💡Alright, so I've got this function here: f(x) = x⁵ + a x³ + b x - 8. And I know that f(-2) equals 10. I need to find f(2). Hmm, okay, let me think about how to approach this.First, I remember that when dealing with functions, especially polynomials, certain properties can be really helpful. Like, if a function is odd or even, that can simplify things. Let me recall: an odd function satisfies f(-x) = -f(x), and an even function satisfies f(-x) = f(x). Looking at f(x) = x⁵ + a x³ + b x - 8, I notice that all the terms except the constant term (-8) are odd-powered terms. So, x⁵ is odd, x³ is odd, and x is odd. The constant term, -8, is even because it's just a number without any x. So, if I remove the constant term, the rest of the function is odd.Let me define a new function, let's say g(x), which is f(x) plus 8. So, g(x) = f(x) + 8. That would make g(x) = x⁵ + a x³ + b x. Now, this g(x) is purely made up of odd-powered terms, so g(x) should be an odd function. That means g(-x) = -g(x).Okay, so since g(x) is odd, we have g(-2) = -g(2). But g(x) is f(x) + 8, so g(-2) = f(-2) + 8. We know f(-2) is 10, so g(-2) = 10 + 8 = 18. Therefore, 18 = -g(2), which means g(2) = -18.But g(2) is also equal to f(2) + 8. So, f(2) + 8 = -18. To find f(2), I just subtract 8 from both sides: f(2) = -18 - 8 = -26.Wait, let me double-check that. If g(-2) = 18 and g(-2) = -g(2), then g(2) must be -18. Then, since g(2) = f(2) + 8, f(2) = g(2) - 8 = -18 - 8 = -26. Yeah, that seems right.I guess another way to think about it is by plugging in x = -2 into f(x). Let's see:f(-2) = (-2)⁵ + a (-2)³ + b (-2) - 8 = -32 - 8a - 2b - 8 = (-32 - 8) + (-8a - 2b) = -40 - 8a - 2b.And we know f(-2) = 10, so:-40 - 8a - 2b = 10Let me solve for -8a - 2b:-8a - 2b = 10 + 40 = 50So, -8a - 2b = 50.Now, let's compute f(2):f(2) = (2)⁵ + a (2)³ + b (2) - 8 = 32 + 8a + 2b - 8 = (32 - 8) + (8a + 2b) = 24 + 8a + 2b.Hmm, so f(2) = 24 + 8a + 2b.But from earlier, we have -8a - 2b = 50. Let me see if I can relate this to f(2). If I take -8a - 2b = 50, then multiplying both sides by -1 gives 8a + 2b = -50.So, 8a + 2b = -50. Then, plugging that into f(2):f(2) = 24 + (8a + 2b) = 24 + (-50) = 24 - 50 = -26.Okay, so that confirms it. Whether I use the odd function approach or directly compute f(-2) and f(2), I end up with f(2) = -26. That makes sense because the odd function property allows us to relate f(-2) and f(2) without needing to know the specific values of a and b.So, I think I'm confident that the answer is -26.**Final Answer**The value of ( f(2) ) is boxed{-26}.

