant of f₁ divided by 1² = 4 -4aB² = discriminant of f₂ divided by 1² = b² +4C² = discriminant of f₃ divided by 2² = [ (6 -b)^2 -8(3a +1) ] /4 = [36 -12b +b² -24a -8]/4 = [b² -12b -24a +28]/4Similarly, D² = discriminant of f₄ divided by 2² = [ (3b -2)^2 -8(-a -3) ] /4 = [9b² -12b +4 +8a +24]/4 = [9b² -12b +8a +28]/4Thus C² - D² = [ (b² -12b -24a +28) - (9b² -12b +8a +28) ] /4 = (-8b² -32a)/4 = -2b² -8aA² - B² = (4 -4a) - (b² +4) = -4a -b²Therefore, the ratio is (-2b² -8a)/(-b² -4a) = [ -2(b² +4a) ] / [ - (b² +4a) ] = 2.Yes, that seems correct. So the ratio is 2. Therefore, the answer is boxed{2}Final Answerboxed{2}</think>Given quadratic trinomials ( f_{1}(x) = x^{2} + 2x + a ), ( f_{2}(x) = x^{2} + bx - 1 ), ( f_{3}(x) = 2x^{2} + (6 - b)x + 3a + 1 ), and ( f_{4}(x) = 2x^{2} + (3b - 2)x - a - 3 ). The differences of their roots are ( A, B, C, ) and ( D ) respectively, and ( |A| neq |B| ). We need to find the ratio ( frac{C^{2} - D^{2}}{A^{2} - B^{2}} ).1. Calculate ( A^{2} ) and ( B^{2} ): - For ( f_{1}(x) ), the discriminant is ( 4 - 4a ), so ( A^{2} = 4 - 4a ). - For ( f_{2}(x) ), the discriminant is ( b^{2} + 4 ), so ( B^{2} = b^{2} + 4 ).2. Calculate ( C^{2} ) and ( D^{2} ): - For ( f_{3}(x) ), the discriminant is ( (6 - b)^{2} - 8(3a + 1) = b^{2} - 12b - 24a + 28 ), so ( C^{2} = frac{b^{2} - 12b - 24a + 28}{4} ). - For ( f_{4}(x) ), the discriminant is ( (3b - 2)^{2} - 8(-a - 3) = 9b^{2} - 12b + 8a + 28 ), so ( D^{2} = frac{9b^{2} - 12b + 8a + 28}{4} ).3. Compute ( C^{2} - D^{2} ): [ C^{2} - D^{2} = frac{b^{2} - 12b - 24a + 28}{4} - frac{9b^{2} - 12b + 8a + 28}{4} = frac{-8b^{2} - 32a}{4} = -2b^{2} - 8a ]4. Compute ( A^{2} - B^{2} ): [ A^{2} - B^{2} = (4 - 4a) - (b^{2} + 4) = -b^{2} - 4a ]5. Find the ratio: [ frac{C^{2} - D^{2}}{A^{2} - B^{2}} = frac{-2b^{2} - 8a}{-b^{2} - 4a} = frac{-2(b^{2} + 4a)}{-(b^{2} + 4a)} = 2 ]Thus, the ratio is (boxed{2}).]

🔑:Answer: 2.Solution. Let alpha x^{2}+beta x+gamma be a quadratic trinomial with a positive discriminant T. Then its roots are determined by the formula x_{1,2}=frac{-b pm sqrt{T}}{2 a}, so left|x_{2}-x_{1}right|=left|frac{-b+sqrt{T}-(-b-sqrt{T})}{2 a}right|= frac{sqrt{T}}{|a|}. Applying this formula four times, we getA=sqrt{4-4 a}, B=sqrt{b^{2}+4}, C=frac{1}{2} sqrt{(6-b)^{2}-8(1+3 a)}, D=frac{1}{2} sqrt{(3 b-2)^{2}+8(a+3)}From this, it follows that C^{2}-D^{2}=frac{1}{4}left(left(b^{2}-12 b-24 a+28right)-left(9 b^{2}-12 b+8 a+28right)right)=-2left(b^{2}+4 aright), A^{2}-B^{2}=-left(b^{2}+4 aright). Therefore, the desired ratio is 2.

