answer：Okay, so I have this problem with two parts. Let me start with part (1). We are given two fixed points A and B, with coordinates A(-√2, 0) and B(√2, 0). There's a moving point P, and its projection on the y-axis is Q. The condition given is 2 times the dot product of vectors PA and PB equals the square of the magnitude of vector PQ. I need to find the equation of the trajectory C of point P.First, let me visualize this. Points A and B are on the x-axis, symmetric about the origin. Point P is somewhere in the plane, and Q is its projection on the y-axis, so Q has the same x-coordinate as P but y-coordinate 0. So if P is (x, y), then Q is (x, 0).The condition is 2 * (PA · PB) = |PQ|². Let me write that in terms of coordinates.Vectors PA and PB can be expressed as:PA = A - P = (-√2 - x, 0 - y) = (-√2 - x, -y)PB = B - P = (√2 - x, 0 - y) = (√2 - x, -y)The dot product PA · PB is:(-√2 - x)(√2 - x) + (-y)(-y)Let me compute that:First, multiply the x-components:(-√2 - x)(√2 - x) = (-√2)(√2) + (-√2)(-x) + (-x)(√2) + (-x)(-x)= (-2) + (√2 x) + (-√2 x) + x²Simplify:-2 + 0 + x² = x² - 2Then, the y-components:(-y)(-y) = y²So the dot product PA · PB is (x² - 2) + y² = x² + y² - 2Now, the left side of the condition is 2 times this:2 * (x² + y² - 2) = 2x² + 2y² - 4The right side is |PQ|². Since Q is (x, 0) and P is (x, y), the vector PQ is (0, -y). So |PQ|² is (0)² + (-y)² = y².So the equation becomes:2x² + 2y² - 4 = y²Subtract y² from both sides:2x² + y² - 4 = 0Bring the 4 to the other side:2x² + y² = 4Divide both sides by 4 to get the standard form:(2x²)/4 + y²/4 = 1Simplify:x²/2 + y²/4 = 1Wait, that doesn't look right. Let me check my calculations again.Wait, no. Let me go back.Wait, when I computed PA · PB, I had:(-√2 - x)(√2 - x) + y²Which I expanded as:(-√2)(√2) + (-√2)(-x) + (-x)(√2) + (-x)(-x) + y²Which is:-2 + √2 x - √2 x + x² + y²So that's x² + y² - 2.Then 2*(PA · PB) is 2x² + 2y² - 4.And |PQ|² is y².So 2x² + 2y² - 4 = y²Subtract y²:2x² + y² - 4 = 0So 2x² + y² = 4Divide both sides by 4:(2x²)/4 + y²/4 = 1Which simplifies to:x²/2 + y²/4 = 1Wait, that's an ellipse with semi-major axis 2 along y and semi-minor axis √2 along x.But in the initial problem statement, the solution given was x²/4 + y²/2 = 1. Hmm, so I must have made a mistake.Let me check the dot product again.PA = (-√2 - x, -y)PB = (√2 - x, -y)Dot product PA · PB:(-√2 - x)(√2 - x) + (-y)(-y)Let me compute (-√2 - x)(√2 - x):Multiply term by term:First term: (-√2)(√2) = -2Second term: (-√2)(-x) = √2 xThird term: (-x)(√2) = -√2 xFourth term: (-x)(-x) = x²So adding them up:-2 + √2 x - √2 x + x² = x² - 2Then, the y-components: (-y)(-y) = y²So total dot product: x² - 2 + y²Thus, 2*(PA · PB) = 2x² + 2y² - 4And |PQ|² is y²So equation: 2x² + 2y² - 4 = y²Simplify: 2x² + y² - 4 = 0So 2x² + y² = 4Divide by 4: (2x²)/4 + y²/4 = 1 → x²/2 + y²/4 = 1Wait, but the solution given was x²/4 + y²/2 = 1. So perhaps I made a mistake in the algebra.Wait, let me check the initial condition again.Wait, maybe I misapplied the projection. Wait, the projection of P on the y-axis is Q, so Q is (0, y) if P is (x, y)? Wait, no, the projection on the y-axis would have the same y-coordinate but x=0. Wait, no, that's the projection on the y-axis. Wait, no, the projection on the y-axis would be (0, y), but in the problem, it's stated as Q being the projection on the y-axis, so if P is (x, y), then Q is (0, y). Wait, but in the initial