💡Okay, so I have this problem about an ellipse, and I need to find the area of a specific triangle formed by a point on the ellipse and the two foci. Let me try to break this down step by step.First, the equation of the ellipse is given as (frac{x^2}{25} + frac{y^2}{9} = 1). I remember that the standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. So in this case, (a^2 = 25) which means (a = 5), and (b^2 = 9) so (b = 3).Now, the foci of an ellipse are located at a distance (c) from the center along the major axis. I recall that (c^2 = a^2 - b^2). Let me compute that: (c^2 = 25 - 9 = 16), so (c = 4). That means the foci are at ((-4, 0)) and ((4, 0)). Let me denote these as (F_1) and (F_2) respectively.The problem states that point (P) is on the ellipse such that the line segments (PF_1) and (PF_2) are perpendicular to each other. I need to find the area of triangle (PF_1F_2).Hmm, okay. So point (P) is somewhere on the ellipse where the two lines from (P) to each focus are perpendicular. I think I can approach this by using coordinates. Let me denote the coordinates of (P) as ((x, y)).Since (PF_1) and (PF_2) are perpendicular, the vectors from (P) to each focus should have a dot product of zero. Let me write that down.The vector from (P) to (F_1) is ((x - (-4), y - 0) = (x + 4, y)). Similarly, the vector from (P) to (F_2) is ((x - 4, y - 0) = (x - 4, y)). The dot product of these two vectors should be zero because they are perpendicular. So:[(x + 4)(x - 4) + y cdot y = 0]Simplifying this:[x^2 - 16 + y^2 = 0][x^2 + y^2 = 16]Okay, so that's one equation. But I also know that (P) lies on the ellipse, so it must satisfy the ellipse equation:[frac{x^2}{25} + frac{y^2}{9} = 1]So now I have two equations:1. (x^2 + y^2 = 16)2. (frac{x^2}{25} + frac{y^2}{9} = 1)I need to solve these simultaneously to find the coordinates of (P). Let me see how to do that.From the first equation, I can express (x^2) as (16 - y^2). Let me substitute this into the second equation.Substituting (x^2 = 16 - y^2) into the ellipse equation:[frac{16 - y^2}{25} + frac{y^2}{9} = 1]Let me compute this step by step. First, write out the equation:[frac{16 - y^2}{25} + frac{y^2}{9} = 1]To combine these fractions, I need a common denominator. The denominators are 25 and 9, so the least common multiple is 225. Let me multiply each term by 225 to eliminate the denominators:[225 cdot frac{16 - y^2}{25} + 225 cdot frac{y^2}{9} = 225 cdot 1]Simplify each term:- (225 / 25 = 9), so the first term becomes (9(16 - y^2))- (225 / 9 = 25), so the second term becomes (25y^2)- The right side is 225So the equation becomes:[9(16 - y^2) + 25y^2 = 225]Let me expand the left side:[144 - 9y^2 + 25y^2 = 225]Combine like terms:[144 + 16y^2 = 225]Subtract 144 from both sides:[16y^2 = 81]Divide both sides by 16:[y^2 = frac{81}{16}]Take the square root of both sides:[y = pm frac{9}{4}]So the y-coordinate of point (P) is either (frac{9}{4}) or (-frac{9}{4}). Now, let's find the corresponding x-coordinates. Recall from the first equation (x^2 + y^2 = 16). Plugging in (y^2 = frac{81}{16}):[x^2 + frac{81}{16} = 16][x^2 = 16 - frac{81}{16}][x^2 = frac{256}{16} - frac{81}{16}][x^2 = frac{175}{16}][x = pm frac{sqrt{175}}{4}][x = pm frac{5sqrt{7}}{4}]So the coordinates of point (P) are (left( pm frac{5sqrt{7}}{4}, pm frac{9}{4} right)). Now, I need to find the area of triangle (PF_1F_2). Let me recall that the area of a triangle can be calculated in several ways. One common method is using the base and