🧠:[<think>Okay, let's try to solve this problem step by step. So, we have a complex number z defined as (a - i)/(1 - i), where a is a real number. We need to find the value of a such that the modulus of z is equal to the integral from 0 to π of (sin x - 1/π) dx. The answer choices are given as A to D. First, let me recall that the modulus of a complex number z = (c + di) is |z| = sqrt(c² + d²). So, I need to compute the modulus of z here. But z is given as a fraction, so maybe I should first simplify z by rationalizing the denominator. Starting with z = (a - i)/(1 - i). To rationalize the denominator, I'll multiply both the numerator and the denominator by the complex conjugate of the denominator, which is (1 + i). So:z = [(a - i)(1 + i)] / [(1 - i)(1 + i)]Let's compute the denominator first. (1 - i)(1 + i) is a difference of squares, so it's 1² - (i)² = 1 - (-1) = 2. So the denominator becomes 2.Now the numerator: (a - i)(1 + i). Let's expand this:= a*(1) + a*(i) - i*(1) - i*(i)= a + ai - i - i²Since i² = -1, this becomes:= a + ai - i - (-1)= a + ai - i + 1Combine like terms. The real parts are a + 1. The imaginary parts are ai - i = (a - 1)i. So the numerator simplifies to:(a + 1) + (a - 1)iTherefore, z = [ (a + 1) + (a - 1)i ] / 2So, the complex number z can be written as ( (a + 1)/2 ) + ( (a - 1)/2 )i.Now, to find the modulus |z|, which is sqrt( [ (a + 1)/2 ]² + [ (a - 1)/2 ]² ).Let me compute that:|z| = sqrt( [ (a + 1)² + (a - 1)² ] / 4 )Factor out 1/4:= sqrt( [ (a + 1)² + (a - 1)² ] ) / 2Now compute the numerator inside the square root:(a + 1)² + (a - 1)²Expanding both:= (a² + 2a + 1) + (a² - 2a + 1)= a² + 2a + 1 + a² - 2a + 1= 2a² + 2So, |z| = sqrt(2a² + 2)/2Simplify sqrt(2(a² + 1))/2 = sqrt(2) * sqrt(a² + 1)/2 = (sqrt(a² + 1))/sqrt(2)Alternatively, |z| = sqrt( (2a² + 2)/4 ) = sqrt( (a² + 1)/2 ) = sqrt(a² + 1)/sqrt(2)So, that's |z|. Now, we need to compute the integral given: ∫₀^π (sin x - 1/π) dx. Then set |z| equal to that integral and solve for a.Let me compute the integral step by step. Let's split the integral into two parts:∫₀^π sin x dx - ∫₀^π (1/π) dxFirst integral: ∫ sin x dx from 0 to π. The integral of sin x is -cos x, so:[-cos x]₀^π = (-cos π) - (-cos 0) = -(-1) - (-1) = 1 + 1 = 2Second integral: ∫₀^π (1/π) dx. The integral of 1/π dx from 0 to π is (1/π)(π - 0) = 1.Therefore, the integral is 2 - 1 = 1.So, |z| = 1. Therefore, we have:sqrt(a² + 1)/sqrt(2) = 1Multiply both sides by sqrt(2):sqrt(a² + 1) = sqrt(2)Square both sides:a² + 1 = 2Subtract 1:a² = 1Therefore, a = ±1Looking back at the answer choices, option A is ±1, which matches this result. So, the answer should be A.But let me double-check my steps to make sure I didn't make a mistake.First, simplifying z: (a - i)/(1 - i). Multiply numerator and denominator by (1 + i). Denominator becomes (1 - i)(1 + i) = 1 + i - i - i² = 1 - (-1) = 2. Correct.Numerator: (a - i)(1 + i) = a*1 + a*i - i*1 - i*i = a + ai - i - i². Since i² = -1, becomes a + ai - i + 1. Combine terms: (a + 1) + (a - 1)i. Correct.Then modulus: sqrt( [(a + 1)/2]² + [(a - 1)/2]² ) = sqrt( (a +1)^2 + (a -1)^2 ) / 2. Expanding: (a^2 + 2a +1) + (a^2 -2a +1) = 2a^2 + 2. So sqrt(2a^2 + 2)/2 = sqrt(2(a^2 +1))/2 = sqrt(a^2 +1)/sqrt(2). Correct.Integral: ∫₀^π sinx dx = 2, ∫₀^π (1/π)dx = 1. So 2 -1 =1. Correct.Set |z| =1: sqrt(a² +1)/sqrt(2) =1 => sqrt(a² +1) = sqrt(2) => a² +1=2 => a²=1 => a=±1. Correct.Therefore, answer is A. The options are A: ±1, B:1, C:-1, D:±1/2. So, A is the correct choice.Final Answerboxed{A}</think>Given the complex number ( z = frac{a - i}{1 - i} ) where ( a in mathbb{R} ), we need to find the value of ( a ) such that ( |z| = int_{0}^{pi} left( sin x - frac{1}{pi} right) dx ).First, we simplify ( z ) by rationalizing the denominator:[z = frac{(a - i)(1 + i)}{(1 - i)(1 + i)} = frac{(a - i)(1 + i)}{2}]Expanding the numerator:[(a - i)(1 + i) = a + ai - i - i^2 = a + ai - i + 1 = (a + 1) + (a - 1)i]Thus,[z = frac{(a + 1) + (a - 1)i}{2} = frac{a + 1}{2} + frac{a - 1}{2}i]The modulus of ( z ) is:[|z| = sqrt{left( frac{a + 1}{2} right)^2 + left( frac{a - 1}{2} right)^2} = sqrt{frac{(a + 1)^2 + (a - 1)^2}{4}} = sqrt{frac{2a^2 + 2}{4}} = sqrt{frac{a^2 + 1}{2}} = frac{sqrt{a^2 + 1}}{sqrt{2}}]Next, we compute the integral:[int_{0}^{pi} left( sin x - frac{1}{pi} right) dx = int_{0}^{pi} sin x , dx - frac{1}{pi} int_{0}^{pi} dx = left[ -cos x right]_{0}^{pi} - frac{1}{pi} left[ x right]_{0}^{pi}][= left( -cos pi + cos 0 right) - frac{1}{pi} (pi - 0) = (1 + 1) - 1 = 2 - 1 = 1]Setting ( |z| = 1 ):[frac{sqrt{a^2 + 1}}{sqrt{2}} = 1 implies sqrt{a^2 + 1} = sqrt{2} implies a^2 + 1 = 2 implies a^2 = 1 implies a = pm 1]Thus, the value of ( a ) is (boxed{A}).,<think>Okay, let me try to solve this problem step by step. So, the question is about a complex number z given by the expression (a - i)/(1 - i), where a is a real number. We need to find the value of a such that the modulus of z is equal to the integral from 0 to π of (sin x - 1/π) dx. The answer choices are given as A to D. Hmm, let's break this down.First, let me recall that the modulus of a complex number z = (x + yi) is sqrt(x² + y²). So, if I can express z in the form of x + yi, then I can compute its modulus. The given z is (a - i)/(1 - i). To simplify this, I think I need to multiply the numerator and denominator by the conjugate of the denominator to rationalize it. The conjugate of (1 - i) is (1 + i). So, multiplying numerator and denominator by (1 + i):z = [(a - i)(1 + i)] / [(1 - i)(1 + i)]Let me compute the denominator first. (1 - i)(1 + i) is 1² - (i)² = 1 - (-1) = 2. So, the denominator is 2.Now the numerator: (a - i)(1 + i). Let's expand this:= a*1 + a*i - i*1 - i*i= a + ai - i - i²Since i² = -1, this becomes:= a + ai - i + 1= (a + 1) + (a - 1)iSo, z = [(a + 1) + (a - 1)i] / 2Therefore, z can be written as ( (a + 1)/2 ) + ( (a - 1)/2 )iTherefore, the real part is (a + 1)/2 and the imaginary part is (a - 1)/2.The modulus |z| is sqrt[ ( (a + 1)/2 )² + ( (a - 1)/2 )² ]Let me compute that:= sqrt[ ( (a + 1)² + (a - 1)² ) / 4 ]Let me expand the numerator:(a + 1)² = a² + 2a + 1(a - 1)² = a² - 2a + 1Adding them together:a² + 2a + 1 + a² - 2a + 1 = 2a² + 2So, |z| = sqrt[ (2a² + 2)/4 ] = sqrt[ (a² + 1)/2 ] = sqrt(a² + 1)/sqrt(2)So, that's the modulus of z. Now, we need this to be equal to the integral from 0 to π of (sin x - 1/π) dx.Let me compute that integral.First, split the integral into two parts:∫₀^π sin x dx - ∫₀^π (1/π) dxCompute the first integral: ∫ sin x dx from 0 to π. The integral of sin x is -cos x, so:[-cos x] from 0 to π = (-cos π) - (-cos 0) = (-(-1)) - (-1) = 1 + 1 = 2Second integral: ∫₀^π (1/π) dx. The integral of 1/π dx is (1/π)x, evaluated from 0 to π:(1/π)(π) - (1/π)(0) = 1 - 0 = 1So, the entire integral is 2 - 1 = 1.Therefore, |z| = 1.Therefore, sqrt(a² + 1)/sqrt(2) = 1Multiply both sides by sqrt(2):sqrt(a² + 1) = sqrt(2)Square both sides:a² + 1 = 2Subtract 1:a² = 1Therefore, a = ±1Looking at the options, A is ±1, B is 1, C is -1, D is ±1/2. So the answer should be A.Wait, but let me double-check all steps to make sure I didn't make a mistake.First, simplifying z = (a - i)/(1 - i). Multiply numerator and denominator by (1 + i):Numerator: (a - i)(1 + i) = a(1) + a(i) - i(1) - i(i) = a + ai - i - i². Since i² = -1, this becomes a + ai - i +1. Then combine like terms: (a + 1) + (a -1)i. So that's correct.Denominator: (1 - i)(1 + i) = 1 - i² = 1 - (-1) = 2. Correct.So z = [(a + 1) + (a - 1)i]/2. Therefore, real part (a + 1)/2, imaginary part (a -1)/2. Then modulus squared is [(a +1)/2]^2 + [(a -1)/2]^2. Expanding both:(a² + 2a +1)/4 + (a² -2a +1)/4 = (2a² + 2)/4 = (a² +1)/2. So modulus is sqrt((a² +1)/2). Correct.Then the integral. Let's check again. The integral of sin x from 0 to π is 2. Then the integral of 1/π from 0 to π is (1/π)*(π -0) = 1. So 2 -1 =1. Therefore, |z| =1. Then sqrt((a² +1)/2)=1. Squared: (a² +1)/2 =1 → a² +1=2→ a²=1→a=±1. Correct.So the answer should be A: ±1. So that's option A. The other options are B:1, which is only the positive, but since a can be both 1 and -1, the correct answer is A. So yeah, I think that's correct.Final Answerboxed{A}</think>Given the complex number ( z = frac{a - i}{1 - i} ) where ( a in mathbb{R} ), we need to find ( a ) such that ( |z| = int_{0}^{pi} left( sin x - frac{1}{pi} right) dx ).First, simplify ( z ):[z = frac{(a - i)(1 + i)}{(1 - i)(1 + i)} = frac{(a + 1) + (a - 1)i}{2}]The modulus of ( z ) is:[|z| = sqrt{left( frac{a + 1}{2} right)^2 + left( frac{a - 1}{2} right)^2} = sqrt{frac{(a + 1)^2 + (a - 1)^2}{4}} = sqrt{frac{2a^2 + 2}{4}} = sqrt{frac{a^2 + 1}{2}}]Next, compute the integral:[int_{0}^{pi} left( sin x - frac{1}{pi} right) dx = int_{0}^{pi} sin x , dx - int_{0}^{pi} frac{1}{pi} , dx][= left[ -cos x right]_{0}^{pi} - frac{1}{pi} left[ x right]_{0}^{pi} = (1 + 1) - frac{1}{pi} (pi - 0) = 2 - 1 = 1]Set the modulus equal to the integral result:[sqrt{frac{a^2 + 1}{2}} = 1]Squaring both sides:[frac{a^2 + 1}{2} = 1 implies a^2 + 1 = 2 implies a^2 = 1 implies a = pm 1]Thus, the value of ( a ) is (boxed{A}).]

🔑:First, calculate |z|= int_{0}^{pi} (sin x- frac {1}{pi})dx=(-cos x- frac {x}{pi})|_{0}^{pi}=(-cos pi-1)-(-cos 0-0)=1.Since z= frac {a-i}{1-i}= frac {(a-i)(1+i)}{2}= frac {a+1}{2}+ frac {a-1}{2}i,then left( frac {a+1}{2}right)^{2}+left( frac {a-1}{2}right)^{2}=1.Solving this equation yields a=pm1.Therefore, the correct choice is: boxed{A}.This problem involves calculating a definite integral to find |z|, then simplifying z using the algebraic operations of complex numbers, and substituting into the formula for the modulus of a complex number to find the value of a. It tests the method of calculating definite integrals and the method of finding the modulus of complex numbers, making it a basic question.