solution, they took Q as (x, 0). Wait, that's conflicting.Wait, perhaps I misread the problem. Let me check again.The problem says: "a moving point P whose projection on the y-axis is Q". So projection on the y-axis would mean that Q is (0, y), since the projection on the y-axis would set x=0. But in the initial solution, they took Q as (x, 0), which is the projection on the x-axis. That seems contradictory.Wait, so perhaps the initial solution was wrong. Let me confirm.Wait, if Q is the projection on the y-axis, then Q is (0, y). So vector PQ would be from P(x, y) to Q(0, y), which is (-x, 0). So |PQ|² is x².Wait, that's different from what I had before. So perhaps that's where the mistake was.So let me redo the problem with this correction.Given that Q is the projection on the y-axis, so Q is (0, y). So vector PQ is Q - P = (0 - x, y - y) = (-x, 0). So |PQ|² is (-x)^2 + 0^2 = x².So the condition is 2*(PA · PB) = |PQ|² = x².So let's compute PA · PB again.PA = A - P = (-√2 - x, 0 - y) = (-√2 - x, -y)PB = B - P = (√2 - x, 0 - y) = (√2 - x, -y)Dot product PA · PB:(-√2 - x)(√2 - x) + (-y)(-y)Compute the x-component:(-√2 - x)(√2 - x) = (-√2)(√2) + (-√2)(-x) + (-x)(√2) + (-x)(-x)= (-2) + (√2 x) + (-√2 x) + x²Simplify: -2 + 0 + x² = x² - 2Then, the y-component:(-y)(-y) = y²So PA · PB = x² - 2 + y²Thus, 2*(PA · PB) = 2x² + 2y² - 4Set equal to |PQ|² = x²:2x² + 2y² - 4 = x²Subtract x²:x² + 2y² - 4 = 0So x² + 2y² = 4Divide by 4:x²/4 + y²/2 = 1Ah, that's the equation given in the initial solution. So I must have misread the projection initially. So the correct equation is x²/4 + y²/2 = 1, which is an ellipse centered at the origin with semi-major axis 2 along the x-axis and semi-minor axis √2 along the y-axis.So part (1) is solved, and the equation is x²/4 + y²/2 = 1.Now, moving on to part (2).We have point F at (1, 0). We draw two perpendicular lines through F, intersecting the ellipse C at points G, H, M, N. Then, E1 and E2 are the midpoints of GH and MN, respectively. We need to prove that the line E1E2 always passes through a fixed point.First, let me understand the setup.We have the ellipse x²/4 + y²/2 = 1.Point F is at (1, 0), which is inside the ellipse since plugging in (1,0) gives 1/4 + 0 = 1/4 < 1.We draw two lines through F that are perpendicular to each other. Each line intersects the ellipse at two points, say G and H for one line, and M and N for the other line. Then, E1 is the midpoint of GH, and E2 is the midpoint of MN. We need to show that the line connecting E1 and E2 passes through a fixed point, regardless of the choice of the two perpendicular lines.Let me consider parametric equations for the lines through F.Let me denote the slope of one line as k, then the slope of the perpendicular line would be -1/k.So, let's define:Line GH: y = k(x - 1)Line MN: y = (-1/k)(x - 1)We can find the points of intersection of these lines with the ellipse, then find the midpoints E1 and E2, and then find the equation of the line E1E2, and show that it passes through a fixed point.Let me proceed step by step.First, find the intersection points of line GH with the ellipse.Substitute y = k(x - 1) into the ellipse equation:x²/4 + [k²(x - 1)²]/2 = 1Multiply both sides by 4 to eliminate denominators:x² + 2k²(x - 1)² = 4Expand 2k²(x - 1)²:2k²(x² - 2x + 1) = 2k²x² - 4k²x + 2k²So the equation becomes:x² + 2k²x² - 4k²x + 2k² = 4Combine like terms:(1 + 2k²)x² - 4k²x + (2k² - 4) = 0This is