height. Another is using coordinates with the shoelace formula. Since I have coordinates for all three points, maybe the shoelace formula would be straightforward.But before that, let me visualize the triangle. Points (F_1) and (F_2) are on the x-axis at ((-4, 0)) and ((4, 0)) respectively. Point (P) is somewhere in the plane, either in the upper or lower half, given the y-coordinate is (pm frac{9}{4}). Wait, actually, since the area is a positive quantity, the sign of the y-coordinate won't affect the area. So whether (P) is above or below the x-axis, the area will be the same. So I can just consider one case, say (y = frac{9}{4}).But let me think if there's a simpler way without using coordinates. Since (F_1) and (F_2) are on the x-axis, the base of the triangle (F_1F_2) is the distance between them, which is (8) units (from (-4) to (4)). The height would be the vertical distance from point (P) to the base, which is the x-axis. Since the y-coordinate of (P) is (frac{9}{4}), that is the height.So the area of the triangle is (frac{1}{2} times text{base} times text{height}).Let me compute that:[text{Area} = frac{1}{2} times 8 times frac{9}{4}][= frac{1}{2} times frac{72}{4}][= frac{1}{2} times 18][= 9]So the area is 9.Wait, that seems straightforward. But just to double-check, maybe I should use the shoelace formula as well to confirm.Let me list the coordinates of the three points:- (F_1 = (-4, 0))- (F_2 = (4, 0))- (P = left( frac{5sqrt{7}}{4}, frac{9}{4} right)) (I'll take the positive x and y for simplicity)Shoelace formula is given by:[text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|]Plugging in the coordinates:[text{Area} = frac{1}{2} | (-4)(0 - frac{9}{4}) + 4(frac{9}{4} - 0) + frac{5sqrt{7}}{4}(0 - 0) |][= frac{1}{2} | (-4)(-frac{9}{4}) + 4(frac{9}{4}) + 0 |][= frac{1}{2} | 9 + 9 + 0 |][= frac{1}{2} times 18][= 9]Same result. So that confirms it. The area is indeed 9.Alternatively, I could have used vectors or determinants, but both methods I tried gave the same answer, so I feel confident.Just to recap:1. Identified the ellipse parameters (a), (b), and (c).2. Found the coordinates of the foci.3. Set up the condition for perpendicular vectors from (P) to each focus.4. Solved the system of equations to find the coordinates of (P).5. Calculated the area using two different methods to ensure correctness.Everything seems consistent. I don't see any mistakes in my steps, so I think the area of triangle (PF_1F_2) is 9.**Final Answer**The area of triangle (△PF_1F_2) is boxed{9}.

💡Okay, so I have this problem about an ellipse. Let me try to understand it step by step. First, part (1) asks for the equation of the ellipse. The ellipse is centered at the origin, and its foci are on the y-axis. That means it's a vertical ellipse, right? So the standard form should be (frac{y^2}{a^2} + frac{x^2}{b^2} = 1), where (a > b).They also mention that both the distance from the foci to the corresponding directrix and the eccentricity are (frac{sqrt{2}}{2}). Hmm, okay. Let me recall some properties of ellipses. The eccentricity (e) is given by (e = frac{c}{a}), where (c) is the distance from the center to each focus. Also, the directrix is located at a distance of (frac{a}{e}) from the center. Wait, the distance from the focus to the directrix is given as (frac{sqrt{2}}{2}). So, if the focus is at ((0, c)) and the directrix is the line (y = frac{a}{e}), then the distance between them should be (frac{a}{e} - c). Let me write that