🧠:[<think>Okay, let me think about this problem step by step. So, we have two dance teams, A and B, both with an average height of 1.65 meters. The question is asking which team has more uniform heights. The key here is probably related to the variances given for each team. Team A has a variance of 1.5, and Team B has a variance of 2.4. First, let me recall what variance means. Variance is a measure of how spread out the numbers in a dataset are. A higher variance indicates that the data points are more spread out from the mean, and a lower variance means they are closer to the mean. Since both teams have the same average height, comparing their variances should directly tell us which team has heights that are more clustered around the mean, which would mean more uniform.So, if Team A has a variance of 1.5 and Team B has a variance of 2.4, then Team A has a lower variance. Lower variance implies that the heights in Team A are less spread out compared to Team B. Therefore, Team A's heights are more uniform because there's less variability among the members' heights. Wait, but just to make sure I didn't mix up variance and standard deviation. Variance is the square of the standard deviation. However, when comparing variances directly, the one with the smaller value still indicates less spread. So even though standard deviation is in the same units as the original data, in this case, since both variances are given in the same units (meters squared), comparing them directly is valid. Therefore, Team A with the lower variance (1.5) has more uniform heights compared to Team B with a higher variance (2.4).Is there anything else I need to consider here? Maybe sample size? The problem doesn't mention the number of members in each team. Typically, variance can be affected by sample size, but since the problem doesn't provide that information, I assume that the variances given are already calculated appropriately, whether they are sample variances or population variances. The symbols used are S_A² and S_B², which usually denote sample variances. However, even if they are sample variances, the comparison between them is still valid for uniformity. Since both are variances calculated the same way (assuming same degrees of freedom), the lower one still indicates more uniformity.Another thought: sometimes people confuse variance with other measures like range or mean absolute deviation. But the problem specifically mentions variance, so I don't need to calculate anything else. The answer should be based purely on which variance is smaller. So, summarizing: Variance measures spread. Lower variance means more uniform. Team A has lower variance. Therefore, Team A has more uniform heights.Final Answerboxed{A}</think>The problem involves determining which dance team, A or B, has more uniform heights. Both teams have the same average height of 1.65 meters, but different variances: Team A has a variance of ( S_{text{A}}^2 = 1.5 ) and Team B has a variance of ( S_{text{B}}^2 = 2.4 ).Variance is a measure of how spread out the data points are from the mean. A lower variance indicates that the data points are closer to the mean, meaning the heights are more uniform. Since Team A has a lower variance (1.5) compared to Team B (2.4), Team A's heights are less spread out and more uniform.Thus, the team with more uniform heights is:[boxed{A}],<think>Okay, so the problem is about two dance teams, A and B, both with an average height of 1.65 meters. The variances are given as 1.5 for team A and 2.4 for team B. The question is asking which team has more uniform heights. I need to figure out whether it's team A or B.First, I remember that variance is a measure of how spread out the data points are. A lower variance means the data points are closer to the mean, which would indicate more uniformity. Conversely, a higher variance means the data is more spread out, so less uniform. Since both teams have the same average height, comparing variances directly should tell us which one is more uniform.So team A has a variance of 1.5 and team B has a variance of 2.4. Comparing these two numbers, 1.5 is less than 2.4. That would mean team A's heights are less spread out compared to the mean, so their heights are more uniform. Therefore, team A should be the answer.Wait, let me make sure I didn't mix up variance with standard deviation. Variance is the square of the standard deviation, but in terms of interpretation, lower variance still means less spread. So even if we took the standard deviation, the square roots of 1.5 and 2.4 would be approximately 1.225 and 1.549 respectively, confirming that team A has a smaller standard deviation as well. So yeah, team A is more uniform. Is there any catch here? The problem mentions "uniform heights," which directly relates to variability. Since both teams have the same mean, comparing variances is appropriate. There's no need to consider other measures like range or interquartile range because variance is already given. Also, since the units are the same (meters), we don't have to worry about unit conversion. Another thought: sometimes people confuse variance with the mean when talking about averages, but in this case, the question is clearly about uniformity, which is about spread, not central tendency. So, no issues there. I think that's all. The key points are that variance measures spread, lower variance means more uniformity, and since 1.5 < 2.4, team A has more uniform heights. So the answer should be A.Final Answerboxed{A}</think>The problem involves two dance teams, A and B, both with an average height of 1.65 meters. The variances of their heights are given as ( S_{text{A}}^2 = 1.5 ) for team A and ( S_{text{B}}^2 = 2.4 ) for team B. To determine which team has more uniform heights, we compare the variances. Variance measures the spread of data points around the mean. A lower variance indicates that the data points are closer to the mean, implying more uniformity. Since ( 1.5 ) (variance of team A) is less than ( 2.4 ) (variance of team B), team A's heights are less spread out and therefore more uniform. This conclusion is consistent when considering the standard deviations as well, where the square roots of the variances (approximately 1.225 for team A and 1.549 for team B) confirm that team A has a smaller spread.Thus, the team with more uniform heights is (boxed{A}).]

🔑:Analysis of the Problem: Variance is a measure used to quantify the amount of variation or dispersion of a set of data values. Therefore, a judgment can be made based on the variances of teams A and B.Solution Analysis: Since S_{text{A}}^2 < S_{text{B}}^2, it indicates that the heights in team A are more uniform.Therefore, the team with more uniform heights is team A.Hence, the answer is boxed{text{A}}.Key Point: Variance.

🧠:[<think>Okay, let's see. I need to solve this problem where there's a function f(x) defined piecewise, and a geometric sequence {a_n} with positive common ratio. They give that a_3*a_4*a_5 = 1, and the sum of f(a_1) + f(a_2) + ... + f(a_6) equals 2a_1. We need to find a_1.First, let me recall what a geometric sequence is. A geometric sequence has each term multiplied by a common ratio r. So, if a_1 is the first term, then a_2 = a_1*r, a_3 = a_1*r^2, and so on. Since the common ratio is greater than 0, all terms will be positive, which is good because the function f(x) is defined for x > 0.Given that a_3*a_4*a_5 = 1. Let me write down those terms:a_3 = a_1*r^2a_4 = a_1*r^3a_5 = a_1*r^4Multiplying them together: (a_1*r^2)*(a_1*r^3)*(a_1*r^4) = a_1^3 * r^(2+3+4) = a_1^3 * r^9 = 1So, a_1^3 * r^9 = 1. That's equation (1).Then, the sum f(a_1) + f(a_2) + ... + f(a_6) = 2a_1.So, we need to compute f(a_1) through f(a_6) using the piecewise function and sum them up. Let's note that the function f(x) is defined as x ln x for x ≥ 1 and (ln x)/x for 0 < x < 1.Since the common ratio r is positive, and the terms of the geometric sequence depend on r and a_1. Depending on the value of r and a_1, each term a_n could be greater than or equal to 1 or less than 1. Therefore, we need to figure out for each term a_n whether it's in [1, ∞) or (0,1) to apply the correct part of the function.This seems tricky because depending on r and a_1, the terms could cross the threshold x=1. So, maybe we need to consider different cases? For example, if r > 1, the sequence is increasing, so if a_1 is less than 1, terms might cross 1. If r < 1, the sequence is decreasing, so if a_1 is greater than 1, terms might drop below 1. If r = 1, all terms are equal, so either all terms are >=1 or all <1. But since the problem states the common ratio is greater than 0, so r can be 1, but perhaps we need to check.Wait, but maybe we can find relations using equation (1) and the sum equation. Let me try to express everything in terms of a_1 and r. Let's see.First, from equation (1): a_1^3 * r^9 = 1. So, (a_1 * r^3)^3 = 1, which implies a_1 * r^3 = 1. So, a_1 = 1 / r^3. That's useful! So, a_1 is 1 over r cubed. Therefore, we can express all terms in terms of r.Therefore, a_1 = 1/r^3a_2 = a_1 * r = 1/r^3 * r = 1/r^2a_3 = a_1 * r^2 = 1/r^3 * r^2 = 1/ra_4 = a_1 * r^3 = 1/r^3 * r^3 = 1a_5 = a_1 * r^4 = 1/r^3 * r^4 = ra_6 = a_1 * r^5 = 1/r^3 * r^5 = r^2So, the terms are:a_1 = 1/r^3a_2 = 1/r^2a_3 = 1/ra_4 = 1a_5 = ra_6 = r^2So, now, let's note that:- a_1 = 1/r^3: whether this is ≥1 or <1 depends on r. If r < 1, then 1/r^3 >1. If r =1, then a_1=1. If r >1, then a_1 <1.Similarly,a_2 = 1/r^2: same as a_1 but one less power. So, if r <1, 1/r^2 >1. If r=1, 1. If r>1, <1.a_3 =1/r: same logic. If r <1, then 1/r >1; if r=1, 1; if r>1, <1.a_4=1, so f(a_4)=1*ln1=0, since ln1=0.a_5=r: if r >=1, then a_5 >=1; if r <1, a_5 <1.a_6=r^2: same as a_5. If r >=1, then r^2 >=1; if r <1, r^2 <1.Therefore, depending on whether r is greater than or less than 1, different terms will fall into different cases of the function f(x).So, we have two possibilities for r: r >=1 or r <1. Let's consider both cases.Case 1: r >=1In this case:a_1 =1/r^3. Since r >=1, 1/r^3 <=1. But since r is positive, 1/r^3 is in (0,1].Similarly, a_2=1/r^2 <=1, a_3=1/r <=1, a_4=1, a_5=r >=1, a_6=r^2 >=1.Therefore, for each term:a_1: 0 < x <1: use f(x)= (ln x)/xa_2: same as a_1a_3: same as a_1a_4: x=1: use x ln x, which is 1*0=0a_5: x >=1: use x ln xa_6: x >=1: use x ln xCase 2: r <1Here:a_1=1/r^3 >1, since r <1a_2=1/r^2 >1a_3=1/r >1a_4=1a_5=r <1a_6=r^2 <1Therefore, for each term:a_1: x >=1: use x ln xa_2: samea_3: samea_4: 0, as beforea_5: 0 <x <1: (ln x)/xa_6: same as a_5Therefore, depending on the case, the f(a_n) will be computed differently.Now, we need to compute the sum f(a1)+f(a2)+...