a quadratic in x. Let me denote it as:A x² + B x + C = 0Where:A = 1 + 2k²B = -4k²C = 2k² - 4The solutions are x = [4k² ± sqrt(16k^4 - 4*(1 + 2k²)*(2k² - 4))]/(2*(1 + 2k²))But instead of finding the roots, since we need the midpoint, we can use the fact that the sum of the roots x1 + x2 = -B/A = (4k²)/(1 + 2k²)Similarly, the product x1 x2 = C/A = (2k² - 4)/(1 + 2k²)The midpoint E1 will have x-coordinate (x1 + x2)/2 = (4k²)/(2*(1 + 2k²)) = (2k²)/(1 + 2k²)To find the y-coordinate, since the line is y = k(x - 1), the midpoint's y-coordinate is k*( (x1 + x2)/2 - 1 ) = k*( (2k²)/(1 + 2k²) - 1 ) = k*( (2k² - (1 + 2k²))/(1 + 2k²) ) = k*( (-1)/(1 + 2k²) ) = -k/(1 + 2k²)So E1 is at (2k²/(1 + 2k²), -k/(1 + 2k²))Similarly, for line MN with slope -1/k, let's find E2.Line MN: y = (-1/k)(x - 1)Substitute into ellipse equation:x²/4 + [ ( (-1/k)(x - 1) )² ] /2 = 1Simplify:x²/4 + [ (1/k²)(x - 1)^2 ] /2 = 1Multiply both sides by 4:x² + 2*(1/k²)(x - 1)^2 = 4Expand 2*(1/k²)(x² - 2x + 1) = (2/k²)x² - (4/k²)x + 2/k²So the equation becomes:x² + (2/k²)x² - (4/k²)x + 2/k² = 4Combine like terms:(1 + 2/k²)x² - (4/k²)x + (2/k² - 4) = 0Multiply numerator and denominator by k² to simplify:(k² + 2)x² - 4x + (2 - 4k²) = 0Wait, let me check that step again.Wait, the equation is:(1 + 2/k²)x² - (4/k²)x + (2/k² - 4) = 0Multiply each term by k² to eliminate denominators:k²*(1 + 2/k²)x² - k²*(4/k²)x + k²*(2/k² - 4) = 0Simplify each term:(k² + 2)x² - 4x + (2 - 4k²) = 0So the quadratic equation is:(k² + 2)x² - 4x + (2 - 4k²) = 0Let me denote this as A x² + B x + C = 0, where:A = k² + 2B = -4C = 2 - 4k²The sum of roots x3 + x4 = -B/A = 4/(k² + 2)The product x3 x4 = C/A = (2 - 4k²)/(k² + 2)The midpoint E2 will have x-coordinate (x3 + x4)/2 = (4/(k² + 2))/2 = 2/(k² + 2)To find the y-coordinate, since the line is y = (-1/k)(x - 1), the midpoint's y-coordinate is (-1/k)*( (x3 + x4)/2 - 1 ) = (-1/k)*( (2/(k² + 2)) - 1 ) = (-1/k)*( (2 - (k² + 2))/(k² + 2) ) = (-1/k)*( (-k²)/(k² + 2) ) = (k)/(k² + 2)So E2 is at (2/(k² + 2), k/(k² + 2))Now, we have E1 at (2k²/(1 + 2k²), -k/(1 + 2k²)) and E2 at (2/(k² + 2), k/(k² + 2))We need to find the equation of the line E1E2 and show that it passes through a fixed point.Let me denote E1 as (x1, y1) and E2 as (x2, y2).x1 = 2k²/(1 + 2k²)y1 = -k/(1 + 2k²)x2 = 2/(k² + 2)y2 = k/(k² + 2)First, let's find the slope of E1E2.Slope m = (y2 - y1)/(x2 - x1)Compute numerator:y2 - y1 = [k/(k² + 2)] - [ -k/(1 + 2k²) ] = k/(k² + 2) + k/(1 + 2k²)Let me find a common denominator, which is (k² + 2)(1 + 2k²)So:= [k(1 + 2k²) + k(k² + 2)] / [(k² + 2)(1 + 2k²)]= [k + 2k³ + k³ + 2k] / [(k² + 2)(1 + 2k²)]= [3k + 3k³] / [(k² + 2)(1 + 2k²)]= 3k(1 + k²) / [(k² + 2)(1 + 2k²)]Denominator:x2 - x1 = [2/(k² + 2)] - [2k²/(1 + 2k²)] = 2/(k² + 2) - 2k²/(1 + 2k²)Again, common denominator is (k² + 2)(1 + 2k²)= [2(1 + 2k²) - 2k²(k² + 2)] / [(k² + 2)(1 + 2k²)]Expand numerator:= [2 + 4k² - 2k^4 - 4k²] / [(k² + 2)(1 + 2k²)]Simplify:= [2 - 2k^4] / [(k² + 2)(1 + 2k²)]= 2(1 - k^4) / [(k² + 2)(1 + 2k²)]Now, the slope m is:[3k(1 + k²)] / [2(1 - k^4)] = [3k(1 + k²)] / [2(1 - k²)(1 + k²)] ) = [3k] / [2(1 - k²)]So m = 3k / [2(1 - k²)]Now, let's write the equation of line E1E2 using point-slope form.Using point E1:y - y1 = m(x - x1)So:y - (-k/(1 + 2k²)) = [3k/(2(1 - k²))](x - 2k²/(1 + 2k²))Simplify:y + k/(1 + 2k²) = [3k/(2(1 - k²))](x - 2k²/(1 + 2k²))Let me try to simplify this equation.First, let me write it as:y = [3k/(2(1 - k²))](x - 2k²/(1 + 2k²)) - k/(1 + 2k²)Let me expand the right-hand side:= [3k/(2(1 - k²))]x - [3k/(2(1 - k²))]*(2k²/(1 + 2k²)) - k/(1 + 2k²)Simplify term by term:First term: [3k/(2(1 - k²))]xSecond term: [3k/(2(1 - k²))]*(2k²/(1 + 2k²)) = [3k * 2k²] / [2(1 - k²)(1 + 2k²)] = [6k³] / [2(1 - k²)(1 + 2k²)] = [3k³]/[(1 - k²)(1 + 2k²)]Third term: -k/(1 + 2k²)So combining:y = [3k/(2(1 - k²))]x - [3k³]/[(1 - k²)(1 + 2k²)] - k/(1 + 2k²)Let me combine the last two terms:= [3k/(2(1 - k²))]x - [3k³ + k(1 - k²)] / [(1 - k²)(1 + 2k²)]Simplify the numerator of the second term:3k³ + k(1 - k²) = 3k³ + k - k³ = 2k³ + kSo:y = [3k/(2(1 - k²))]x - [2k³ + k]/[(1 - k²)(1 + 2k²)]Factor numerator:= [3k/(2(1 - k²))]x - [k(2k² + 1)] / [(1 - k²)(1 + 2k²)]Notice that 2k² + 1 = 1 + 2k², so:= [3k/(2(1 - k²))]x - [k(1 + 2k²)] / [(1 - k²)(1 + 2k²)]Simplify the second term:= [3k/(2(1 - k²))]x - k/(1 - k²)So:y = [3k/(2(1 - k²))]x - k/(1 - k²)Factor out k/(1 - k²):= [k/(1 - k²)] * [ (3/2)x - 1 ]Let me write this as:y = [k/(1 - k²)] * ( (3x)/2 - 1 )Now, let me see if this line passes through a fixed point. Let's assume that for some x and y, this equation holds for all k. Let me set x = 2/3 and see what y is.If x = 2/3, then:y = [k/(1 - k²)] * ( (3*(2/3))/2 - 1 ) = [k/(1 - k²)] * ( (2)/2 - 1 ) = [k/(1 - k²)] * (1 - 1) = 0So when x = 2/3, y = 0 regardless of k. Therefore, the line E1E2 passes through the point (2/3, 0).To confirm, let me check another value. Suppose k = 1.Then, E1 is at (2*1/(1 + 2*1), -1/(1 + 2*1)) = (2/3, -1/3)E2 is at (2/(1 + 2), 1/(1 + 2)) = (2/3, 1/3)So the line E1E2 is from (2/3, -1/3) to (2/3, 1/3), which is a vertical line x = 2/3, passing through (2/3, 0).Another example, let k = 0. Then line GH is horizontal, y=0, intersecting the ellipse at x²/4 + 0 =1 → x=±2. So G and H are (-2,0) and (2,0). Midpoint E1 is (0,0).Line MN is vertical, x=1, since slope is undefined. Substitute x=1 into ellipse: 1/4 + y²/2 =1 → y²= 3/2 → y=±√(3/2). So M and N are (1, √(3/2)) and (1, -√(3/2)). Midpoint E2 is (1,0).So E1 is (0,0) and E2 is (1,0). The line E1E2 is the x-axis from (0,0) to (1,0), which passes through (2/3, 0) as well.Wait, but in this case, the line is the x-axis, which passes through (2/3, 0).Another test: let k approach infinity, meaning the line GH becomes vertical. Then, E1 would approach ( (2k²)/(2k²) , -k/(2k²) ) = (1, 0). Similarly, E2 would approach (2/(k² + 2), k/(k² + 2)) → (0,0). So the line E1E2 would be from (1,0) to (0,0), which is the x-axis, passing through (2/3, 0).Wait, but when k approaches infinity, the line GH becomes vertical, x=1, which intersects the ellipse at (1, y) where y² = 2*(1 - 1/4) = 2*(3/4) = 3/2, so y=±√(3/2). So points G and H are (1, √(3/2)) and (1, -√(3/2)), midpoint E1 is (1,0). Similarly, line MN becomes horizontal, y=0, intersecting the ellipse at x=±2, midpoint E2 is (0,0). So line E1E2 is from (1,0) to (0,0), which is the x-axis, passing through (2/3, 0).Thus, in all cases, the line E1E2 passes through (2/3, 0). Therefore, the fixed point is (2/3, 0).So, summarizing:Part (1): The equation of the trajectory is x²/4 + y²/2 = 1.Part (2): The line E1E2 always passes through the fixed point (2/3, 0).