🧠:[<think>Okay, let's see. I need to calculate 2016 multiplied by the expression (1/7 minus 5 divided by (6 times 8)). Hmm. Let me break this down step by step. First, maybe I should handle the part inside the parentheses. The expression is (1/7 - 5/(6×8)). Let me compute the denominators first. So, 6 times 8 is 48. That makes the second term 5/48. Then I need to subtract 5/48 from 1/7. To subtract fractions, they need a common denominator. What's the least common denominator of 7 and 48? Let's see, 7 is a prime number and 48 is 16 times 3, so their least common multiple should be 7×48=336. Hmm, that's a bit big, but okay.Convert 1/7 to 48/336 because 1×48=48 and 7×48=336. Then 5/48 converted to 35/336 since 5×7=35 and 48×7=336. So now the expression becomes (48/336 - 35/336) = (48-35)/336 = 13/336. Wait, let me check that again. 1/7 is equal to 48/336 because 336 divided by 7 is 48. And 5/48 is 35/336 because 336 divided by 48 is 7, so multiply numerator and denominator by 7. So 5×7=35, 48×7=336. So 48/336 - 35/336 is indeed 13/336. Okay, that seems right.So now the original expression simplifies to 2016 multiplied by 13/336. Let me compute that. 2016 divided by 336 times 13. Let's see, 336 times 6 is 2016, right? Because 300×6=1800 and 36×6=216, so 1800+216=2016. Yes, so 2016 divided by 336 is 6. So then 6 times 13 is 78. Therefore, the result should be 78. Wait, let me verify that again. 336 × 6 = 2016, correct. So 2016/336 = 6. Then 6 ×13 =78. Yes. So the answer is 78. Alternatively, maybe there's another way to approach this. Let's see. Maybe factor 2016 and see if there's cancellation. Let's try that. 2016 divided by 336 is 6, as before. But if I didn't see that immediately, maybe break down 2016 into prime factors. 2016 is 2^5 × 3^2 ×7. 336 is 2^4 ×3×7. So dividing those gives (2^(5-4)) ×(3^(2-1)) ×(7^(1-1)) = 2×3×1=6. So yeah, 2016/336=6. Then 6×13=78. So same result.Alternatively, maybe compute 2016×(1/7 -5/48) directly. Let's compute each term separately. 2016×1/7 is 2016 divided by 7. 2016 ÷7. 7×288=2016 because 7×200=1400, 7×88=616, so 1400+616=2016. So 2016/7=288. Then the other term is 2016×5/(6×8). Let's compute 6×8=48 as before. So 2015×5/48. Wait, no, it's 2016×5/48. So compute 2016 divided by 48 first. 48×42=2016, since 48×40=1920 and 48×2=96, so 1920+96=2016. Therefore, 2016/48=42. Then multiply by 5: 42×5=210. So the second term is 210. So the original expression is 288 (from 2016×1/7) minus 210 (from 2016×5/48). So 288-210=78. Yep, same answer. So both methods give me 78, which seems consistent. So I think 78 is the correct answer. Let me just check once more. First method: (1/7 -5/48)= (48-35)/336=13/336. Then 2016×(13/336)= (2016/336)×13=6×13=78. Second method: 2016×1/7=288. 2016×5/48=210. Then 288-210=78. Yes, both methods confirm that. Therefore, I'm confident the result is 78.Final AnswerThe result of the expression is boxed{78}.