+f(a6) in each case and set equal to 2a1, then solve for a1 (or r first, since a1 is expressed in terms of r).Given that a1=1/r^3, so once we find r, we can find a1.So, let's handle Case 1 and Case 2 separately.Case 1: r >=1Compute f(a1) + f(a2) + f(a3) + f(a4) + f(a5) + f(a6)Each term:f(a1)= (ln a1)/a1But a1=1/r^3, so ln(1/r^3)/(1/r^3) = (-3 ln r)/(1/r^3) = -3 r^3 ln rSimilarly, f(a2)= (ln a2)/a2. a2=1/r^2, so ln(1/r^2)/(1/r^2) = (-2 ln r)/(1/r^2) = -2 r^2 ln rf(a3)= (ln a3)/a3. a3=1/r, so ln(1/r)/(1/r) = (-ln r)/ (1/r) = -r ln rf(a4)=0, as before.f(a5)=a5 ln a5 = r ln rf(a6)=a6 ln a6 = r^2 ln r^2 = 2 r^2 ln rTherefore, summing all these:f(a1)+f(a2)+f(a3)+f(a4)+f(a5)+f(a6)=(-3 r^3 ln r) + (-2 r^2 ln r) + (-r ln r) + 0 + (r ln r) + (2 r^2 ln r)Let me compute term by term:First term: -3 r^3 ln rSecond term: -2 r^2 ln rThird term: -r ln rFourth term: 0Fifth term: +r ln rSixth term: +2 r^2 ln rSo, combining:-3 r^3 ln r -2 r^2 ln r - r ln r + r ln r + 2 r^2 ln rSimplify:-3 r^3 ln r + (-2 r^2 ln r + 2 r^2 ln r) + (-r ln r + r ln r) = -3 r^3 ln r + 0 + 0 = -3 r^3 ln rTherefore, in Case 1 (r >=1), the sum is -3 r^3 ln r. According to the problem, this sum equals 2 a1. Since a1=1/r^3, so 2a1=2/r^3.Therefore, equation:-3 r^3 ln r = 2 / r^3Multiply both sides by r^3:-3 r^6 ln r = 2But r >=1 here. Let's check if possible.But let's see, if r >=1, then ln r >=0, so left side is -3 r^6 ln r, which is non-positive (since ln r >=0 and multiplied by -3). But the right side is 2, which is positive. So, equality is impossible here. Therefore, no solution in Case 1.Therefore, Case 1 (r >=1) leads to a contradiction. So, no solution here.Case 2: r <1Now, in this case, the terms are:a1=1/r^3 >1, a2=1/r^2 >1, a3=1/r >1, a4=1, a5=r <1, a6=r^2 <1.Therefore, for f(a1), f(a2), f(a3): since they are >=1, we use x ln x.For a4=1: x ln x = 0.For a5 and a6: 0 <x <1, use (ln x)/x.So, compute each term:f(a1)=a1 ln a1 = (1/r^3) ln(1/r^3) = (1/r^3)*(-3 ln r) = -3 (ln r)/r^3Similarly, f(a2)=a2 ln a2 = (1/r^2) ln(1/r^2) = (1/r^2)*(-2 ln r) = -2 (ln r)/r^2f(a3)=a3 ln a3 = (1/r) ln(1/r) = (1/r)*(-ln r) = - (ln r)/rf(a4)=0f(a5)= (ln a5)/a5 = (ln r)/rf(a6)= (ln a6)/a6 = (ln r^2)/r^2 = (2 ln r)/r^2Therefore, summing all:f(a1) + f(a2) + f(a3) + f(a4) + f(a5) + f(a6) =[-3 (ln r)/r^3] + [-2 (ln r)/r^2] + [- (ln r)/r] + 0 + [(ln r)/r] + [2 (ln r)/r^2]Let's compute term by term:First term: -3 (ln r)/r^3Second term: -2 (ln r)/r^2Third term: - (ln r)/rFourth term: 0Fifth term: + (ln r)/rSixth term: +2 (ln r)/r^2Combine like terms:-3 (ln r)/r^3 + (-2 (ln r)/r^2 + 2 (ln r)/r^2) + (- (ln r)/r + (ln r)/r) + 0Simplify:-3 (ln r)/r^3 + 0 + 0 + 0 = -3 (ln r)/r^3Therefore, in Case 2 (r <1), the sum is -3 (ln r)/r^3. According to the problem, this sum equals 2a1. Since a1=1/r^3, so 2a1=2/r^3.Therefore, the equation is:-3 (ln r)/r^3 = 2 / r^3Multiply both sides by r^3 (since r >0, r^3 >0):-3 ln r = 2Therefore:ln r = -2/3Then, exponentiating both sides:r = e^{-2/3}Which is approximately e^{-0.6667} ≈ 0.5134. Since e^{-2/3} is less than 1, this is valid in Case 2.Therefore, r = e^{-2/3}, so a1 = 1/r^3 = (e^{-2/3})^{-3} = e^{(2/3)*3} = e^{2} ≈ 7.389Therefore, a1= e^2.Wait, but let me check:Wait, r = e^{-2/3}, so a1 =1/r^3 = (e^{-2/3})^{-3} = e^{(2/3)*3} = e^{2}, yes. So, a1= e^2.But let me check if this satisfies all the conditions.First, check a_3*a_4*a_5:a_3=1/r= e^{2/3}a_4=1a_5=r= e^{-2/3}So, a_3*a_4*a_5= e^{2/3} *1* e^{-2/3}=1, which is correct.Then, check the sum:Compute f(a1) + ... +f(a6). Let's compute each term with a1= e^2, r= e^{-2/3}First, compute the terms:a1= e^2a2= a1*r= e^2 * e^{-2/3}= e^{2 - 2/3}= e^{4/3}a3= a2*r= e^{4/3} * e^{-2/3}= e^{2/3}a4=1a5= r= e^{-2/3}a6= r^2= e^{-4/3}Now, compute f(a1)=a1 ln a1= e^2 * ln(e^2)= e^2 *2= 2 e^2f(a2)=a2 ln a2= e^{4/3} * ln(e^{4/3})= e^{4/3}*(4/3)= (4/3) e^{4/3}f(a3)=a3 ln a3= e^{2/3} * ln(e^{2/3})= e^{2/3}*(2/3)= (2/3) e^{2/3}f(a4)=1*ln1=0f(a5)= (ln a5)/a5= ln(e^{-2/3}) / e^{-2/3}= (-2/3)/e^{-2/3}= (-2/3) e^{2/3}f(a6)= (ln a6)/a6= ln(e^{-4/3}) / e^{-4/3}= (-4/3)/e^{-4/3}= (-4/3) e^{4/3}Sum them up:2 e^2 + (4/3) e^{4/3} + (2/3) e^{2/3} + 0 + (-2/3) e^{2/3} + (-4/3) e^{4/3}Let's compute term by term:2 e^2+ (4/3 e^{4/3} -4/3 e^{4/3}) = 0+ (2/3 e^{2/3} -2/3 e^{2/3}) =0So total sum=2 e^2Which is equal to 2a1, since a1= e^2. 2a1=2 e^2. So, yes, the sum equals 2a1. Therefore, this checks out.Therefore, the solution is a1= e^2.But wait, let me check if this is correct. Because in the problem statement, the answer is requested to be boxed. So, maybe it's e squared, which is approximately 7.389, but the exact answer is e².But let me check if there was a miscalculation when solving for r.We had in Case 2:-3 (ln r)/r^3 = 2 / r^3Multiply both sides by r³:-3 ln r =2So, ln r = -2/3Thus, r= e^{-2/3}Hence, a1=1/r³= e^{2}Yes, so the answer is e squared. So, boxed, it is boxed{e^2}But maybe the problem expects a different form? Let me check once again.Wait, let's check if there's an alternative approach.Alternatively, given that a1=1/r³, and r= e^{-2/3}, so a1= e^{2}. Therefore, that's correct. So, the answer is e squared, boxed.Final Answerboxed{e^2}</think>Given the function ( f(x) = begin{cases} x ln x & text{if } x geq 1 frac{ln x}{x} & text{if } 0 < x < 1 end{cases} ) and a geometric sequence ({a_n}) with a common ratio ( r > 0 ) such that ( a_3 a_4 a_5 = 1 ) and ( f(a_1) + f(a_2) + ldots + f(a_6) = 2a_1 ), we need to find ( a_1 ).First, we express the terms of the geometric sequence in terms of ( a_1 ) and ( r ):- ( a_3 = a_1 r^2 )- ( a_4 = a_1 r^3 )- ( a_5 = a_1 r^4 )Given ( a_3 a_4 a_5 = 1 ), we have:[(a_1 r^2)(a_1 r^3)(a_1 r^4) = a_1^3 r^9 = 1 implies a_1^3 r^9 = 1 implies a_1 r^3 = 1 implies a_1 = frac{1}{r^3}]Next, we consider the terms ( a_1, a_2, ldots, a_6 ):- ( a_1 = frac{1}{r^3} )- ( a_2 = frac{1}{r^2} )- ( a_3 = frac{1}{r} )- ( a_4 = 1 )- ( a_5 = r )- ( a_6 = r^2 )We analyze the function ( f ) for each term depending on whether ( r geq 1 ) or ( r < 1 ). For ( r < 1 ):- ( a_1, a_2, a_3 geq 1 )- ( a_5, a_6 < 1 )Calculating ( f(a_i) ) for each term:- ( f(a_1) = a_1 ln a_1 = frac{1}{r^3} ln left( frac{1}{r^3} right) = -frac{3 ln r}{r^3} )- ( f(a_2) = a_2 ln a_2 = frac{1}{r^2} ln left( frac{1}{r^2} right) = -frac{2 ln r}{r^2} )- ( f(a_3) = a_3 ln a_3 = frac{1}{r} ln left( frac{1}{r} right) = -frac{ln r}{r} )- ( f(a_4) = 0 )- ( f(a_5) = frac{ln a_5}{a_5} = frac{ln r}{r} )- ( f(a_6) = frac{ln a_6}{a_6} = frac{2 ln r}{r^2} )Summing these contributions:[- frac{3 ln r}{r^3} - frac{2 ln r}{r^2} - frac{ln r}{r} + 0 + frac{ln r}{r} + frac{2 ln r}{r^2} = - frac{3 ln r}{r^3}]Given ( f(a_1) + f(a_2) + ldots + f(a_6) = 2a_1 ):[- frac{3 ln r}{r^3} = 2 cdot frac{1}{r^3} implies -3 ln r = 2 implies ln r = -frac{2}{3} implies r = e^{-2/3}]Thus, ( a_1 = frac{1}{r^3} = left( e^{-2/3} right)^{-3} = e^2 ).The final answer is:[boxed{e^2}],<think>Okay, so I need to solve this problem where there's a function f(x) defined piecewise, and there's a geometric sequence {a_n} with a common ratio greater than 0. The problem gives that a_3a_4a_5 = 1, and the sum f(a_1) + f(a_2) + ... + f(a_6) = 2a_1. We need to find a_1.First, let's recall what a geometric sequence is. A geometric sequence has each term after the first obtained by multiplying the previous term by a constant called the common ratio, which in this case is greater than 0. Let's denote the common ratio by r. So, the terms of the sequence are: a_1, a_1*r, a_1*r^2, a_1*r^3, a_1*r^4, a_1*r^5, etc. Since we need terms up to a_6, that would be a_1*r^5.Given that a_3a_4a_5 = 1, let's express these terms in terms of a_1 and r. So, a_3 = a_1*r^2, a_4 = a_1*r^3, a_5 = a_1*r^4. Therefore, multiplying these together:a_3*a_4*a_5 = (a_1*r^2)*(a_1*r^3)*(a_1*r^4) = (a_1)^3 * r^(2+3+4) = (a_1)^3 * r^9 = 1.So, (a_1)^3 * r^9 = 1. Let me note that down as equation (1).Now, the other condition is f(a_1) + f(a_2) + ... + f(a_6) = 2a_1. Let's first figure out each term in this sum.Since the function f(x) is defined piecewise:- For x ≥ 1, f(x) = x ln x- For 0 < x < 1, f(x) = (ln x)/xSo, for each term a_i (i from 1 to 6), we need to check whether a_i is ≥1 or between 0 and 1, then apply the corresponding formula.But since {a_n} is a geometric sequence with common ratio r > 0, and a_1 is the first term, the terms are a_1, a_1*r, a_1*r^2, a_1*r^3, a_1*r^4, a_1*r^5.Depending on the value of r and a_1, these terms could be increasing (if r >1) or decreasing (if 0 < r <1). Also, since all terms are positive (as r >0 and a_1 is part of the sequence, which is defined for 0 < x <1 and x ≥1, so a_1 must be positive).So, we need to consider whether each term a_1*r^{k} (for k from 0 to 5) is ≥1 or <1. That depends on the values of a_1 and r.But since r is positive, and the product a_3*a_4*a_5 =1, maybe there's a symmetry here. Let's see.From equation (1): (a_1)^3 * r^9 = 1. So, a_1^3 = 1 / r^9 ⇒ a_1 = (1 / r^9)^{1/3} = 1 / r^3. So, a_1 = r^{-3}. Hmm, that seems useful. So a_1 is r^{-3}.Therefore, we can express a_1 in terms of r, which might simplify things.So, let's let a_1 = r^{-3}. Then, the terms of the sequence are:a_1 = r^{-3}a_2 = a_1 * r = r^{-3} * r = r^{-2}a_3 = a_1 * r^2 = r^{-3} * r^2 = r^{-1}a_4 = a_1 * r^3 = r^{-3} * r^3 = 1a_5 = a_1 * r^4 = r^{-3} * r^4 = r^{1}a_6 = a_1 * r^5 = r^{-3} * r^5 = r^{2}Therefore, the terms from a_1 to a_6 are: r^{-3}, r^{-2}, r^{-1}, 1, r, r^2.So, each term is a power of r, starting from r^{-3} up to r^2.Now, depending on whether r is greater than 1 or less than 1, the terms will be increasing or decreasing.Case 1: r >1. Then, the terms progress from r^{-3} (which is less than 1) up to r^2 (which is greater than 1). So, some terms may be less than 1, some equal to 1, and some greater than 1.Case 2: 0 < r <1. Then, the terms progress from r^{-3} (which is greater than 1, since r <1) down to r^2 (which is less than 1). So again, some terms may be greater than 1, some equal to 1, some less than 1.But given that a_4 =1, regardless of r. So, the fourth term is always 1. So, a_4=1. Therefore, f(a_4) = f(1) = 1 * ln 1 = 0, since 1 ≥1. But ln 1 is 0. So, f(a_4) =0.So, in the sum f(a1) + f(a2) + ... + f(a6), the fourth term is zero. So, we can ignore that term.So, the sum reduces to f(a1) + f(a2) + f(a3) + f(a5) + f(a6) = 2a1.But let's check the other terms. Let's list all terms:a1 = r^{-3}a2 = r^{-2}a3 = r^{-1}a4 =1a5 = ra6 = r^2So, depending on whether each term is ≥1 or <1, we need to apply the corresponding f(x).So, let's analyze each term:a1 = r^{-3}:If r >1, then r^{-3} <1, so f(a1) = (ln a1)/a1If r <1, then r^{-3} >1, so f(a1) = a1 ln a1Similarly for a2, a3, a5, a6.But since we don't know whether r is greater than or less than 1, perhaps we need to consider both cases. However, maybe there is a specific r that satisfies the conditions. Let's try to proceed step by step.Alternatively, perhaps we can consider r =1. But if r=1, then the common ratio is 1, so all terms are equal to a1. Then, a3a4a5 = (a1)^3 =1 ⇒ a1=1. Then, the sum f(a1)+...+f(a6) would be 6*f(1) =6*0=0, which would have to equal 2a1=2*1=2. But 0≠2. Hence, r≠1.Therefore, r is not 1. So, r>1 or 0<r<1.Let me try to proceed by assuming that r>1. Then, a1 = r^{-3} <1, since r>1.a2 = r^{-2} <1a3 = r^{-1} <1a4=1a5=r >1a6=r^2 >1Therefore, in this case, a1, a2, a3 <1; a5, a6 >1.Therefore, applying the function f to each:f(a1) = (ln a1)/a1Similarly, f(a2) = (ln a2)/a2f(a3) = (ln a3)/a3f(a4) =0f(a5) =a5 ln a5f(a6) =a6 ln a6So, sum S = [ (ln a1)/a1 + (ln a2)/a2 + (ln a3)/a3 ] + [a5 ln a5 + a6 ln a6] =2a1But since a1 = r^{-3}, a2 = r^{-2}, a3 = r^{-1}, a5 =r, a6 =r^2.Therefore, substituting:First, compute (ln a1)/a1:ln(a1)/a1 = ln(r^{-3}) / r^{-3} = (-3 ln r) / r^{-3} = (-3 ln r) * r^3 = -3 r^3 ln rSimilarly, (ln a2)/a2 = ln(r^{-2}) / r^{-2} = (-2 ln r)/r^{-2} = (-2 ln r) * r^2 = -2 r^2 ln r(ln a3)/a3 = ln(r^{-1}) / r^{-1} = (-1 ln r)/ r^{-1} = (-ln r)*r = -r ln rThen, the terms from a5 and a6:a5 ln a5 = r * ln ra6 ln a6 = r^2 * ln r^2 = r^2 * 2 ln rSo, sum S = (-3 r^3 ln r -2 r^2 ln r - r ln r) + (r ln r + 2 r^2 ln r) =2a1=2 r^{-3}Simplify the sum:First, group the terms:Negative terms:-3 r^3 ln r -2 r^2 ln r - r ln rPositive terms:+ r ln r + 2 r^2 ln rSo, combining:-3 r^3 ln r -2 r^2 ln r - r ln r + r ln r +2 r^2 ln r= -3 r^3 ln r + (-2 r^2 ln r +2 r^2 ln r) + (-r ln r + r ln r)Simplify term by term:-3 r^3 ln r + 0 + 0 = -3 r^3 ln rThus, S = -3 r^3 ln r =2a1=2 r^{-3}So, set up the equation:-3 r^3 ln r = 2 r^{-3}Multiply both sides by r^3 to eliminate denominator:-3 r^6 ln r = 2Thus,-3 r^6 ln r =2 ⇒ 3 r^6 ln r = -2But ln r is negative here? Wait, if r>1, then ln r >0, so left side would be positive, right side negative. Contradiction.Therefore, there's no solution in the case where r>1.Hmm, so maybe my assumption that r>1 is invalid. Therefore, we need to check the other case where 0 < r <1.So, if 0 < r <1, then:a1 = r^{-3} >1 (since r <1)a2 = r^{-2} >1a3 = r^{-1} >1a4=1a5=r <1a6=r^2 <1Therefore, applying the function f:f(a1) =a1 ln a1f(a2)=a2 ln a2f(a3)=a3 ln a3f(a4)=0f(a5)=(ln a5)/a5f(a6)=(ln a6)/a6Therefore, sum S = [a1 ln a1 + a2 ln a2 + a3 ln a3] + [(ln a5)/a5 + (ln a6)/a6] =2a1Again, substitute the values:a1 =r^{-3}, a2=r^{-2}, a3=r^{-1}, a5=r, a6=r^2First, compute the terms from a1, a2, a3:a1 ln a1 = r^{-3} ln(r^{-3}) = r^{-3}*(-3 ln r) = -3 