down:[frac{a}{e} - c = frac{sqrt{2}}{2}]But since (e = frac{c}{a}), we can substitute that into the equation:[frac{a}{frac{c}{a}} - c = frac{sqrt{2}}{2}][frac{a^2}{c} - c = frac{sqrt{2}}{2}]Also, the eccentricity (e) is given as (frac{sqrt{2}}{2}), so:[e = frac{c}{a} = frac{sqrt{2}}{2}][c = frac{sqrt{2}}{2}a]Now, substitute (c) back into the first equation:[frac{a^2}{frac{sqrt{2}}{2}a} - frac{sqrt{2}}{2}a = frac{sqrt{2}}{2}]Simplify the first term:[frac{a^2}{frac{sqrt{2}}{2}a} = frac{a}{frac{sqrt{2}}{2}} = frac{2a}{sqrt{2}} = sqrt{2}a]So the equation becomes:[sqrt{2}a - frac{sqrt{2}}{2}a = frac{sqrt{2}}{2}]Factor out (sqrt{2}a):[sqrt{2}a left(1 - frac{1}{2}right) = frac{sqrt{2}}{2}][sqrt{2}a cdot frac{1}{2} = frac{sqrt{2}}{2}][frac{sqrt{2}a}{2} = frac{sqrt{2}}{2}]Multiply both sides by 2:[sqrt{2}a = sqrt{2}]Divide both sides by (sqrt{2}):[a = 1]Okay, so (a = 1). Then, since (c = frac{sqrt{2}}{2}a), we have:[c = frac{sqrt{2}}{2} times 1 = frac{sqrt{2}}{2}]Now, since (c^2 = a^2 - b^2), we can solve for (b):[left(frac{sqrt{2}}{2}right)^2 = 1^2 - b^2][frac{2}{4} = 1 - b^2][frac{1}{2} = 1 - b^2][b^2 = 1 - frac{1}{2} = frac{1}{2}]So, (b = frac{sqrt{2}}{2}) as well. Therefore, the equation of the ellipse is:[frac{y^2}{1^2} + frac{x^2}{left(frac{sqrt{2}}{2}right)^2} = 1]Simplify the denominator for (x^2):[frac{x^2}{frac{1}{2}} = 2x^2]So, the equation becomes:[y^2 + 2x^2 = 1]Alright, that seems straightforward. I think that's part (1) done.Moving on to part (2). The line (l) intersects the y-axis at point (P(0, m)) and intersects the ellipse at two distinct points (A) and (B). The vector condition is given as (overrightarrow{AP} = lambda overrightarrow{PB}). Additionally, it's given that (overrightarrow{OA} + lambda overrightarrow{OB} = 4overrightarrow{OP}). We need to find the range of values for (m).Let me parse this. First, the vector condition (overrightarrow{AP} = lambda overrightarrow{PB}) suggests that point (P) divides the segment (AB) in the ratio (lambda:1). So, (P) is a weighted average of points (A) and (B). Given that, we can express this in terms of coordinates. Let me denote (A = (x_1, y_1)) and (B = (x_2, y_2)). Then, the vector equation translates to:[overrightarrow{AP} = lambda overrightarrow{PB}]Which means:[overrightarrow{P} - overrightarrow{A} = lambda (overrightarrow{B} - overrightarrow{P})]So, in coordinates:[(0 - x_1, m - y_1) = lambda (x_2 - 0, y_2 - m)]Which gives two equations:[-x_1 = lambda x_2 quad text{(1)}][m - y_1 = lambda (y_2 - m) quad text{(2)}]From equation (1), we can express (x_1 = -lambda x_2). Similarly, from equation (2):[m - y_1 = lambda y_2 - lambda m][m + lambda m = y_1 + lambda y_2][m(1 + lambda) = y_1 + lambda y_2]So, that's another relationship between (y_1) and (y_2).Now, the other condition is (overrightarrow{OA} + lambda overrightarrow{OB} = 4overrightarrow{OP}). Let me write that out in coordinates:[overrightarrow{OA} + lambda overrightarrow{OB} = (x_1 + lambda x_2, y_1 + lambda y_2)]And (4overrightarrow{OP} = (0, 4m)). So, we have:[(x_1 + lambda x_2, y_1 + lambda y_2) = (0, 4m)]Which gives two equations:[x_1 + lambda x_2 = 0 quad text{(3)}][y_1 + lambda y_2 = 4m quad text{(4)}]From equation (3), since we already have (x_1 = -lambda x_2) from equation (1), substituting into equation (3):[-lambda x_2 + lambda x_2 = 0]Which is (0 = 0), so it doesn't give new information. From equation (4), we had earlier:[m(1 + lambda) = y_1 + lambda y_2]But equation (4) says:[y_1 + lambda y_2 = 4m]Therefore, combining