answer：Alright, so I've got this equation to solve: |x|² - 3|x| - 10 = 0. Hmm, okay. Absolute value equations can sometimes be tricky, but I think I can handle this by making a substitution. Let me think... If I let y = |x|, then |x|² is just y², right? So the equation becomes y² - 3y - 10 = 0. That looks like a quadratic equation in terms of y, which I can solve using factoring or the quadratic formula.Let me try factoring first. I need two numbers that multiply to -10 and add up to -3. Hmm, 2 and -5? Yeah, because 2 * (-5) = -10 and 2 + (-5) = -3. So, the equation factors to (y + 2)(y - 5) = 0. Setting each factor equal to zero gives y + 2 = 0 or y - 5 = 0, so y = -2 or y = 5.Wait a second, y = |x|, and since absolute values can't be negative, y = -2 doesn't make sense in this context. So, the only valid solution is y = 5. That means |x| = 5. So, x can be either 5 or -5 because both have an absolute value of 5.Now, the question asks about the sum and product of the roots. The roots are 5 and -5. Let's calculate the sum: 5 + (-5) = 0. And the product: 5 * (-5) = -25. So, the sum of the roots is 0, and the product is -25.Looking at the options given:A) Sum of the roots is 0B) Product of the roots is -25C) Sum of the roots is 10D) Product of the roots is 25So, both A and B are correct. C and D don't match because the sum isn't 10 and the product isn't 25. It's important to remember that when dealing with absolute values, sometimes you get extraneous solutions, but in this case, only y = 5 was valid, leading to x = 5 and x = -5. Their sum and product are straightforward from there.I think I've covered all the steps. Substituted y for |x|, solved the quadratic, discarded the negative solution, found the roots, and then calculated their sum and product. Everything seems to check out. I don't think I made any mistakes here, but just to be sure, let me plug the roots back into the original equation.For x = 5: |5|² - 3|5| - 10 = 25 - 15 - 10 = 0. That works.For x = -5: |-5|² - 3|-5| - 10 = 25 - 15 - 10 = 0. That also works.Yep, both roots satisfy the equation. So, I'm confident that A and B are the correct answers.**Final Answer**boxed{textbf{(A)} text{ and } textbf{(B)}}

answer：Okay, so I have this function f(x) = a sin x + x², and I know that f(1) = 2. I need to find f(-1). Hmm, let's see. First, I should probably figure out what the value of 'a' is because it's a constant that affects the function. Alright, let's plug in x = 1 into the function. So, f(1) = a sin(1) + (1)². That simplifies to a sin(1) + 1. And we know that f(1) is equal to 2, so I can set up the equation: a sin(1) + 1 = 2. Now, I can solve for a sin(1). Subtracting 1 from both sides gives me a sin(1) = 1. Okay, so that tells me that a sin(1) is equal to 1. I'll keep that in mind because I might need it later.Next, I need to find f(-1). Let's plug x = -1 into the function. So, f(-1) = a sin(-1) + (-1)². I know that sin(-1) is equal to -sin(1) because sine is an odd function. So, sin(-1) = -sin(1). That means a sin(-1) is equal to -a sin(1). From earlier, I found that a sin(1) = 1, so substituting that in, a sin(-1) = -1. Now, let's not forget the other part of the function, which is (-1) squared. That's just 1. So, putting it all together, f(-1) = -1 + 1. Wait, that simplifies to 0. So, f(-1) is 0. That seems straightforward, but let me double-check to make sure I didn't make any mistakes. First, I found that a sin(1) = 1, which seems correct because f(1) = 2 and 1 squared is 1, so 2 - 1 = 1. Then, for f(-1), I used the property of sine being odd, so sin(-1) = -sin(1), which is correct. Then, substituting a sin(1) = 1 into that gives me -1. Adding the squared term, which is 1, gives me 0. Yeah, that all checks out. I don't think I made any calculation errors. So, I'm pretty confident that f(-1) is 0.