r^{-3} ln ra2 ln a2 = r^{-2} ln(r^{-2}) = r^{-2}*(-2 ln r) = -2 r^{-2} ln ra3 ln a3 = r^{-1} ln(r^{-1}) = r^{-1}*(-1 ln r) = - r^{-1} ln rThen, the terms from a5 and a6:(ln a5)/a5 = ln(r)/r(ln a6)/a6 = ln(r^2)/r^2 = 2 ln r / r^2So, sum S = [ -3 r^{-3} ln r -2 r^{-2} ln r - r^{-1} ln r ] + [ ln r / r + 2 ln r / r^2 ] =2a1=2 r^{-3}Let me factor out ln r:Sum S = ln r [ -3 r^{-3} -2 r^{-2} - r^{-1} + r^{-1} + 2 r^{-2} ] =2 r^{-3}Simplify inside the brackets:-3 r^{-3} -2 r^{-2} - r^{-1} + r^{-1} +2 r^{-2}Combine like terms:-3 r^{-3} + (-2 r^{-2} +2 r^{-2}) + (-r^{-1} + r^{-1}) = -3 r^{-3}So, S = ln r * (-3 r^{-3}) =2 r^{-3}Therefore:-3 r^{-3} ln r =2 r^{-3}Multiply both sides by r^3 to eliminate the denominator:-3 ln r =2Thus, ln r= -2/3Therefore, r= e^{-2/3}Therefore, since r is found, then a1= r^{-3}= (e^{-2/3})^{-3}= e^{2}So, a1= e^2≈7.389, but since we need an exact answer, it's e squared. However, let me check if that's correct.Wait, let me verify.If r = e^{-2/3}, then a1= r^{-3}= (e^{-2/3})^{-3}= e^{(2/3)*3}= e^2. Correct.So, a1= e^2.But let's check if this satisfies the original equation.Compute S= f(a1)+f(a2)+f(a3)+f(a4)+f(a5)+f(a6)= sum of these terms.Given r= e^{-2/3}, so a1= e^2, a2= e^{2}*(e^{-2/3})= e^{2 -2/3}= e^{4/3}a3= e^{4/3}*e^{-2/3}= e^{2/3}a4=1a5= e^{-2/3}a6= e^{-4/3}Now, compute f(a1)= a1 ln a1= e^2 * ln(e^2)= e^2*2=2 e^2Similarly, f(a2)= a2 ln a2= e^{4/3} * ln(e^{4/3})= e^{4/3}*(4/3)f(a3)= a3 ln a3= e^{2/3} * ln(e^{2/3})= e^{2/3}*(2/3)f(a4)=0f(a5)= (ln a5)/a5= ln(e^{-2/3}) / e^{-2/3}= (-2/3)/e^{-2/3}= (-2/3) e^{2/3}f(a6)= (ln a6)/a6= ln(e^{-4/3}) / e^{-4/3}= (-4/3)/e^{-4/3}= (-4/3) e^{4/3}Sum S=2 e^2 + (4/3)e^{4/3} + (2/3)e^{2/3} +0 + (-2/3)e^{2/3} + (-4/3)e^{4/3}Let's compute term by term:2 e^2+ (4/3)e^{4/3}+ (2/3)e^{2/3}- (2/3)e^{2/3}- (4/3)e^{4/3}So, the (4/3)e^{4/3} and -(4/3)e^{4/3} cancel each other.Similarly, (2/3)e^{2/3} - (2/3)e^{2/3}=0.Thus, the sum S=2 e^2 +0 +0=2 e^2.But according to the problem, the sum S=2 a1=2 e^2. So, 2 e^2=2 e^2. That works. So, yes, it checks out.Therefore, the value a1= e^2 is the correct solution.Therefore, the answer is boxed{e^2}Final Answerboxed{e^2}</think>Given the function ( f(x) = begin{cases} x ln x & text{if } x geq 1 frac{ln x}{x} & text{if } 0 < x < 1 end{cases} ) and a geometric sequence ({a_n}) with a common ratio ( r > 0 ) such that ( a_3 a_4 a_5 = 1 ) and ( f(a_1) + f(a_2) + ldots + f(a_6) = 2a_1 ), we need to find ( a_1 ).First, express the terms of the geometric sequence in terms of ( a_1 ) and ( r ):- ( a_3 = a_1 r^2 )- ( a_4 = a_1 r^3 )- ( a_5 = a_1 r^4 )Given ( a_3 a_4 a_5 = 1 ):[(a_1 r^2)(a_1 r^3)(a_1 r^4) = a_1^3 r^9 = 1 implies a_1^3 = frac{1}{r^9} implies a_1 = frac{1}{r^3}]Next, consider the terms ( a_1, a_2, ldots, a_6 ) in terms of ( r ):- ( a_1 = r^{-3} )- ( a_2 = r^{-2} )- ( a_3 = r^{-1} )- ( a_4 = 1 )- ( a_5 = r )- ( a_6 = r^2 )Analyzing the function ( f ) for each term:- For ( 0 < r < 1 ), ( a_1, a_2, a_3 > 1 ) and ( a_5, a_6 < 1 )- For ( r > 1 ), ( a_1, a_2, a_3 < 1 ) and ( a_5, a_6 > 1 )Considering ( 0 < r < 1 ):- ( f(a_1) = a_1 ln a_1 )- ( f(a_2) = a_2 ln a_2 )- ( f(a_3) = a_3 ln a_3 )- ( f(a_4) = 0 )- ( f(a_5) = frac{ln a_5}{a_5} )- ( f(a_6) = frac{ln a_6}{a_6} )Substituting the values:- ( a_1 = r^{-3} ), ( a_2 = r^{-2} ), ( a_3 = r^{-1} ), ( a_5 = r ), ( a_6 = r^2 )Calculating each term:- ( f(a_1) = -3 r^{-3} ln r )- ( f(a_2) = -2 r^{-2} ln r )- ( f(a_3) = -r^{-1} ln r )- ( f(a_5) = frac{ln r}{r} )- ( f(a_6) = frac{2 ln r}{r^2} )Summing these terms:[-3 r^{-3} ln r - 2 r^{-2} ln r - r^{-1} ln r + frac{ln r}{r} + frac{2 ln r}{r^2} = 2 r^{-3}]Simplifying:[ln r left( -3 r^{-3} - 2 r^{-2} - r^{-1} + r^{-1} + 2 r^{-2} right) = 2 r^{-3}][-3 r^{-3} ln r = 2 r^{-3}][-3 ln r = 2 implies ln r = -frac{2}{3} implies r = e^{-2/3}]Thus, ( a_1 = r^{-3} = left( e^{-2/3} right)^{-3} = e^2 ).The final answer is (boxed{e^2}).]

🔑:If x > 1, then 0 1, then a_1ln a_1=2a_1, thus a_1=boxed{e^2}; If 0 < a_1 < 1, then frac {ln a_1}{a_1} < 0, hence no solution; Therefore, the answer is: boxed{e^2}. From the problem, we can derive that f(x)+fleft(frac {1}{x}right)=0; hence, f(a_2)+ldots+f(a_6)=f(a_2)+f(a_6)+f(a_3)+f(a_5)+f(a_4)=0, thus simplifying f(a_1)+f(a_2)+ldots+f(a_6)=f(a_1)=2a_1, from which the solution is obtained. This problem examines the definition of a geometric sequence and the application of piecewise functions, and is considered a medium-level question.

🧠:[<think>Okay, so I need to prove that if A is a 2x2 real matrix and x is a positive real number such that the determinant of (A² + xI₂) is zero, then the determinant of (A² + A + xI₂) is equal to x. Hmm, let's start by recalling some properties of 2x2 matrices and determinants.First, for any 2x2 matrix A, the characteristic polynomial is given by λ² - tr(A)λ + det(A) = 0, where tr(A) is the trace and det(A) is the determinant. By the Cayley-Hamilton theorem, a matrix satisfies its own characteristic equation, so A² - tr(A)A + det(A)I₂ = 0. Maybe this will be useful.Given that det(A² + xI₂) = 0, this means that A² + xI₂ is a singular matrix. So there exists a non-zero vector v such that (A² + xI₂)v = 0, which implies A²v = -xv. So -x is an eigenvalue of A². Since A is a 2x2 real matrix, its eigenvalues are either both real or complex conjugates. Similarly, the eigenvalues of A² will be the squares of the eigenvalues of A. If -x is an eigenvalue of A², then the eigenvalues of A must be either purely imaginary or complex numbers whose squares are real and negative. But since x is positive, the eigenvalues of A² must be -x. Let me think more carefully.Let’s denote the eigenvalues of A as λ₁ and λ₂. Then the eigenvalues of A² are λ₁² and λ₂². The determinant of A² + xI₂ is the product of (λ₁² + x) and (λ₂² + x). Since this determinant is zero, at least one of these factors must be zero. Therefore, either λ₁² = -x or λ₂² = -x. Since x is a positive real number, this implies that λ₁ and λ₂ are purely imaginary numbers. Because if λ is a real number, then λ² is non-negative, which can't equal -x. So A must have eigenvalues that are purely imaginary, say λ₁ = i√x and λ₂ = -i√x, since they have to be conjugates if the matrix is real. Wait, but if x is fixed, does that mean that both eigenvalues squared are -x? Let me check.Suppose λ₁² = -x. Then λ₂² is also -x because the determinant is the product (λ₁² + x)(λ₂² + x) = 0. If only one of them was -x, the other term would be (λ₂² + x), but since x is positive, unless λ₂² is also -x, which isn't possible for real numbers, then both factors must be zero. But since eigenvalues of a real matrix must be complex conjugates, if λ₁ is i√x, then λ₂ must be -i√x. So both eigenvalues squared are -x. Therefore, the eigenvalues of A are ±i√x. Therefore, the trace of A, which is the sum of eigenvalues, is i√x - i√x = 0. So tr(A) = 0. The determinant of A is the product of the eigenvalues, which is (i√x)(-i√x) = -i²x = x. So det(A) = x. So by Cayley-Hamilton theorem, A² - tr(A)A + det(A)I₂ = A² + xI₂ = 0. Wait, but the original problem states that det(A² + xI₂) = 0, which would be true if A² + xI₂ is the zero matrix? But if A² + xI₂ is the zero matrix, then its determinant is zero, which is given. But if A² + xI₂ = 0, then A² = -xI₂. Then A² + A + xI₂ would be (-xI₂) + A + xI₂ = A. Therefore, det(A² + A + xI₂) = det(A) = x. But wait, if that's the case, then the determinant is x, which is exactly what we need to prove. But is this true?Wait, hold on. If A² + xI₂ is the zero matrix, then A² = -xI₂. Then A² + A + xI₂ would be A. Therefore, det(A) = x. But in this case, det(A² + A + xI₂) = det(A) = x, which matches the conclusion. But is A² + xI₂ necessarily the zero matrix? Because Cayley-Hamilton theorem says that A² - tr(A)A + det(A)I₂ = 0. Since we found tr(A) = 0 and det(A) = x, so A² + xI₂ = 0. Therefore, indeed A² = -xI₂. Therefore, A² + A + xI₂ = A, so determinant is x. So that works out. But is this the only case?Wait, maybe the problem is more general. Suppose that det(A² + xI₂) = 0, but A² + xI₂ is not necessarily the zero matrix. But for a 2x2 matrix, if the determinant is zero, then the matrix is rank 1 or 0. So if A² + xI₂ is rank 1, then it's not the zero matrix, but we can still have determinant zero. But from the eigenvalues perspective, if det(A² + xI₂) = 0, then as I thought before, the eigenvalues of A² must be -x. But for a 2x2 matrix, the eigenvalues of A² are both -x. So then A² is similar to a diagonal matrix with entries -x, -x. Therefore, A² is diagonalizable and equal to -xI₂. Wait, but is that necessarily the case? Suppose A² is a Jordan block. If A² is a Jordan block, then it's upper triangular with eigenvalues -x on the diagonal. But if it's a Jordan block, then the only 2x2 Jordan block with eigenvalue -x is actually the diagonal matrix because if it had a 1 in the superdiagonal, then the matrix would not be diagonalizable, but since A² has eigenvalues -x, -x, and if it's diagonalizable, then A² = -xI₂. If it's not diagonalizable, then A² would be similar to a Jordan block, but in that case, the determinant of A² + xI₂ would be zero only if all eigenvalues are -x. Wait, even if it's a Jordan block, the determinant is still the product of the diagonal entries, so ( (-x + x) ) * ( (-x + x) ) = 0*0 = 0. So even if A² is a Jordan block, the determinant would still be zero. But actually, in that case, if A² is a Jordan block, then A² + xI₂ would have zeros on the diagonal, but a 1 in the superdiagonal. Wait, no. If A² is a Jordan block with eigenvalue -x, then A² + xI₂ would have zeros on the diagonal and possibly a 1 in the superdiagonal. So the determinant of a Jordan block matrix with zeros on the diagonal and 1s above is zero, but the determinant is the product of the eigenvalues, which are both zero. So in any case, whether A² is diagonalizable or not, det(A² + xI₂) = 0. But in our case, since we need to link this to A² + A + xI₂, perhaps using Cayley-Hamilton is the key.Wait, maybe there's another approach. Let's try to compute det(A² + A + xI₂). Maybe express A² in terms of A and I₂ using Cayley-Hamilton. From Cayley-Hamilton, we know that A² - tr(A)A + det(A)I₂ = 0. If we can express A² in terms of A and I₂, then we can substitute into A² + A + xI₂. Let's denote tr(A) as t and det(A) as d. Then A² = tA - dI₂. Then A² + A + xI₂ = tA - dI₂ + A + xI₂ = (t + 1)A + (x - d)I₂. Therefore, det(A² + A + xI₂) = det( (t + 1)A + (x - d)I₂ ). Hmm, not sure if this helps directly.But maybe we need to find t and d in terms of x. From the given condition, det(A² + xI₂) = 0. Let's compute det(A² + xI₂). Using the same substitution, since A² = tA - dI₂, then A² + xI₂ = tA - dI₂ + xI₂ = tA + (x - d)I₂. Therefore, det(A² + xI₂) = det(tA + (x - d)I₂). But this is supposed to be zero. So det(tA + (x - d)I₂) = 0. Hmm, so maybe we have two equations here: one from Cayley-Hamilton (A² = tA - dI₂) and the given condition det(tA + (x - d)I₂) = 0. But how can we connect these to find det(A² + A + xI₂) = x?Alternatively, maybe using eigenvalues. If A has eigenvalues λ and μ, then A² has eigenvalues λ² and μ². The condition det(A² + xI₂) = 0 implies that (λ² + x)(μ² + x) = 0. Therefore, either λ² = -x or μ² = -x. Since A is a real matrix, its eigenvalues must be complex conjugates. So if λ is a complex eigenvalue, then μ is its conjugate. Therefore, if λ² = -x, then μ² = (conjugate(λ))² = conjugate(λ²) = conjugate(-x) = -x, since x is real. So both λ² and μ² are equal to -x. Therefore, λ and μ are ±i√x. Therefore, the eigenvalues of A are i√x and -i√x. Therefore, trace of A is 0, determinant is (i√x)(-i√x) = x. Therefore, by Cayley-Hamilton, A² - 0*A + xI₂ = 0 ⇒ A² = -xI₂. Then, A² + A + xI₂ = -xI₂ + A + xI₂ = A. Therefore, det(A² + A + xI₂) = det(A) = x. So that works out. Therefore, this seems to be the way.But wait, this relies on the fact that if A is a real 2x2 matrix with eigenvalues i√x and -i√x, then A² = -xI₂. Is that always the case? Let me verify with an example. Let’s take A = [[0, -√x], [√x, 0]]. Then eigenvalues of A can be found by solving det(A - λI) = 0 ⇒ λ² + x = 0 ⇒ λ = ±i√x. Then A² = [[0, -√x], [√x, 0]] * [[0, -√x], [√x, 0]] = [[-x, 0], [0, -x]] = -xI₂. Therefore, A² + xI₂ = 0, so det(A² + xI₂) = 0. Then A² + A + xI₂ = -xI₂ + A + xI₂ = A, so det(A) = (0)(0) - (-√x)(√x) = x. So indeed, the determinant is x. Therefore, in this case, the statement holds.But the problem states "A is a square 2×2 matrix with real entries such that det(A² + xI₂) = 0". So if A is such a matrix, then regardless of its specific form, as long as it satisfies det(A² + xI₂) = 0, then det(A² + A + xI₂) = x. But from the eigenvalue analysis, if det(A² + xI₂) = 0, then the eigenvalues of A must be ±i√x, hence A² = -xI₂, so A² + A + xI₂ = A, and det(A) = x. Therefore, the conclusion follows.But wait, maybe there's a case where A² + xI₂ is not the zero matrix, but still has determinant zero. For example, suppose A² + xI₂ is a non-zero matrix with rank 1. But if the eigenvalues of A² are both -x, then A² is diagonalizable and similar to -xI₂. However, if A² is not diagonalizable, then it could be a Jordan block, but in 2x2, the only Jordan blocks for eigenvalue -x would be the matrix [[-x, 1], [0, -x]], but then A² + xI₂ would be [[0,1],[0,0]], which has determinant zero. However, in this case, can such a matrix A exist?Wait, let's suppose that A² is similar to the Jordan block [[-x,1],[0,-x]]. Then A² + xI₂ is [[0,1],[0,0]], which has determinant zero. However, in this case, can there exist a real matrix A such that A² is [[-x,1],[0,-x]]? Let's see. Let's take x = 1 for simplicity, so we need a matrix A such that A² = [[-1,1],[0,-1]]. Let's attempt to find such a matrix. Let A be [[a,b],[c,d]]. Then A² = [[a² + bc, ab + bd], [ac + cd, bc + d²]]. Setting this equal to [[-1,1],[0,-1]]. Therefore, we have the equations:1. a² + bc = -1,2. ab + bd = 1,3. ac + cd = 0,4. bc + d² = -1.From equation 3: c(a + d) = 0. So either c = 0 or a + d = 0.Case 1: c = 0.Then equation 1: a² = -1, which is impossible since a is real. Therefore, Case 1 is invalid.Case 2: a + d = 0.So d = -a.Then equation 1: a² + bc = -1.Equation 4: bc + d² = bc + a² = -1. So equation 1 and 4 are the same, which is consistent.Equation 2: ab + bd = b(a + d) = b(0) = 0 = 1. But 0 = 1 is impossible. Therefore, no solution in this case.Therefore, there is no real matrix A such that A² is the Jordan block [[-1,1],[0,-1]]. Therefore, in the real case, if A² has eigenvalues -x, then A² must be diagonalizable, hence similar to -xI₂. Therefore, A² = -xI₂. Therefore, in the real 2x2 case, if det(A² + xI₂) = 0, then A² + xI₂ is the zero matrix. Therefore, A² = -xI₂. Then, A² + A + xI₂ = A, so determinant is det(A) = x. Therefore, the result follows.Alternatively, even if there existed a non-diagonalizable A², but as shown, such A cannot exist in real matrices. Therefore, in all cases, A² = -xI₂, so determinant of A² + A + xI₂ is determinant of A, which is x. Hence proved.Alternatively, another approach: Since det(A² + xI₂) = 0, then A² + xI₂ is singular. Therefore, there exists a non-zero vector v such that (A² + xI₂)v = 0 ⇒ A²v = -xv. Let’s compute (A² + A + xI₂)v = A²v + Av + xIv = (-xv) + Av + xv = Av. Therefore, (A² + A + xI₂)v = Av. Similarly, if we can find another vector w such that (A² + A + xI₂)w = something, but maybe we need to consider the action on two eigenvectors.But since A is a 2x2 real matrix with complex eigenvalues, it can be decomposed into a real Jordan form. Since the eigenvalues are ±i√x, the real Jordan form is [[0, -√x], [√x, 0]]. Therefore, in this basis, A is similar to that matrix. Therefore, determinant is preserved under similarity transformation. So if we can show that the determinant of (A² + A + xI₂) is x in this case, then it holds for all similar matrices. As shown in the example earlier, when A is [[0, -√x], [√x, 0]], then A² = -xI₂, so A² + A + xI₂ = A, and det(A) = x. Therefore, regardless of the basis, determinant is x.Alternatively, another method: Let's compute det(A² + A + xI₂). Let’s note that if A² = -xI₂, then A² + A + xI₂ = A. So determinant is det(A). But det(A) is the product of eigenvalues, which are i√x and -i√x, so det(A) = (i√x)(-i√x) = -i²x = x. Therefore, det(A) = x, so det(A² + A + xI₂) = x. Hence proved.Alternatively, use polynomial relations. Since A² = -xI₂, then we can substitute into A² + A + xI₂ to get -xI₂ + A + xI₂ = A, so determinant is det(A) = x. Therefore, the result follows.But the key step is recognizing that det(A² + xI₂) = 0 implies that A² = -xI₂. This is because in a 2x2 real matrix, if the determinant of (A² + xI₂) is zero, then both eigenvalues of A² must be -x, and since A is real, A² must be similar to -xI₂, hence A² = -xI₂. Therefore, the conclusion follows directly.So overall, the proof is as follows:Given that A is a real 2x2 matrix with det(A² + xI₂) = 0. This implies that -x is an eigenvalue of A². Since A is real, its non-real eigenvalues must come in complex conjugate pairs. Therefore, the eigenvalues of A are ±i√x, leading to tr(A) = 0 and det(A) = x. By Cayley-Hamilton theorem, A² - tr(A)A + det(A)I₂ = A² + xI₂ = 0. Thus, A² = -xI₂. Substituting into A² + A + xI₂ gives A, whose determinant is det(A) = x. Hence, det(A² + A + xI₂) = x.Final Answerboxed{x}</think>Given a real number ( x > 0 ) and a ( 2 times 2 ) real matrix ( A ) such that ( det(A^2 + xI_2) = 0 ), we need to prove that ( det(A^2 + A + xI_2) = x ).1. Eigenvalue Analysis: - The determinant condition ( det(A^2 + xI_2) = 0 ) implies that ( -x ) is an eigenvalue of ( A^2 ). - Since ( A ) is a real matrix, its eigenvalues must be complex conjugates. Let the eigenvalues of ( A ) be ( lambda ) and ( mu ). Then the eigenvalues of ( A^2 ) are ( lambda^2 ) and ( mu^2 ). - Given ( lambda^2 = -x ) and ( mu^2 = -x ), the eigenvalues of ( A ) must be ( pm isqrt{x} ).2. Trace and Determinant: - The trace of ( A ), ( operatorname{tr}(A) ), is the sum of its eigenvalues, which is ( isqrt{x} + (-isqrt{x}) = 0 ). - The determinant of ( A ), ( det(A) ), is the product of its eigenvalues, which is ( (isqrt{x})(-isqrt{x}) = x ).3. Cayley-Hamilton Theorem: - By the Cayley-Hamilton theorem, ( A ) satisfies its characteristic polynomial: ( A^2 - operatorname{tr}(A)A + det(A)I_2 = 0 ). - Substituting the trace and determinant values, we get ( A^2 + xI_2 = 0 ), hence ( A^2 = -xI_2 ).4. Substitution and Determinant Calculation: - Substitute ( A^2 = -xI_2 ) into ( A^2 + A + xI_2 ): [ A^2 + A + xI_2 = -xI_2 + A + xI_2 = A ] - Therefore, ( det(A^2 + A + xI_2) = det(A) ).5. Conclusion: - Since ( det(A) = x ), we conclude that ( det(A^2 + A + xI_2) = x ).Thus, the final answer is (boxed{x}).,<think>Okay, let me try to figure out how to prove that determinant of (A² + A + xI₂) is equal to x given that determinant of (A² + xI₂) is zero. Hmm. So we have a 2x2 matrix A, and x is a positive real number. The condition is that det(A² + xI₂) = 0. We need to show that det(A² + A + xI₂) = x.First, I remember that for 2x2 matrices, there are some properties related to determinants and characteristic equations. Maybe Cayley-Hamilton theorem could be useful here. Let's recall that for any 2x2 matrix A, the characteristic polynomial is λ² - tr(A)λ + det(A) = 0, and by Cayley-Hamilton theorem, A² - tr(A)A + det(A)I₂ = 0. So A² = tr(A)A - det(A)I₂. Maybe this can help us express A² in terms of A and I₂.But wait, in the problem, we have A² + xI₂. The determinant of that is zero. So det(A² + xI₂) = 0. Let me denote B = A². Then det(B + xI₂) = 0. Which means that -x is an eigenvalue of B. Since B is a 2x2 matrix, if -x is an eigenvalue, then the characteristic polynomial of B has a root at -x. So the characteristic polynomial of B is det(B - λI₂) = (λ + x)(λ - μ) for some μ, but since the product of the eigenvalues is det(B). Wait, maybe that's a bit too abstract.Alternatively, since det(B + xI₂) = 0, then there exists a non-zero vector v such that (B + xI₂)v = 0, so Bv = -xv. Therefore, -x is an eigenvalue of B = A². But eigenvalues of A² are the squares of the eigenvalues of A. Wait, not exactly. If A has eigenvalues λ₁ and λ₂, then A² has eigenvalues λ₁² and λ₂². So if -x is an eigenvalue of A², then there exists some eigenvalue λ of A such that λ² = -x. But x is a positive real number, so λ² = -x implies that λ is purely imaginary, because the square of a real number can't be negative. So this suggests that the eigenvalues of A are purely imaginary or complex? Wait, but the matrix A has real entries. So if A has complex eigenvalues, they must come in complex conjugate pairs. So if λ is an eigenvalue of A, then its conjugate is also an eigenvalue.But since λ² = -x, which is negative real, then λ must be purely imaginary. Let me check that. Suppose λ is a complex number such that λ² = -x. Then λ = ±i√x. Therefore, the eigenvalues of A are ±i√x. But since A is a real matrix, its complex eigenvalues must come in conjugate pairs, so if i√x is an eigenvalue, then -i√x must also be an eigenvalue. Therefore, the eigenvalues of A are i√x and -i√x. Then the eigenvalues of A² would be (i√x)² = -x and (-i√x)² = -x. So both eigenvalues of A² are -x. Therefore, A² is similar to a matrix with both eigenvalues -x, but since A is real, is A² diagonalizable?Wait, if A has eigenvalues ±i√x, then A is similar to a diagonal matrix with entries i√x and -i√x. But since A is real, it can't be diagonalized over the reals, but it can be diagonalized over the complex numbers. However, A² would then be similar to a diagonal matrix with entries -x and -x, which is just -x times the identity matrix. So A² is similar to -xI₂. But similarity preserves determinant and trace. So if A² is similar to -xI₂, then det(A²) = det(-xI₂) = x², and tr(A²) = tr(-xI₂) = -2x.But actually, since A is a 2x2 real matrix with eigenvalues ±i√x, then its trace is zero (since trace is sum of eigenvalues), and determinant is (i√x)(-i√x) = x. So by Cayley-Hamilton theorem, A² - tr(A)A + det(A)I₂ = A² + xI₂ = 0. Wait, that's interesting. So Cayley-Hamilton theorem would tell us that A² + xI₂ = 0? But the problem states that det(A² + xI₂) = 0. Wait, if A² + xI₂ = 0, then the determinant would be det(0 matrix) = 0, which is consistent. But in that case, A² = -xI₂. Then A² + A + xI₂ = -xI₂ + A + xI₂ = A. Therefore, det(A² + A + xI₂) = det(A) = x. Because earlier we saw that determinant of A is x. So that would give us the result.Wait, that seems too straightforward. But let's check. If A² + xI₂ = 0, then A² = -xI₂. Then A² + A + xI₂ = (-xI₂) + A + xI₂ = A. Therefore, determinant of A is det(A). But from the eigenvalues of A, which are ±i√x, determinant is product of eigenvalues, which is (i√x)(-i√x) = -i²x = x. So det(A) = x. Therefore, det(A² + A + xI₂) = det(A) = x. So that's exactly what we needed to prove.But wait, the problem only states that det(A² + xI₂) = 0. So does this imply that A² + xI₂ is the zero matrix, or just that it's a singular matrix? Because determinant zero just means it's singular, not necessarily the zero matrix. So my previous reasoning assumes that A² + xI₂ = 0, but actually, the problem only gives that determinant is zero. So maybe there's a different approach needed here.Let me see. Let's recall that if A is a 2x2 real matrix, then the determinant of (A² + xI₂) = 0. So A² + xI₂ is a singular matrix. Which means there exists a non-zero vector v such that (A² + xI₂)v = 0, so A²v = -xv. So -x is an eigenvalue of A². As before, eigenvalues of A² are the squares of eigenvalues of A. So if A has eigenvalues λ and μ, then A² has eigenvalues λ² and μ². So if det(A² + xI₂) = 0, then at least one of λ² + x or μ² + x is zero. So λ² = -x or μ² = -x. Therefore, similar to before, the eigenvalues of A must be ±i√x.But since A is a real matrix, the eigenvalues must be complex conjugates. Therefore, if one eigenvalue is i√x, the other must be -i√x. Therefore, the trace of A is zero (i√x + (-i√x) = 0), and determinant is (i√x)(-i√x) = x. Therefore, by