these two:[m(1 + lambda) = 4m]Assuming (m neq 0), we can divide both sides by (m):[1 + lambda = 4]So, (lambda = 3).If (m = 0), then from equation (4), (y_1 + 3y_2 = 0), but since the line intersects the ellipse at two points, (m = 0) might be a special case. However, let's see if (m = 0) is allowed. If (m = 0), the line passes through the origin, but the ellipse is centered at the origin, so the line would pass through the center, and the points (A) and (B) would be symmetric with respect to the origin. But in that case, the ratio (lambda) would be 1, but we found (lambda = 3), so (m) cannot be zero. So, (m neq 0), and (lambda = 3).So, now we know (lambda = 3). Let's write down the relationships again:From (overrightarrow{AP} = 3overrightarrow{PB}):[x_1 = -3x_2][m - y_1 = 3(y_2 - m)]From the second equation:[m - y_1 = 3y_2 - 3m][4m = y_1 + 3y_2]Which is consistent with equation (4).So, now, let's think about the line (l). It intersects the ellipse at points (A) and (B), and passes through (P(0, m)). Let me denote the equation of line (l) as (y = kx + m), where (k) is the slope.We can substitute this into the ellipse equation:[y^2 + 2x^2 = 1]Substitute (y = kx + m):[(kx + m)^2 + 2x^2 = 1]Expand the square:[k^2x^2 + 2kmx + m^2 + 2x^2 = 1]Combine like terms:[(k^2 + 2)x^2 + 2kmx + (m^2 - 1) = 0]This is a quadratic in (x). Let me denote this as:[A x^2 + B x + C = 0]Where:- (A = k^2 + 2)- (B = 2km)- (C = m^2 - 1)Since the line intersects the ellipse at two distinct points, the discriminant must be positive:[Delta = B^2 - 4AC > 0]Compute (Delta):[Delta = (2km)^2 - 4(k^2 + 2)(m^2 - 1)]Simplify:[Delta = 4k^2m^2 - 4(k^2 + 2)(m^2 - 1)]Factor out 4:[Delta = 4left[k^2m^2 - (k^2 + 2)(m^2 - 1)right]]Expand the second term:[(k^2 + 2)(m^2 - 1) = k^2m^2 - k^2 + 2m^2 - 2]So, substitute back:[Delta = 4left[k^2m^2 - (k^2m^2 - k^2 + 2m^2 - 2)right]]Simplify inside the brackets:[k^2m^2 - k^2m^2 + k^2 - 2m^2 + 2 = k^2 - 2m^2 + 2]Thus:[Delta = 4(k^2 - 2m^2 + 2) > 0]So, the condition is:[k^2 - 2m^2 + 2 > 0 quad text{(Condition 1)}]Now, going back to the quadratic equation, the roots are (x_1) and (x_2), which correspond to points (A) and (B). From Vieta's formulas, we know:[x_1 + x_2 = -frac{B}{A} = -frac{2km}{k^2 + 2}][x_1 x_2 = frac{C}{A} = frac{m^2 - 1}{k^2 + 2}]But we also have from the vector condition that (x_1 = -3x_2). Let me write that:[x_1 = -3x_2]So, substitute into the sum:[-3x_2 + x_2 = -2x_2 = -frac{2km}{k^2 + 2}]Thus:[-2x_2 = -frac{2km}{k^2 + 2}]Divide both sides by -2:[x_2 = frac{km}{k^2 + 2}]Similarly, since (x_1 = -3x_2), we have:[x_1 = -3 cdot frac{km}{k^2 + 2} = -frac{3km}{k^2 + 2}]Now, let's use the product of roots:[x_1 x_2 = frac{m^2 - 1}{k^2 + 2}]Substitute (x_1) and (x_2):[left(-frac{3km}{k^2 + 2}right) left(frac{km}{k^2 + 2}right) = frac{m^2 - 1}{k^2 + 2}]Simplify the left side:[-frac{3k^2m^2}{(k^2 + 2)^2} = frac{m^2 - 1}{k^2 + 2}]Multiply both sides by ((k^2 + 2)^2):[-3k^2m^2 = (m^2 - 1)(k^2 + 2)]Expand the right side:[-3k^2m^2 = m^2k^2 + 2m^2 - k^2 - 2]Bring all terms to the left side:[-3k^2m^2 - m^2k^2 - 2m^2 + k^2 + 2 = 0]Combine like terms:- For (k^2m^2): (-3k^2m^2 - m^2k^2 = -4k^2m^2)- For (m^2): (-2m^2)- For (k^2): (k^2)- Constants: (+2)So, the equation becomes:[-4k^2m^2 - 2m^2 + k^2 + 2 = 0]Let me rearrange terms:[(-4m^2 + 1)k^2 + (-2m^2 + 2) = 0]Factor out terms:[(-4m^2 + 1)k^2 + (-2m^2 + 2) = 0]Let me write this as:[(1 - 4m^2)k^2 + (2 - 2m^2) = 0]Solve for (k^2):[(1 - 4m^2)k^2 = 2m^2 - 2][k^2 = frac{2m^2 - 2}{1 - 4m^2}]Simplify