answer：Alright, so I have this function f(x) that's defined piecewise. It means it has different expressions depending on the value of x. Let me write it down again to make sure I have it right:f(x) = { x² + 4x + 3, when x < 2 ln(x - 1) + 1, when x ≥ 2}Okay, so the function has two parts. For values of x less than 2, it's a quadratic function, and for x equal to or greater than 2, it's a natural logarithm function shifted up by 1.First, I need to find f(e + 1). Hmm, e is approximately 2.718, so e + 1 is about 3.718. That's definitely greater than 2, so I should use the second part of the function, the logarithmic one.So, f(e + 1) = ln((e + 1) - 1) + 1. Simplifying inside the logarithm, (e + 1) - 1 is just e. So, f(e + 1) = ln(e) + 1. I remember that ln(e) is 1 because the natural logarithm of e is 1. So, 1 + 1 is 2. Therefore, f(e + 1) is 2. That seems straightforward.Now, the second part is finding the range of the function f(x). The range is all the possible output values (y-values) the function can take. Since the function is piecewise, I need to find the range for each piece separately and then combine them.Starting with the first piece: f(x) = x² + 4x + 3 for x < 2. This is a quadratic function, and its graph is a parabola. Since the coefficient of x² is positive (1), the parabola opens upwards, meaning it has a minimum point.To find the range, I can complete the square for the quadratic expression. Let's do that:x² + 4x + 3To complete the square, take the coefficient of x, which is 4, divide it by 2 to get 2, and square it to get 4. So, add and subtract 4:= (x² + 4x + 4) - 4 + 3= (x + 2)² - 1So, f(x) = (x + 2)² - 1 for x < 2. The vertex of this parabola is at (-2, -1). Since the parabola opens upwards, the minimum value is -1, and it goes to infinity as x increases or decreases.But wait, in this case, x is only less than 2. So, I need to see what values f(x) can take when x is less than 2.As x approaches 2 from the left, let's see what f(x) approaches. Plugging x = 2 into the quadratic: (2 + 2)² - 1 = 16 - 1 = 15. But since x is less than 2, f(x) approaches 15 but doesn't include it.As x goes to negative infinity, (x + 2)² grows without bound, so f(x) goes to infinity. Therefore, the range for the quadratic part is [-1, 15). Wait, no, that's not quite right. Because as x approaches negative infinity, (x + 2)² goes to infinity, so f(x) goes to infinity. But the minimum is -1, so the range should be [-1, ∞). But wait, when x is less than 2, the maximum value is approaching 15, but since x can go to negative infinity, the function can take any value from -1 upwards. Hmm, maybe I was confused earlier.Wait, let's think again. The quadratic function f(x) = (x + 2)² - 1 has a minimum at y = -1, and as x moves away from -2 in either direction, y increases. So, for x < 2, the function can take all values from -1 upwards, because even though x is restricted to less than 2, the function can still go to infinity as x approaches negative infinity. So, the range for the first part is [-1, ∞).But hold on, when x approaches 2 from the left, f(x) approaches (2 + 2)² - 1 = 16 - 1 = 15. So, does that mean the function can take values up to 15? But since x can be any number less than 2, including numbers much less