Cayley-Hamilton theorem, A² - tr(A)A + det(A)I₂ = A² + xI₂ = 0. So in this case, A² = -xI₂. Therefore, the determinant condition implies A² = -xI₂. So then, as before, A² + A + xI₂ = A, so determinant is x. Therefore, the conclusion holds.But wait, does determinant(A² + xI₂) = 0 necessarily imply that A² = -xI₂? Because in general, for a 2x2 matrix, if the determinant is zero, it could have rank 1 or rank 0. So if A² + xI₂ has determinant zero, it could be rank 1 or zero matrix. However, since A is a real matrix and A² + xI₂ has determinant zero, which for a 2x2 matrix means that it's either the zero matrix or rank 1. But can A² + xI₂ be rank 1?Suppose A² + xI₂ has rank 1. Then there exists a non-zero vector v such that (A² + xI₂)v = 0, as before. But A² + xI₂ is a real matrix. If A² + xI₂ is rank 1, then its image is a one-dimensional subspace. However, since A² is a real matrix, and x is real, then A² + xI₂ is real. However, in the case where A² + xI₂ has rank 1, how does that affect the determinant of A² + A + xI₂?Alternatively, maybe we can use the fact that if A² + xI₂ is singular, then (A² + xI₂) is not invertible. So there exists a non-zero vector v such that (A² + xI₂)v = 0. So A²v = -xv. Then, let's try to compute (A² + A + xI₂)v. Let's see:(A² + A + xI₂)v = A²v + Av + xI₂v = (-xv) + Av + xv = Av.So (A² + A + xI₂)v = Av.If Av is non-zero, then Av is an eigenvector of A² + A + xI₂ with eigenvalue 0? Wait, not necessarily. Wait, no, it's not necessarily an eigenvector. Wait, perhaps we can think in terms of eigenvalues.Alternatively, maybe consider that since A satisfies its own characteristic equation. If A has eigenvalues ±i√x, then as we saw earlier, the characteristic equation is λ² + x = 0, so A² + xI₂ = 0. Therefore, A² = -xI₂. Then, as before, A² + A + xI₂ = A, so determinant is det(A) = x.But this requires that A² + xI₂ = 0, which only holds if the minimal polynomial of A divides λ² + x. However, if A² + xI₂ is not the zero matrix but just singular, then this approach may not hold. Wait, but if the determinant of A² + xI₂ is zero, then the characteristic polynomial of A must have roots whose squares are -x. Which, for a 2x2 matrix, would require that both eigenvalues squared are -x, leading to eigenvalues ±i√x, hence A² = -xI₂. Because if the eigenvalues of A are i√x and -i√x, then A is diagonalizable (over the complex numbers) and A² would be diagonal with entries -x and -x, so A² = -xI₂.Wait, but is it possible for a real matrix A to have eigenvalues i√x and -i√x but not satisfy A² = -xI₂? Let's think. For example, consider the matrix [0, -x; 1, 0]. Its eigenvalues satisfy λ² + x = 0, so λ = ±i√x. Then A² would be [0, -x; 1, 0]^2 = [-x, 0; 0, -x] = -xI₂. So in this case, A² = -xI₂. So perhaps all such matrices A with eigenvalues ±i√x must satisfy A² = -xI₂? Wait, but let's take another example. Suppose A is a Jordan block. But since the eigenvalues are distinct (i√x and -i√x), the matrix is diagonalizable, so it cannot be in Jordan form unless it's diagonal. But over the complex numbers, it's diagonalizable. However, over the real numbers, it would be similar to a rotation matrix. For example, a rotation by 90 degrees matrix would satisfy A² = -I₂. But in that case, yes, A² = -xI₂ if scaled appropriately.Wait, let's think of a general real matrix with eigenvalues ±i√x. Such a matrix is similar to a rotation matrix scaled by √x. For example, consider the matrix A = [0, -√x; √x, 0]. Then A² = [ -x, 0; 0, -x ] = -xI₂. So indeed, A² = -xI₂. So in general, any real 2x2 matrix with eigenvalues ±i√x must satisfy A² = -xI₂. Therefore, det(A² + xI₂) = 0 implies that A² = -xI₂. Therefore, replacing in the expression A² + A + xI₂ gives A, whose determinant is x. Therefore, det(A² + A + xI₂) = det(A) = x.But wait, is that always true? If A is a real 2x2 matrix with det(A² + xI₂) = 0, then does A² necessarily equal -xI₂?Suppose there exists a matrix A where A² + xI₂ is not the zero matrix, but has determinant zero. Then, since A² + xI₂ is a 2x2 matrix of determinant zero, it has rank 1. But can such a matrix A exist?Let’s try to construct a matrix A where A² + xI₂ has rank 1 but is not the zero matrix. For example, let’s take A such that A² = [ -x, 1; 0, -x ]. Then A² + xI₂ = [ 0, 1; 0, 0 ], which has determinant zero and rank 1. However, can such a matrix A exist? Let's check if [ -x, 1; 0, -x ] is a square of some real matrix.Suppose there exists a matrix B such that B² = [ -x, 1; 0, -x ]. Let’s attempt to find B. Let B = [ a, b; c, d ]. Then B² = [ a² + bc, ab + bd; ac + cd, bc + d² ]. Setting this equal to [ -x, 1; 0, -x ].So, we have the equations:1. a² + bc = -x2. ab + bd = 13. ac + cd = 04. bc + d² = -xFrom equation 3: c(a + d) = 0. So either c = 0 or a + d = 0.Case 1: c = 0.Then equation 1: a² = -x. Since x > 0 and a is real, this is impossible. So this case is invalid.Case 2: a + d = 0.Then d = -a.From equation 2: ab + bd = b(a + d) = b(0) = 0. But equation 2 requires this to be 1. Therefore, 0 = 1, which is impossible. Therefore, there is no real matrix B such that B² = [ -x, 1; 0, -x ]. Therefore, such a matrix A cannot exist. Therefore, any real matrix A with det(A² + xI₂) = 0 must satisfy A² = -xI₂. Therefore, the previous reasoning holds, and det(A² + A + xI₂) = det(A) = x.Hence, the determinant is x. Therefore, proved.But let me try to approach this problem from another angle, maybe using properties of determinants and matrices. Suppose we can write A² + A + xI₂ as (A² + xI₂) + A. Since det(A² + xI₂) = 0, maybe there is a relationship between A² + xI₂ and A that allows us to compute the determinant of their sum.Alternatively, perhaps use the fact that if M and N are matrices such that MN = NM, then det(M + N) = det(M) + det(N) + ... but this is generally not true. However, in some cases, if M and N commute, we can use certain properties.Alternatively, perhaps consider that since det(A² + xI₂) = 0, then A² + xI₂ is a rank 1 or rank 0 matrix. If it's rank 0, then A² = -xI₂. If it's rank 1, then there exists vectors u, v such that A² + xI₂ = uv^T. But since A² + xI₂ is a real matrix, uv^T is a rank 1 matrix. But I don't know if that helps.Alternatively, maybe use polynomial equations. Let’s denote B = A² + xI₂, which is given to be singular, so rank(B) ≤ 1. Then, we need to compute det(B + A). If B has rank 1, maybe we can express B as a outer product of vectors, and then use matrix determinant lemma: det(B + A) = det(A + B). But matrix determinant lemma applies when B is rank 1, so det(A + uv^T) = det(A) + v^T adjugate(A) u. But not sure if that helps here.Alternatively, let's consider the eigenvalues. If A has eigenvalues λ and μ, then A² has eigenvalues λ² and μ². The determinant det(A² + xI₂) = (λ² + x)(μ² + x) = 0. Therefore, either λ² + x = 0 or μ² + x = 0. Therefore, as before, eigenvalues of A are ±i√x. Therefore, determinant of A is (i√x)(-i√x) = x. Then, the determinant of A² + A + xI₂ would be the product of (λ² + λ + x) and (μ² + μ + x). Let's compute this.Since λ = i√x, then λ² = -x. Therefore, λ² + λ + x = (-x) + λ + x = λ. Similarly, μ = -i√x, so μ² + μ + x = (-x) + (-i√x) + x = -i√x. Therefore, the determinant would be λ * μ = (i√x)(-i√x) = x. So that gives the determinant as x. Therefore, this also works.So, generalizing, if λ is an eigenvalue of A, then det(A² + xI₂) = 0 implies that λ² = -x, so the eigenvalues are ±i√x. Then, det(A² + A + xI₂) is the product of (λ² + λ + x) and (μ² + μ + x) where μ is the other eigenvalue. Plugging in λ = i√x and μ = -i√x gives ( -x + i√x + x ) and ( -x - i√x + x ), which simplifies to (i√x) and (-i√x). Multiplying them gives (i√x)(-i√x) = -i²x = x. Hence, determinant is x.This seems to confirm the result. Therefore, regardless of the approach, the determinant of (A² + A + xI₂) is x. So the key idea is that the condition det(A² + xI₂) = 0 forces the eigenvalues of A to be ±i√x, which leads to the determinant of (A² + A + xI₂) being the product of (i√x) and (-i√x), which is x.Another way could be to use the Cayley-Hamilton theorem. Since A is a 2x2 matrix, it satisfies its own characteristic equation. If det(A² + xI₂) = 0, then as we found earlier, the characteristic equation of A is λ² + x = 0, so by Cayley-Hamilton, A² + xI₂ = 0. Therefore, A² = -xI₂. Then, substituting into A² + A + xI₂ gives -xI₂ + A + xI₂ = A. Therefore, determinant is det(A) = x. Since the determinant of A is the product of its eigenvalues, which are ±i√x, det(A) = x.Therefore, all approaches lead to the same conclusion. Hence, the determinant of (A² + A + xI₂) is x.Final Answerboxed{x}</think>Given a real number ( x > 0 ) and a ( 2 times 2 ) real matrix ( A ) such that ( det(A^2 + xI_2) = 0 ), we need to prove that ( det(A^2 + A + xI_2) = x ).1. Eigenvalues of ( A ): - Since ( det(A^2 + xI_2) = 0 ), the matrix ( A^2 + xI_2 ) is singular, implying that (-x) is an eigenvalue of ( A^2 ). - The eigenvalues of ( A^2 ) are the squares of the eigenvalues of ( A ). Therefore, the eigenvalues of ( A ) must be ( pm isqrt{x} ).2. Characteristic Polynomial and Cayley-Hamilton Theorem: - The characteristic polynomial of ( A ) is ( lambda^2 + x ), since the eigenvalues are ( pm isqrt{x} ). - By the Cayley-Hamilton theorem, ( A ) satisfies its characteristic polynomial: ( A^2 + xI_2 = 0 ). Hence, ( A^2 = -xI_2 ).3. Substituting ( A^2 ) into the Desired Determinant: - Substitute ( A^2 = -xI_2 ) into ( A^2 + A + xI_2 ): [ A^2 + A + xI_2 = -xI_2 + A + xI_2 = A ] - Therefore, ( det(A^2 + A + xI_2) = det(A) ).4. Determinant of ( A ): - The eigenvalues of ( A ) are ( pm isqrt{x} ), so the determinant of ( A ) is the product of its eigenvalues: [ det(A) = (isqrt{x})(-isqrt{x}) = -i^2 x = x ]Thus, we conclude that ( det(A^2 + A + xI_2) = x ).[boxed{x}]]

🔑:1. Given that ( det(A^2 + xI_2) = 0 ), we start by noting that ( A^2 + xI_2 ) is a ( 2 times 2 ) matrix. Since its determinant is zero, ( A^2 + xI_2 ) is singular, meaning it has at least one eigenvalue equal to zero.2. Let ( lambda ) be an eigenvalue of ( A ). Then, ( lambda^2 + x = 0 ) because ( det(A^2 + xI_2) = 0 ). Solving for ( lambda ), we get: [ lambda^2 = -x implies lambda = pm isqrt{x} ] Therefore, the eigenvalues of ( A ) are ( isqrt{x} ) and ( -isqrt{x} ).3. The trace of ( A ), which is the sum of its eigenvalues, is: [ text{tr}(A) = isqrt{x} + (-isqrt{x}) = 0 ]4. The determinant of ( A ), which is the product of its eigenvalues, is: [ det(A) = (isqrt{x})(-isqrt{x}) = -i^2 x = x ]5. The characteristic polynomial of ( A ) is given by: [ lambda^2 - (text{tr}(A))lambda + det(A) = lambda^2 + x ] By the Cayley-Hamilton theorem, ( A ) satisfies its own characteristic polynomial: [ A^2 + xI_2 = 0 implies A^2 = -xI_2 ]6. We need to prove that ( det(A^2 + A + xI_2) = x ). Using the result from the Cayley-Hamilton theorem: [ A^2 + xI_2 = 0 implies A^2 = -xI_2 ] Substituting ( A^2 = -xI_2 ) into ( A^2 + A + xI_2 ): [ A^2 + A + xI_2 = -xI_2 + A + xI_2 = A ]7. Therefore, we have: [ det(A^2 + A + xI_2) = det(A) ] From step 4, we know that ( det(A) = x ).Thus, we have shown that:[det(A^2 + A + xI_2) = x]The final answer is ( boxed{ x } )