numerator and denominator:Factor numerator: (2(m^2 - 1))Factor denominator: (-(4m^2 - 1)) or (1 - 4m^2 = -(4m^2 - 1))So,[k^2 = frac{2(m^2 - 1)}{-(4m^2 - 1)} = frac{2(1 - m^2)}{4m^2 - 1}]Thus,[k^2 = frac{2(1 - m^2)}{4m^2 - 1}]Since (k^2) must be non-negative, the right-hand side must be non-negative as well. So,[frac{2(1 - m^2)}{4m^2 - 1} geq 0]Let me analyze the numerator and denominator:Numerator: (2(1 - m^2)). This is positive when (1 - m^2 > 0), i.e., (|m| < 1), and negative when (|m| > 1).Denominator: (4m^2 - 1). This is positive when (|m| > frac{1}{2}), and negative when (|m| < frac{1}{2}).So, the fraction is non-negative when numerator and denominator have the same sign.Case 1: Both positive.Numerator positive: (|m| < 1)Denominator positive: (|m| > frac{1}{2})So, intersection: (frac{1}{2} < |m| < 1)Case 2: Both negative.Numerator negative: (|m| > 1)Denominator negative: (|m| < frac{1}{2})But (|m| > 1) and (|m| < frac{1}{2}) cannot happen simultaneously, so no solution here.Thus, the only valid case is (frac{1}{2} < |m| < 1). So, (m) is in ((-1, -frac{1}{2})) or ((frac{1}{2}, 1)).But we also need to ensure that the discriminant condition (Condition 1) is satisfied:[k^2 - 2m^2 + 2 > 0]We have (k^2 = frac{2(1 - m^2)}{4m^2 - 1}), so substitute:[frac{2(1 - m^2)}{4m^2 - 1} - 2m^2 + 2 > 0]Let me compute this expression step by step.First, let me write all terms with a common denominator. The denominator is (4m^2 - 1). So, express each term accordingly:1. (frac{2(1 - m^2)}{4m^2 - 1}) is already over the denominator.2. (-2m^2 = frac{-2m^2(4m^2 - 1)}{4m^2 - 1})3. (2 = frac{2(4m^2 - 1)}{4m^2 - 1})So, combining all terms:[frac{2(1 - m^2) - 2m^2(4m^2 - 1) + 2(4m^2 - 1)}{4m^2 - 1} > 0]Simplify the numerator:Compute each part:- (2(1 - m^2) = 2 - 2m^2)- (-2m^2(4m^2 - 1) = -8m^4 + 2m^2)- (2(4m^2 - 1) = 8m^2 - 2)Combine all together:[2 - 2m^2 - 8m^4 + 2m^2 + 8m^2 - 2]Simplify term by term:- Constants: (2 - 2 = 0)- (m^2) terms: (-2m^2 + 2m^2 + 8m^2 = 8m^2)- (m^4) term: (-8m^4)So, numerator simplifies to:[-8m^4 + 8m^2 = -8m^2(m^2 - 1)]Thus, the inequality becomes:[frac{-8m^2(m^2 - 1)}{4m^2 - 1} > 0]Factor out the negative sign:[frac{-8m^2(m^2 - 1)}{4m^2 - 1} = frac{8m^2(1 - m^2)}{4m^2 - 1}]So, the inequality is:[frac{8m^2(1 - m^2)}{4m^2 - 1} > 0]Since (8m^2) is always non-negative, and (m neq 0) (as we saw earlier), we can divide both sides by (8m^2) without changing the inequality direction:[frac{1 - m^2}{4m^2 - 1} > 0]So, the inequality reduces to:[frac{1 - m^2}{4m^2 - 1} > 0]Let me analyze this fraction.The numerator (1 - m^2) is positive when (|m| < 1), negative when (|m| > 1).The denominator (4m^2 - 1) is positive when (|m| > frac{1}{2}), negative when (|m| < frac{1}{2}).So, the fraction is positive when numerator and denominator have the same sign.Case 1: Both positive.Numerator positive: (|m| < 1)Denominator positive: (|m| > frac{1}{2})Intersection: (frac{1}{2} < |m| < 1)Case 2: Both negative.Numerator negative: (|m| > 1)Denominator negative: (|m| < frac{1}{2})No overlap, so no solution.Thus, the inequality holds when (frac{1}{2} < |m| < 1), which is consistent with our earlier finding.Therefore, combining all conditions, (m) must satisfy (frac{1}{2} < |m| < 1). So, the range of (m) is ((-1, -frac{1}{2}) cup (frac{1}{2}, 1)).I think that's the answer. Let me just recap to make sure I didn't miss anything.We found the ellipse equation correctly, then for part (2), we used vector conditions to find (lambda = 3), set up the quadratic equation for the line intersecting the ellipse, used Vieta's formulas with the given ratio to relate (x_1) and (x_2), solved for (k^2) in terms of (m), ensured (k^2) is non-negative, and then checked the discriminant condition which led us to the same interval for (m). So, I think it's consistent.**Final Answer**(1) The equation of the ellipse is (boxed{y^2 + 2x^2 = 1}).(2) The range of values for (m) is (boxed{(-1, -frac{1}{2}) cup (frac{1}{2}, 1)}).