than 2, the function can actually go beyond 15 as x becomes more negative. So, the range is still [-1, ∞).Now, moving on to the second piece: f(x) = ln(x - 1) + 1 for x ≥ 2. The natural logarithm function ln(x) is defined for x > 0, and it increases without bound as x increases, but it approaches negative infinity as x approaches 0 from the right.In this case, x - 1 must be greater than 0, so x > 1. But since our domain here is x ≥ 2, x - 1 is at least 1. So, ln(x - 1) is defined and increases as x increases.At x = 2, ln(2 - 1) = ln(1) = 0, so f(2) = 0 + 1 = 1.As x increases beyond 2, ln(x - 1) increases, so f(x) increases without bound. Therefore, the range for this part is [1, ∞).Now, combining both parts:- The first part gives us a range of [-1, ∞).- The second part gives us a range of [1, ∞).So, the overall range of f(x) is the union of these two ranges. The first part covers from -1 to infinity, and the second part covers from 1 to infinity. So, combining them, the overall range is [-1, ∞).Wait, but let me double-check. The first part goes from -1 to infinity, and the second part starts at 1 and goes to infinity. So, together, they cover from -1 to infinity. There's no gap between -1 and 1 because the first part already covers that. So, yes, the overall range is [-1, ∞).Just to make sure, let's see if there are any overlaps or gaps. The first part goes from -1 upwards, and the second part starts at 1, which is already included in the first part's range. So, no gaps, and the union is just [-1, ∞).Therefore, f(e + 1) is 2, and the range of f(x) is [-1, ∞).**Final Answer**The value of ( f(e+1) ) is (boxed{2}), and the range of the function ( f(x) ) is (boxed{[-1, +infty)}).

answer：Okay, so I have this quadratic equation: 4x² - 6x√3 + k = 0. I need to find the value of k and determine the nature of the roots. They told me that the discriminant is 18. Hmm, let me recall what the discriminant is. The discriminant of a quadratic equation ax² + bx + c = 0 is given by b² - 4ac. It tells us about the nature of the roots. If the discriminant is positive, there are two real and distinct roots; if it's zero, there's exactly one real root (a repeated root); and if it's negative, the roots are complex or imaginary.Alright, so in this equation, a is 4, b is -6√3, and c is k. The discriminant is given as 18. Let me write down the discriminant formula with these values.Discriminant (Δ) = b² - 4acΔ = (-6√3)² - 4 * 4 * kLet me compute (-6√3)² first. Squaring -6√3 would be (-6)² * (√3)², which is 36 * 3 = 108. So, the discriminant becomes:Δ = 108 - 16kThey told me that the discriminant is 18, so I can set up the equation:108 - 16k = 18Now, I need to solve for k. Let me subtract 108 from both sides:-16k = 18 - 108-16k = -90Now, divide both sides by -16:k = (-90)/(-16)k = 90/16Simplify that fraction. Both numerator and denominator are divisible by 2:90 ÷ 2 = 4516 ÷ 2 = 8So, k = 45/8.Now, since the discriminant is 18, which is positive, the roots are real and distinct. So, the equation has two different real roots.Let me double-check my calculations to make sure I didn't make any mistakes. Starting from the discriminant:Δ = (-6√3)² - 4*4*k = 108 - 16kSet equal to 18:108 - 16k = 18Subtract 108:-16k = -90Divide by -16:k = 90/16 = 45/8Yes, that seems correct. And since 18 is positive, the roots are real and distinct. So, the answer should be option A: k = 45/8, real and distinct roots.