</think>To calculate the result of the expression (2016 times left( frac{1}{7} - frac{5}{6 times 8} right)), we can break it down step by step:1. Compute the denominator for the second term: [ 6 times 8 = 48 ] So, the expression inside the parentheses becomes: [ frac{1}{7} - frac{5}{48} ]2. Find a common denominator for the fractions (frac{1}{7}) and (frac{5}{48}). The least common multiple of 7 and 48 is 336. Convert the fractions: [ frac{1}{7} = frac{48}{336} quad text{and} quad frac{5}{48} = frac{35}{336} ]3. Subtract the fractions: [ frac{48}{336} - frac{35}{336} = frac{13}{336} ]4. Multiply the result by 2016: [ 2016 times frac{13}{336} ]5. Simplify the multiplication. Since (2016 div 336 = 6), we have: [ 6 times 13 = 78 ]Thus, the result of the expression is (boxed{78}).,<think>Okay, let me try to figure out this problem. So, the expression is 2016 multiplied by (1/7 minus 5 divided by (6 times 8)). Hmm. Let me write that down step by step. First, I need to compute the part inside the parentheses: 1/7 minus 5/(6*8). Let me handle the denominators first. The second term in the parentheses is 5 divided by (6 multiplied by 8). Let me calculate 6 times 8 first. 6*8 is 48, right? So, that term becomes 5/48. So now the expression inside the parentheses is 1/7 minus 5/48.Now, to subtract these two fractions, I need a common denominator. What's the least common denominator of 7 and 48? Well, 7 is a prime number, and 48 is 16*3, so their LCM would be 7*48, which is 336. So, converting both fractions to have denominator 336.1/7 is equal to (1*48)/(7*48) = 48/336. Similarly, 5/48 is equal to (5*7)/(48*7) = 35/336. So, subtracting these gives 48/336 - 35/336 = (48 - 35)/336 = 13/336. So, the expression inside the parentheses simplifies to 13/336.Now, the entire expression is 2016 multiplied by 13/336. Let me compute that. Maybe there's a way to simplify this before multiplying. Let's see. 2016 divided by 336. Hmm. Let's see how many times 336 fits into 2016.Well, 336 times 6 is 2016. Let me check that. 300*6=1800, 36*6=216, so 1800+216=2016. Yes, that's right. So, 2016 divided by 336 is 6. Therefore, 2016*(13/336) is the same as 6*13. 6 times 13 is... 78. So, the result of the entire expression should be 78. Let me verify my steps to make sure I didn't make any mistakes.First, inside the parentheses: 1/7 - 5/(6*8). Calculated 6*8 as 48. Then 5/48. Then found common denominator 336. Converted 1/7 to 48/336 and 5/48 to 35/336. Subtracted to get 13/336. Then multiplied by 2016. Since 2016 divided by 336 is 6, multiplied 6 by 13 to get 78. Seems correct. Wait, let me check if 2016 divided by 336 is indeed 6. Let's divide 2016 by 336. 336 times 5 is 1680. 2016 minus 1680 is 336. So, that's 5 + 1 = 6. Correct. So, 2016/336 = 6. Multiply by 13: 6*13=78. Hmm, seems all steps check out. So, the final answer should be 78.Final Answerboxed{78}</think>To solve the expression (2016 timesleft(frac{1}{7}-frac{5}{6 times 8}right)), we start by simplifying the terms inside the parentheses.First, calculate the denominator of the second term:[6 times 8 = 48]So, the expression inside the parentheses becomes:[frac{1}{7} - frac{5}{48}]Next, we find a common denominator for the fractions (frac{1}{7}) and (frac{5}{48}). The least common multiple of 7 and 48 is 336. Converting both fractions to have this common denominator:[frac{1}{7} = frac{48}{336} quad text{and} quad frac{5}{48} = frac{35}{336}]Subtracting these fractions:[frac{48}{336} - frac{35}{336} = frac{13}{336}]Now, we multiply this result by 2016:[2016 times frac{13}{336}]Simplifying the multiplication by noting that (2016 div 336 = 6):[6 times 13 = 78]Thus, the result of the expression is:[boxed{78}]]

🔑:【Solution】Solve: 2016 timesleft(frac{1}{7}-frac{5}{6 times 8}right)begin{array}{l}=2016 times frac{1}{7}-frac{5}{6 times 8} times 2016 =288-210 =78 ;end{array}Therefore, the answer is: 78 .