💡Okay, so I have this equation: (x^4 - 4x^3 + 6x^2 - 4x = 1007). I need to find the product of the nonreal roots. Hmm, let me think about how to approach this.First, I remember that for polynomials, the product of all roots can be found using Vieta's formulas. But wait, Vieta's formulas give the product of all roots, not just the nonreal ones. So maybe I need another approach.Looking at the left-hand side of the equation, (x^4 - 4x^3 + 6x^2 - 4x), it seems familiar. It looks similar to the expansion of ((x - 1)^4). Let me check that:[(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1]Oh, right! So the left-hand side is almost ((x - 1)^4), but it's missing the constant term +1. So if I add 1 to both sides of the equation, I get:[x^4 - 4x^3 + 6x^2 - 4x + 1 = 1007 + 1][(x - 1)^4 = 1008]Okay, so now the equation is ((x - 1)^4 = 1008). To find the roots, I can take the fourth root of both sides. Remember, taking roots of complex numbers can give multiple solutions, especially when dealing with even roots.So, the fourth roots of 1008 are going to be the real and imaginary roots. Let me denote (sqrt[4]{1008}) as a real number. Then the solutions to the equation are:[x - 1 = sqrt[4]{1008}, quad x - 1 = -sqrt[4]{1008}, quad x - 1 = isqrt[4]{1008}, quad x - 1 = -isqrt[4]{1008}]Therefore, the roots are:[x = 1 + sqrt[4]{1008}, quad x = 1 - sqrt[4]{1008}, quad x = 1 + isqrt[4]{1008}, quad x = 1 - isqrt[4]{1008}]So, out of these four roots, two are real and two are nonreal. The nonreal roots are (1 + isqrt[4]{1008}) and (1 - isqrt[4]{1008}).Now, I need to find the product of these nonreal roots. Let me denote them as (a) and (b), where:[a = 1 + isqrt[4]{1008}, quad b = 1 - isqrt[4]{1008}]The product (a times b) is:[(1 + isqrt[4]{1008})(1 - isqrt[4]{1008})]This looks like the product of a binomial and its conjugate, which is a difference of squares. So, applying that formula:[(1)^2 - (isqrt[4]{1008})^2 = 1 - (i^2)(sqrt[4]{1008})^2]I know that (i^2 = -1), so substituting that in:[1 - (-1)(sqrt[4]{1008})^2 = 1 + (sqrt[4]{1008})^2]Now, ((sqrt[4]{1008})^2) is the same as (sqrt{1008}), because taking the square of the fourth root is the same as taking the square root.So, simplifying:[1 + sqrt{1008}]Therefore, the product of the nonreal roots is (1 + sqrt{1008}).Wait, let me double-check my steps to make sure I didn't make a mistake. I started by recognizing the left-hand side as a binomial expansion, which seems correct. Then, I added 1 to both sides to complete the fourth power, which makes sense. Solving for (x) by taking the fourth root gives four roots, two real and two nonreal. Then, I correctly identified the nonreal roots and multiplied them using the conjugate product formula. That step also seems correct because when you multiply a complex number by its conjugate, you get the sum of squares, which in this case led to (1 + sqrt{1008}). I think that's all correct. So, the product of the nonreal roots is indeed (1 + sqrt{1008}).**Final Answer**The product of the nonreal roots is boxed{1 + sqrt{1008}}.