🧠:[<think>Okay, so I have this problem here with two parts, I and II. Let me start by tackling part I first. First, they want the ordinary equation of line l. The parametric equations given are x = 2 + t and y = -1 + t, where t is the parameter. To convert parametric equations to an ordinary (Cartesian) equation, I need to eliminate the parameter t. Looking at both equations:x = 2 + ty = -1 + tIf I solve both equations for t, I get:From the first equation: t = x - 2From the second equation: t = y + 1Since both are equal to t, I can set them equal to each other:x - 2 = y + 1Then, rearranging terms to get the standard form:x - y - 3 = 0Or, written as y = x - 3. So that should be the ordinary equation for line l. Let me double-check. If x = 2 + t, then t = x - 2. Substitute into y = -1 + t gives y = -1 + (x - 2) = x - 3. Yep, that works.Now, part I also asks for the rectangular coordinate equation of curve C. The polar equation given is ρ = 2√2 a cos(θ + π/4), with a > 5/6. Hmm, converting polar equations to rectangular coordinates. Remember that in polar coordinates, ρ is the radius, θ is the angle, and the conversions are:x = ρ cos θy = ρ sin θand ρ² = x² + y²Also, trigonometric identities might be helpful here. The given equation is ρ = 2√2 a cos(θ + π/4). Let's expand cos(θ + π/4) using the cosine addition formula:cos(θ + π/4) = cosθ cos(π/4) - sinθ sin(π/4)Since cos(π/4) = sin(π/4) = √2/2, so:cos(θ + π/4) = (√2/2)(cosθ - sinθ)Therefore, the polar equation becomes:ρ = 2√2 a * (√2/2)(cosθ - sinθ)Simplify the constants: 2√2 * √2/2 = (2√2)(√2)/2 = (2*2)/2 = 2. So:ρ = 2a (cosθ - sinθ)Now, in rectangular coordinates, ρ cosθ = x and ρ sinθ = y. Let's write the equation as:ρ = 2a (cosθ - sinθ)Multiply both sides by ρ:ρ² = 2a (ρ cosθ - ρ sinθ)Which translates to:x² + y² = 2a (x - y)So the rectangular coordinate equation of curve C is x² + y² = 2a(x - y). Let me check that again. Starting from ρ = 2√2 a cos(θ + π/4), expand using the identity, simplify to get ρ = 2a(cosθ - sinθ). Multiply both sides by ρ, replace ρ² with x² + y², and ρ cosθ with x, ρ sinθ with y. Yep, so x² + y² = 2a(x - y). Maybe rearrange terms to standard form? Let's see:x² + y² - 2a x + 2a y = 0But unless they specify a particular form, this might be acceptable. So part I is done. Now moving to part II.Given point P(2, -1). Line l intersects curve C at points M and N. If |MN|² = 6|PM|•|PN|, find the value of a.So, need to find the points M and N where line l intersects curve C. Then compute the distance between M and N squared, and set it equal to 6 times the product of the distances from P to M and P to N. Then solve for a.First, let me note that line l is y = x - 3. Curve C is x² + y² = 2a(x - y). So substituting y = x - 3 into the equation of C.Substitute y = x - 3 into x² + y² = 2a(x - y):x² + (x - 3)² = 2a(x - (x - 3))Compute left-hand side:x² + (x² - 6x + 9) = 2x² - 6x + 9Right-hand side:2a(x - x + 3) = 2a(3) = 6aSo equation becomes:2x² - 6x + 9 = 6aBring all terms to left:2x² - 6x + 9 - 6a = 0Simplify:2x² - 6x + (9 - 6a) = 0Divide both sides by 2 to make it simpler:x² - 3x + (9/2 - 3a) = 0So quadratic equation in x: x² - 3x + (9/2 - 3a) = 0.Let me write it as:x² - 3x + (9/2 - 3a) = 0Let me denote coefficients as A=1, B=-3, C=(9/2 - 3a)To find the points M and N, we need to find the roots of this quadratic. Let's compute the discriminant first to check if real roots exist.Discriminant D = B² - 4AC = (-3)² - 4*(1)*(9/2 - 3a) = 9 - 4*(9/2 - 3a)Calculate 4*(9/2) = 18, and 4*(-3a) = -12a. So:D = 9 - (18 - 12a) = 9 - 18 + 12a = -9 + 12aGiven that a > 5/6, so 12a > 10, so D = -9 + 12a > -9 + 10 = 1 > 0. So two distinct real roots, which makes sense as line intersects the curve at two points M and N.Let me find the roots. The solutions are:x = [3 ± √(D)] / 2 = [3 ± √(12a - 9)] / 2So x-coordinates of M and N are [3 + √(12a -9)]/2 and [3 - √(12a -9)]/2. Then the corresponding y-coordinates, since y = x - 3, will be:For x1 = [3 + √(12a -9)]/2, y1 = [3 + √(12a -9)]/2 - 3 = [3 + √(12a -9) - 6]/2 = [-3 + √(12a -9)]/2Similarly, for x2 = [3 - √(12a -9)]/2, y2 = [3 - √(12a -9)]/2 - 3 = [3 - √(12a -9) - 6]/2 = [-3 - √(12a -9)]/2So points M and N are:M: ([3 + √(12a -9)]/2, [-3 + √(12a -9)]/2)N: ([3 - √(12a -9)]/2, [-3 - √(12a -9)]/2)Now, need to compute |MN|² and |PM|•|PN|.First, let's compute |MN|². The distance between M and N squared.Coordinates of M and N:M: ( (3 + s)/2 , (-3 + s)/2 )N: ( (3 - s)/2 , (-3 - s)/2 )Where s = √(12a -9)Let me denote s for simplicity. Then:x coordinates: (3 + s)/2 and (3 - s)/2. Difference in x: [(3 + s)/2 - (3 - s)/2] = (2s)/2 = sSimilarly, y coordinates: (-3 + s)/2 and (-3 - s)/2. Difference in y: [(-3 + s)/2 - (-3 - s)/2] = (2s)/2 = sTherefore, the distance between M and N is sqrt( (s)^2 + (s)^2 ) = sqrt(2s²) = s√2Therefore, |MN|² = (s√2)^2 = 2s². Since s = √(12a -9), then s² = 12a -9. Therefore:|MN|² = 2*(12a -9) = 24a -18Now, need to compute |PM|•|PN|. Where P is (2, -1).First, compute |PM| and |PN|.Coordinates of P: (2, -1)Coordinates of M: ( (3 + s)/2 , (-3 + s)/2 )Coordinates of N: ( (3 - s)/2 , (-3 - s)/2 )Compute |PM|:Distance from P to M:sqrt[ ( ( (3 + s)/2 - 2 )² + ( (-3 + s)/2 - (-1) )² ) ]Simplify the terms:x-coordinate difference: (3 + s)/2 - 2 = (3 + s - 4)/2 = (s -1)/2y-coordinate difference: (-3 + s)/2 + 1 = (-3 + s + 2)/2 = (s -1)/2Therefore, |PM| = sqrt[ ( (s -1)/2 )² + ( (s -1)/2 )² ] = sqrt[ 2*( (s -1)/2 )² ] = sqrt[ ( (s -1)² )/2 ] = |s -1| / √2Similarly, compute |PN|:Distance from P to N:sqrt[ ( ( (3 - s)/2 - 2 )² + ( (-3 - s)/2 - (-1) )² ) ]Compute the differences:x-coordinate: (3 - s)/2 - 2 = (3 - s -4)/2 = (-s -1)/2y-coordinate: (-3 - s)/2 +1 = (-3 - s +2)/2 = (-s -1)/2Thus, |PN| = sqrt[ ( (-s -1)/2 )² + ( (-s -1)/2 )² ] = sqrt[ 2*( (s +1)/2 )² ] = sqrt[ (s +1)^2 / 2 ] = |s +1| / √2Since s = √(12a -9). Since a > 5/6, 12a -9 > 12*(5/6) -9 = 10 -9 =1. So s >1. Therefore, s -1 is positive, and s +1 is positive. So absolute values can be removed:|PM| = (s -1)/√2|PN| = (s +1)/√2Therefore, the product |PM|•|PN| = (s -1)(s +1)/ (√2 * √2 ) = (s² -1)/2But s² = 12a -9, so:|PM|•|PN| = (12a -9 -1)/2 = (12a -10)/2 = 6a -5According to the problem, |MN|² = 6|PM|•|PN|From earlier, |MN|² =24a -18Set equal to 6*(6a -5):24a -18 = 6*(6a -5)Compute RHS: 36a -30So equation:24a -18 = 36a -30Subtract 24a from both sides:-18 = 12a -30Add 30 to both sides:12 =12aDivide both sides by 12:a =1But check if this satisfies the initial condition a >5/6. Indeed, 1>5/6, so it's valid.But let me verify all steps to ensure no mistakes.First, converting parametric to ordinary equation: x -2 = y +1 => y =x -3. Correct.For the polar equation: ρ =2√2 a cos(θ + π/4). Expanded using cos(A+B), correct. Then multiplied by ρ, replaced with x and y. Got x² + y² =2a(x - y). Correct.Then for part II, substituted y =x -3 into x² + y² =2a(x - y). Got 2x² -6x +9=6a. Then quadratic equation x² -3x + (9/2 -3a)=0. Correct.Computed discriminant: D=9 -4*(9/2 -3a)= -9 +12a. Correct.Expressed roots in terms of s=√(12a-9). Then found coordinates of M and N. Then computed |MN|² as 24a -18. Correct.Then computed |PM| and |PN|. Calculated distances by plugging into coordinates. Got |PM|=(s-1)/√2 and |PN|=(s+1)/√2. Then product is (s² -1)/2 = (12a -10)/2=6a -5. Then set |MN|²=6*(6a -5). So equation 24a -18=36a -30. Solved to a=1. Correct. Checked that a=1>5/6. All steps seem correct. Therefore, the answer is a=1.Final AnswerThe value of ( a ) is boxed{1}.</think>(I) To find the ordinary equation of line ( l ), we start with the parametric equations:[begin{cases}x = 2 + t y = -1 + tend{cases}]Solving for ( t ) in both equations, we get ( t = x - 2 ) and ( t = y + 1 ). Equating these, we find:[x - 2 = y + 1 implies x - y - 3 = 0 implies y = x - 3]Thus, the ordinary equation of line ( l ) is ( y = x - 3 ).For the polar equation ( rho = 2sqrt{2}acosleft(theta + frac{pi}{4}right) ), we use the cosine addition formula:[cosleft(theta + frac{pi}{4}right) = costhetacosfrac{pi}{4} - sinthetasinfrac{pi}{4} = frac{sqrt{2}}{2}(costheta - sintheta)]Substituting this into the polar equation, we get:[rho = 2sqrt{2}a cdot frac{sqrt{2}}{2}(costheta - sintheta) = 2a(costheta - sintheta)]Multiplying both sides by ( rho ) and converting to rectangular coordinates, we have:[rho^2 = 2a(rhocostheta - rhosintheta) implies x^2 + y^2 = 2a(x - y)]Thus, the rectangular coordinate equation of curve ( C ) is ( x^2 + y^2 = 2a(x - y) ).(II) Substituting ( y = x - 3 ) into ( x^2 + y^2 = 2a(x - y) ), we get:[x^2 + (x - 3)^2 = 2a(3) implies 2x^2 - 6x + 9 = 6a implies x^2 - 3x + frac{9}{2} - 3a = 0]The discriminant of this quadratic equation is:[D = (-3)^2 - 4 cdot 1 cdot left(frac{9}{2} - 3aright) = 9 - 18 + 12a = 12a - 9]The roots are:[x = frac{3 pm sqrt{12a - 9}}{2}]Let ( s = sqrt{12a - 9} ). The points ( M ) and ( N ) are:[Mleft(frac{3 + s}{2}, frac{-3 + s}{2}right), quad Nleft(frac{3 - s}{2}, frac{-3 - s}{2}right)]The distance ( |MN|^2 ) is:[|MN|^2 = 2s^2 = 2(12a - 9) = 24a - 18]The distances ( |PM| ) and ( |PN| ) are:[|PM| = frac{s - 1}{sqrt{2}}, quad |PN| = frac{s + 1}{sqrt{2}}]Their product is:[|PM| cdot |PN| = frac{(s - 1)(s + 1)}{2} = frac{12a - 10}{2} = 6a - 5]Given ( |MN|^2 = 6|PM| cdot |PN| ), we solve:[24a - 18 = 6(6a - 5) implies 24a - 18 = 36a - 30 implies 12a = 12 implies a = 1]The value of ( a ) is boxed{1}.,<think>Okay, let's try to tackle this problem step by step. It's part I and II. Let me start with part I.Part I: Converting parametric to ordinary equation and polar to rectangular coordinatesFirst, the line l has parametric equations:x = 2 + ty = -1 + tTo find the ordinary equation, I need to eliminate the parameter t. Let me solve both equations for t and set them equal.From the x equation: t = x - 2From the y equation: t = y + 1So, setting them equal: x - 2 = y + 1Then, rearranging: x - y - 3 = 0So, the ordinary equation of line l is x - y - 3 = 0. That seems straightforward.Now, for the curve C: polar equation is ρ = 2√2 a cos(θ + π/4), with a > 5/6.I need to convert this polar equation to rectangular coordinates. Remember, polar to rectangular conversions:ρ = sqrt(x² + y²)x = ρ cosθy = ρ sinθcos(θ + π/4) can be expanded using cosine addition formula.First, expand cos(θ + π/4):cos(θ + π/4) = cosθ cos(π/4) - sinθ sin(π/4) = (cosθ - sinθ)/√2Therefore, the polar equation becomes:ρ = 2√2 a * (cosθ - sinθ)/√2Simplify that: 2√2 / √2 = 2, so:ρ = 2a (cosθ - sinθ)Now, multiply both sides by ρ:ρ² = 2a (ρ cosθ - ρ sinθ)But ρ² = x² + y², ρ cosθ = x, ρ sinθ = y.So substituting:x² + y² = 2a (x - y)So, the rectangular coordinate equation of curve C is x² + y² = 2a(x - y)Let me check that again. Yes, starting from the polar equation, expanding the cosine term, simplifying, multiplying both sides by ρ to get rid of the denominator, then substituting x and y. Seems right.So, part I answers:Line l: x - y - 3 = 0Curve C: x² + y² - 2a x + 2a y = 0Alternatively, written as x² + y² = 2a(x - y)Part II: Finding the value of a given |MN|² = 6 |PM|•|PN|Given point P(2, -1). Line l intersects curve C at points M and N. The condition is that |MN|² = 6 |PM|•|PN|.So, first, need to find points M and N by solving the intersection of line l and curve C.Given line l is x - y - 3 = 0, which we can write as y = x - 3.Wait, but from part I, line l's equation is x - y - 3 = 0, which is equivalent to y = x - 3. But the parametric equations were x = 2 + t, y = -1 + t. Let me check that.If x = 2 + t and y = -1 + t, then t = x - 2, so y = -1 + x - 2 = x - 3. Yes, that's correct. So line l is y = x - 3.Now, substituting this into curve C's equation x² + y² = 2a(x - y).So substitute y = x - 3 into C's equation:x² + (x - 3)^2 = 2a [x - (x - 3)]Compute left side:x² + (x² - 6x + 9) = 2x² - 6x + 9Right side:2a [x - x + 3] = 2a * 3 = 6aSo, equation becomes:2x² - 6x + 9 = 6aBring all terms to left:2x² - 6x + 9 - 6a = 0Divide both sides by 2:x² - 3x + (9/2 - 3a) = 0So, quadratic equation in x: x² - 3x + (9/2 - 3a) = 0Let me write that as x² - 3x + (9/2 - 3a) = 0Let me compute discriminant D to check the roots.D = 9 - 4 * 1 * (9/2 - 3a) = 9 - 18 + 12a = -9 + 12aSince a > 5/6, then 12a > 10, so -9 + 12a > 1, so D > 0. So, two real roots. Good.Let me denote the solutions as x₁ and x₂.Then, the roots are:x = [3 ± sqrt(D)] / 2 = [3 ± sqrt(12a - 9)] / 2Therefore, the points M and N have x-coordinates x₁ and x₂, and y-coordinates y = x - 3.So, coordinates:M: ( [3 + sqrt(12a - 9)] / 2 , [3 + sqrt(12a - 9)] / 2 - 3 )Simplify y-coordinate:= [3 + sqrt(12a - 9)] / 2 - 6/2 = [ -3 + sqrt(12a - 9) ] / 2Similarly, N: ( [3 - sqrt(12a - 9)] / 2 , [ -3 - sqrt(12a - 9) ] / 2 )So, points M and N are:M: ( (3 + sqrt(12a - 9))/2 , (-3 + sqrt(12a - 9))/2 )N: ( (3 - sqrt(12a - 9))/2 , (-3 - sqrt(12a - 9))/2 )Now, need to compute |MN|² and |PM|•|PN|, then set |MN|² = 6 |PM|•|PN| and solve for a.First, let's compute |MN|².Coordinates of