answer：Alright, so I need to find the inverse function of ( f(x) = x^2 ) but only for the domain where ( x < -2 ). Hmm, okay. Let me think about how to approach this step by step.First, I remember that an inverse function essentially undoes what the original function does. So, if ( f(x) = x^2 ), then the inverse function should take a value from the range of ( f ) and give me back the original ( x ) that produced it. But wait, ( f(x) = x^2 ) isn't one-to-one over all real numbers because both positive and negative ( x ) values can give the same ( y ). That means it doesn't have an inverse unless we restrict its domain.In this case, the domain is restricted to ( x < -2 ). That makes sense because if we only consider negative ( x ) values less than -2, the function becomes one-to-one. So, for every ( y ) in the range, there will be exactly one ( x ) in the domain ( x < -2 ) that maps to it. Good, so now I can proceed.Let me write down the function again:( y = x^2 )I need to solve for ( x ) in terms of ( y ) to find the inverse function. So, starting with:( y = x^2 )To solve for ( x ), I can take the square root of both sides. But wait, since ( x ) is less than -2, which is negative, the square root should be negative. So, I need to take the negative square root. That gives me:( x = -sqrt{y} )Okay, so that's the inverse function. But I should also consider the domain of the inverse function. The original function ( f(x) = x^2 ) with ( x < -2 ) will have a range. Let me figure out what that range is.Since ( x < -2 ), squaring ( x ) will give me positive values. The smallest value ( x ) can take is just below -2, so squaring that gives just above 4. As ( x ) becomes more negative (like -3, -4, etc.), ( y ) will increase beyond 9, 16, and so on. So, the range of ( f(x) ) is ( y > 4 ). Therefore, the domain of the inverse function will be ( y > 4 ).Putting it all together, the inverse function is:( f^{-1}(y) = -sqrt{y} ) with the domain ( y > 4 ).But usually, we write functions in terms of ( x ), so replacing ( y ) with ( x ) gives:( f^{-1}(x) = -sqrt{x} ) with the domain ( x > 4 ).Let me double-check my steps to make sure I didn't make a mistake. Starting with ( y = x^2 ), solving for ( x ) gives ( x = pmsqrt{y} ). But since ( x < -2 ), we discard the positive root and take the negative one, so ( x = -sqrt{y} ). Then, the range of the original function is indeed ( y > 4 ), so the domain of the inverse is ( x > 4 ). Yeah, that seems correct.I also remember that the graph of the inverse function is the reflection of the original function over the line ( y = x ). So, if I graph ( f(x) = x^2 ) for ( x < -2 ), it's a parabola opening upwards, starting from ( x = -2 ) and going to the left. Reflecting that over ( y = x ) should give me a function that looks like the left half of the square root function, but only for ( x > 4 ). That makes sense because square roots are defined for non-negative numbers, and since our original function's range was ( y > 4 ), the inverse should only take inputs greater than 4.Wait, let me visualize this. The original function ( f(x) = x^2 ) for ( x < -2 ) is a decreasing function because as ( x ) becomes more negative, ( y ) increases. So, the inverse function should also be decreasing, which it is because as ( x ) increases, ( -sqrt{x} ) becomes more negative. That seems consistent.Is there any other way to verify this? Maybe by plugging in some numbers. Let's take ( x = -3 ). Plugging into ( f(x) ), we get ( y = (-3)^2 = 9 ). Now, applying the inverse function to ( y = 9 ), we should get back ( x = -3 ). Let's see: ( f^{-1}(9) = -sqrt{9} = -3 ). Perfect, that works.Another test: ( x = -4 ). Then, ( y = (-4)^2 = 16 ). Applying the inverse: ( f^{-1}(16) = -sqrt{16} = -4 ). That also works. How about a value just above 4? Let's say ( y = 5 ). Then, ( f^{-1}(5) = -sqrt{5} approx -2.236 ). But wait, the original function was defined for ( x < -2 ), so ( -2.236 ) is indeed less than -2. That checks out.What if I tried a value less than or equal to 4? For example, ( y = 4 ). Then, ( f^{-1}(4) = -sqrt{4} = -2 ). But the original function was defined for ( x < -2 ), not including -2. So, ( y = 4 ) isn't in the range of the original function, which is why the inverse function isn't defined for ( x leq 4 ). That makes sense.I think I've covered all the bases here. I've derived the inverse function, checked the domain and range, tested it with specific values, and even considered the graphical interpretation. Everything seems consistent. So, I'm confident that the inverse function is ( y = -sqrt{x} ) with the domain ( x > 4 ).**Final Answer**The inverse function is boxed{y = -sqrt{x}} for ( x > 4 ).