💡Alright, so I've got this problem here: If ( a ) is a root of the equation ( x^2 + x - 1 = 0 ), then I need to find the value of the expression ( 2021 - 2a^2 - 2a ). The options are A: 2019, B: 2021, C: 2022, D: 2023.Hmm, okay. Let me think about this step by step. First, since ( a ) is a root of the equation ( x^2 + x - 1 = 0 ), that means when I plug ( a ) into the equation, it should satisfy it. So, substituting ( a ) into the equation gives:[ a^2 + a - 1 = 0 ]Right? That's the fundamental property of roots of equations. So, from this equation, I can express ( a^2 ) in terms of ( a ). Let me rearrange the equation:[ a^2 = -a + 1 ]So, ( a^2 ) is equal to ( -a + 1 ). That might be useful because the expression I need to evaluate has ( a^2 ) in it. Let me write down the expression again:[ 2021 - 2a^2 - 2a ]I notice that both ( a^2 ) and ( a ) are present here. Since I have an expression for ( a^2 ) in terms of ( a ), maybe I can substitute that in and simplify.Let me try that. Substitute ( a^2 = -a + 1 ) into the expression:[ 2021 - 2(-a + 1) - 2a ]Okay, let's simplify this step by step. First, distribute the -2 into the parentheses:[ 2021 - 2(-a) - 2(1) - 2a ]Which simplifies to:[ 2021 + 2a - 2 - 2a ]Wait, let me double-check that. The first term is 2021, then -2 times -a is +2a, and -2 times 1 is -2. So, yes, that gives:[ 2021 + 2a - 2 - 2a ]Now, let's combine like terms. The ( 2a ) and ( -2a ) terms will cancel each other out:[ 2021 - 2 ]Which is:[ 2019 ]So, the value of the expression ( 2021 - 2a^2 - 2a ) is 2019. Looking back at the options, that's option A.But wait, let me make sure I didn't make any mistakes. Maybe I should go through the steps again.Starting with the equation ( a^2 + a - 1 = 0 ), so ( a^2 = -a + 1 ). Substitute into the expression:[ 2021 - 2a^2 - 2a = 2021 - 2(-a + 1) - 2a ]Expanding the terms:[ 2021 + 2a - 2 - 2a ]Combine like terms:2021 - 2 = 2019, and 2a - 2a = 0. So, yes, it does simplify to 2019.Alternatively, maybe I can approach this differently. Since ( a^2 + a = 1 ), perhaps I can factor the expression ( 2021 - 2a^2 - 2a ) in terms of ( a^2 + a ).Let me see:[ 2021 - 2a^2 - 2a = 2021 - 2(a^2 + a) ]But from the equation, ( a^2 + a = 1 ), so substituting that in:[ 2021 - 2(1) = 2021 - 2 = 2019 ]Oh, that's a quicker way to see it. So, that confirms the same result.I think that's solid. I don't see any errors in the reasoning. So, the answer should be 2019, which is option A.**Final Answer**boxed{A}

💡First, I need to determine the total runs scored by the batsman in the first set of matches. Given that the average runs scored in 10 matches is 40, I can calculate the total runs by multiplying the average by the number of matches.Next, I'll calculate the total runs scored in the next set of 10 matches, where the average is 30 runs per match. Again, I'll multiply the average by the number of matches to find the total runs.After obtaining the total runs for both sets of matches, I'll add them together to find the overall total runs scored in all 20 matches.Finally, to find the average runs scored per match across all 20 matches, I'll divide the total runs by the total number of matches.