M and N:M: ( (3 + d)/2, (-3 + d)/2 )N: ( (3 - d)/2, (-3 - d)/2 )Where d = sqrt(12a - 9)Compute the distance squared between M and N:Δx = [(3 + d)/2 - (3 - d)/2] = [ (3 + d - 3 + d)/2 ] = (2d)/2 = dΔy = [ (-3 + d)/2 - (-3 - d)/2 ] = [ (-3 + d + 3 + d)/2 ] = (2d)/2 = dTherefore, |MN|² = (Δx)^2 + (Δy)^2 = d² + d² = 2d²But d = sqrt(12a -9), so d² = 12a -9Therefore, |MN|² = 2*(12a -9) = 24a - 18Now, need to compute |PM|•|PN|, where P is (2, -1)First, compute |PM| and |PN|.Let me compute |PM|:Coordinates of P: (2, -1)Coordinates of M: ( (3 + d)/2, (-3 + d)/2 )So, vector PM: ( (3 + d)/2 - 2, (-3 + d)/2 - (-1) )Compute x-coordinate: (3 + d)/2 - 4/2 = (-1 + d)/2y-coordinate: (-3 + d)/2 + 1 = (-3 + d + 2)/2 = (-1 + d)/2So, |PM| is the distance:sqrt( [ (-1 + d)/2 ]^2 + [ (-1 + d)/2 ]^2 ) = sqrt( 2 * [ ( (-1 + d)^2 ) /4 ] ) = sqrt( ( (-1 + d)^2 ) / 2 ) = | -1 + d | / sqrt(2 )Similarly, |PN|:Coordinates of N: ( (3 - d)/2, (-3 - d)/2 )Vector PN: ( (3 - d)/2 - 2, (-3 - d)/2 - (-1) )Compute x-coordinate: (3 - d)/2 - 4/2 = (-1 - d)/2y-coordinate: (-3 - d)/2 + 1 = (-3 - d + 2)/2 = (-1 - d)/2Therefore, |PN| = sqrt( [ (-1 - d)/2 ]^2 + [ (-1 - d)/2 ]^2 ) = sqrt(2 * [ ( (-1 - d)^2 ) /4 ]) = sqrt( ( (-1 - d)^2 ) / 2 ) = | -1 - d | / sqrt(2 )Since a > 5/6, let's check if d is positive.d = sqrt(12a -9). Since a >5/6, 12a >10, so 12a -9 >1, so d >1. Therefore, d >1.So, |-1 + d|: since d >1, this is (d -1)Similarly, |-1 - d| = (d +1 )Therefore, |PM| = (d -1)/sqrt(2 )|PN| = (d +1)/sqrt(2 )So, |PM|•|PN| = [ (d -1)(d +1) ] / (2 )= (d² -1)/2But d² =12a -9, so:|PM|•|PN| = (12a -9 -1)/2 = (12a -10)/2 =6a -5Therefore, the condition |MN|² =6 |PM|•|PN| becomes:24a -18 =6*(6a -5)Compute RHS: 36a -30Set equation:24a -18 =36a -30Subtract 24a both sides:-18 =12a -30Add 30 both sides:12=12aTherefore, a=1But we need to check if a=1 satisfies the condition a>5/6. Since 1>5/6, it's valid.Wait, but let me verify all steps to make sure.Let me recap:We had |MN|² =24a -18Computed |PM|•|PN| =6a -5Set 24a -18=6*(6a -5)24a -18=36a -3024a -36a= -30 +18-12a= -12So, a=1. Correct.Therefore, the value of a is 1.But wait, to be thorough, let me verify with a=1.Check if substituting a=1 into curve C equation gives the correct intersections.For a=1, curve C is x² + y² =2*1*(x - y) =>x² + y² =2x -2yBring all terms to left: x² -2x + y² +2y=0Complete the square:x² -2x +1 + y² +2y +1 =2So, (x -1)^2 + (y +1)^2 =2So, it's a circle with center (1,-1) and radius sqrt(2)Line l is y=x -3Find intersection points between y=x -3 and circle (x -1)^2 + (y +1)^2=2Substitute y=x -3 into the circle equation:(x -1)^2 + ( (x -3) +1 )^2 =2Simplify:(x -1)^2 + (x -2)^2 =2Compute each term:(x² -2x +1) + (x² -4x +4) =2Combine like terms:2x² -6x +5=22x² -6x +3=0Divide by 1: 2x² -6x +3=0Quadratic equation: x=(6±sqrt(36-24))/4=(6±sqrt(12))/4=(6±2*sqrt(3))/4=(3±sqrt(3))/2Therefore, x=(3±sqrt(3))/2, y= x -3= (3±sqrt(3))/2 -6/2= (-3±sqrt(3))/2Thus, points M and N are ((3+sqrt(3))/2, (-3+sqrt(3))/2) and ((3-sqrt(3))/2, (-3-sqrt(3))/2)Compute |MN|²:Δx= sqrt(3), Δy= sqrt(3)So, |MN|²= (sqrt(3))² + (sqrt(3))²=3+3=6Compute |PM| and |PN|:Point P is (2, -1)Compute |PM|:Distance from (2,-1) to ((3+sqrt(3))/2, (-3+sqrt(3))/2 )Δx= (3+sqrt(3))/2 - 2= (3+sqrt(3)-4)/2= (-1 + sqrt(3))/2Δy= (-3 + sqrt(3))/2 - (-1)= (-3 + sqrt(3) +2)/2= (-1 + sqrt(3))/2So, |PM|=sqrt( [(-1 + sqrt(3))/2]^2 + [(-1 + sqrt(3))/2]^2 )=sqrt(2*( (1 -2sqrt(3) +3)/4 ))=sqrt(2*(4 -2sqrt(3))/4)=sqrt( (4 -2sqrt(3))/2 )=sqrt(2 - sqrt(3))Similarly, |PN|:Distance from (2,-1) to ((3 - sqrt(3))/2, (-3 - sqrt(3))/2 )Δx= (3 - sqrt(3))/2 -2= (3 - sqrt(3) -4)/2= (-1 - sqrt(3))/2Δy= (-3 - sqrt(3))/2 -(-1)= (-3 - sqrt(3) +2)/2= (-1 - sqrt(3))/2So, |PN|=sqrt( [(-1 - sqrt(3))/2]^2 + [(-1 - sqrt(3))/2]^2 )=sqrt(2*( (1 +2sqrt(3) +3)/4 ))=sqrt(2*(4 +2sqrt(3))/4 )=sqrt( (4 +2sqrt(3))/2 )=sqrt(2 + sqrt(3))Thus, |PM|•|PN|=sqrt(2 - sqrt(3)) * sqrt(2 + sqrt(3))=sqrt( (2 - sqrt(3))(2 + sqrt(3)) )=sqrt(4 -3)=sqrt(1)=1Therefore, 6 |PM|•|PN|=6*1=6And |MN|²=6So, 6=6. Hence, condition holds. Therefore, a=1 is correct.So, the answer is a=1.Final AnswerThe value of ( a ) is boxed{1}.</think>(I) To find the ordinary equation of line ( l ) given the parametric equations:[begin{cases}x = 2 + t y = -1 + tend{cases}]we eliminate the parameter ( t ). From ( x = 2 + t ) and ( y = -1 + t ), we get ( t = x - 2 ) and ( t = y + 1 ). Equating these, we have ( x - 2 = y + 1 ), which simplifies to ( x - y - 3 = 0 ).For the polar equation of curve ( C ), given by ( rho = 2sqrt{2}acosleft(theta + frac{pi}{4}right) ), we use the cosine addition formula:[cosleft(theta + frac{pi}{4}right) = frac{costheta - sintheta}{sqrt{2}}]Substituting this into the polar equation, we get:[rho = 2sqrt{2}a cdot frac{costheta - sintheta}{sqrt{2}} = 2a(costheta - sintheta)]Multiplying both sides by ( rho ) and converting to rectangular coordinates (( rho^2 = x^2 + y^2 ), ( rhocostheta = x ), ( rhosintheta = y )), we obtain:[x^2 + y^2 = 2a(x - y)]Thus, the ordinary equation of line ( l ) is ( x - y - 3 = 0 ) and the rectangular coordinate equation of curve ( C ) is ( x^2 + y^2 = 2a(x - y) ).(II) Given point ( P(2, -1) ), line ( l ) intersects curve ( C ) at points ( M ) and ( N ). We need to find ( a ) such that ( |MN|^2 = 6|PM| cdot |PN| ).Substituting ( y = x - 3 ) (from the line equation) into the curve equation ( x^2 + y^2 = 2a(x - y) ), we get:[x^2 + (x - 3)^2 = 2a cdot 3]Simplifying, we find:[2x^2 - 6x + 9 = 6a implies x^2 - 3x + frac{9}{2} - 3a = 0]The roots ( x_1 ) and ( x_2 ) are given by:[x = frac{3 pm sqrt{12a - 9}}{2}]The distance ( |MN|^2 ) is calculated as:[|MN|^2 = 2(12a - 9)]For points ( M ) and ( N ), we find the distances ( |PM| ) and ( |PN| ):[|PM| = frac{sqrt{(d - 1)^2}}{sqrt{2}}, quad |PN| = frac{sqrt{(d + 1)^2}}{sqrt{2}}]where ( d = sqrt{12a - 9} ). Thus,[|PM| cdot |PN| = frac{(d^2 - 1)}{2} = frac{12a - 10}{2} = 6a - 5]Setting ( |MN|^2 = 6|PM| cdot |PN| ), we solve:[24a - 18 = 6(6a - 5) implies 24a - 18 = 36a - 30 implies 12a = 12 implies a = 1]Thus, the value of ( a ) is boxed{1}.]

🔑:(I) Eliminating t from begin{cases} overset{x=2+t}{y=-1+t}end{cases}, we get x - y - 3 = 0.From ρ = 2sqrt {2}acos(θ + frac {π}{4}) (a > frac {5}{6}), we get ρ^2 = 2aρcosθ - 2aρsinθ, which leads to x^2 + y^2 = 2ax - 2ay.(II) Since point P(2, -1) is on line l, the parametric equation of line l is: begin{cases} overset{x=2+tcosalpha }{t=-1+tsinalpha }end{cases} (t is the parameter).Substituting it into the equation of curve C, we get: 2t^2 + 2t + 5 - 6a = 0.Let t1 and t2 be the parameters corresponding to points M and N respectively.Then, t1 + t2 = -1 and t1t2 = frac {5-6a}{2}.From |MN|^2 = 6|PM|•|PN|, we get: (t1 - t2)^2 = 6|t1t2|.Therefore, (t1 + t2)^2 - 4t1t2 = 6|t1t2|.Since a > frac {5}{6}, we have 1 - 2(5 - 6a) = 3(6a - 5). Solving for a, we get a = 1.In conclusion, boxed{a = 1}.

🧠:[<think>Okay, so I need to solve these three parts of the problem. Let's start with part (1).Problem (1): When a = 2, find the maximum value of the function f(x) = ln x - (1/2)a x² + x.Alright, so when a is 2, the function becomes f(x) = ln x - (1/2)*2*x² + x. Simplifying that, the (1/2)*2 cancels out, so it's ln x - x² + x. Now, to find the maximum value, I need to take the derivative of f(x) with respect to x, set it equal to zero, and solve for x. Then check if that's a maximum by using the second derivative test or analyzing the sign changes.First derivative: f'(x) = d/dx [ln x - x² + x] = 1/x - 2x + 1.Set this equal to zero: 1/x - 2x + 1 = 0.Multiply both sides by x to eliminate the denominator (assuming x ≠ 0, which it isn't since ln x is only defined for x > 0):1 - 2x² + x = 0.Rearranging terms: -2x² + x + 1 = 0. Multiply both sides by -1 to make it easier: 2x² - x - 1 = 0.Now, solve this quadratic equation. The quadratic formula is x = [b ± √(b² - 4ac)] / (2a). Here, a = 2, b = -1, c = -1.Discriminant: (-1)² - 4*2*(-1) = 1 + 8 = 9. So sqrt(9) = 3.Solutions: x = [1 ± 3]/4.So x = (1 + 3)/4 = 4/4 = 1, and x = (1 - 3)/4 = -2/4 = -0.5. But x must be positive since we have ln x, so we discard x = -0.5. So critical point at x = 1.Now, check if this is a maximum. Second derivative test: f''(x) = derivative of f'(x) which is derivative of (1/x - 2x + 1) = -1/x² - 2.At x = 1, f''(1) = -1/1² - 2 = -1 - 2 = -3. Since f''(1) is negative, the function is concave down here, so x=1 is a local maximum.Since the function has only one critical point and as x approaches 0+ ln x tends to -infinity and as x approaches infinity, -x² dominates so f(x) tends to -infinity. Therefore, the local maximum at x=1 is indeed the global maximum.Compute f(1): ln 1 - (1)^2 + 1 = 0 -1 +1 = 0. So maximum value is 0.Wait, that's interesting. Let me check that again. ln 1 is 0, minus 1 squared which is 1, plus 1: 0 -1 +1 = 0. Yeah, that's correct. So the maximum value when a=2 is 0. Hmm, seems straightforward. Let me make sure there are no calculation errors. The steps look okay.Problem (2): Let F(x) = f(x) + (1/2)a x² - x + a/x for 0 < x ≤ 3. If the slope k of the tangent line at any point P(x0, y0) on its graph satisfies k ≤ 1/2 always holds, find the range of the real number a.First, let's write out F(x). Given that f(x) is ln x - (1/2)a x² + x, so substituting into F(x):F(x) = [ln x - (1/2)a x² + x] + (1/2)a x² - x + a/x.Simplify term by term:- The -(1/2)a x² and +(1/2)a x² cancel each other.- The +x and -x also cancel each other.So, F(x) simplifies to ln x + a/x.So F(x) = ln x + (a)/x.Now, the problem states that the slope of the tangent line at any point (i.e., the derivative of F(x)) must be ≤ 1/2 for all x in (0, 3]. So we need F'(x) ≤ 1/2 for all x in (0, 3].First, compute F'(x):F'(x) = derivative of ln x + a/x = 1/x - a/(x²).So F'(x) = (1/x) - (a)/(x²).We need (1/x - a/x²) ≤ 1/2 for all x in (0, 3].So, 1/x - a/x² ≤ 1/2 for all 0 < x ≤ 3.We can rearrange this inequality:1/x - a/x² ≤ 1/2.Multiply both sides by x² (since x > 0, x² is positive, so inequality direction remains the same):x - a ≤ (1/2)x².So, (1/2)x² - x + a ≥ 0 for all x in (0, 3].So, the inequality (1/2)x² - x + a ≥ 0 must hold for all x in (0, 3]. So we need to find a such that a ≥ - (1/2)x² + x for all x in (0, 3].In other words, a must be greater than or equal to the maximum value of the function g(x) = - (1/2)x² + x on the interval (0, 3]. Because if a is greater than or equal to the maximum of g(x), then a ≥ g(x) for all x in the interval.So, let's find the maximum of g(x) = - (1/2)x² + x on (0, 3].First, find critical points by taking derivative:g'(x) = -x + 1.Set to zero: -x +1 = 0 => x=1.Check if x=1 is a maximum. Second derivative: g''(x) = -1 < 0, so it's a local maximum.Compute g(1) = - (1/2)(1)^2 +1 = -1/2 +1 = 1/2.Check endpoints:At x approaching 0+, g(x) approaches -0 +0 = 0.At x=3, g(3) = - (1/2)(9) + 3 = -4.5 +3 = -1.5.Therefore, the maximum of g(x) on (0,3] is 1/2 at x=1. Hence, to satisfy a ≥ g(x) for all x in (0,3], we need a ≥ 1/2.But let's verify this because sometimes endpoints can have maximums, but in this case, the maximum is at x=1. So if a is at least 1/2, then (1/2)x² - x + a ≥ 0 for all x in (0,3].Wait, but let's test for a=1/2. Then the inequality becomes (1/2)x² -x +1/2 ≥0.Let’s check if this is non-negative on (0,3].(1/2)x² -x +1/2 = (x² - 2x +1)/2 = (x -1)^2 /2 ≥0, which is always true because squares are non-negative. So equality holds when x=1. So if a=1/2, the inequality is satisfied. If a >1/2, then (x -1)^2 /2 + (a -1/2) ≥0, which is also true. If a <1/2, then at x=1, the expression becomes (1/2)(1) -1 +a = (1/2 -1) +a = -1/2 +a. If a <1/2, this is negative, which violates the inequality. Therefore, the minimal a is 1/2, and the range is a ≥1/2.Hence, the range of real number a is [1/2, ∞). Wait, but let me check if the inequality is correctly transformed.Original inequality: 1/x - a/x² ≤ 1/2.We rearranged to (1/2)x² -x +a ≥0.But let's check with a=1/2. Then (1/2)x² -x +1/2 = (x² - 2x +1)/2 = (x-1)^2 /2 ≥0, which is always true, correct. So if a=1/2, equality holds at x=1, and for a>1/2, the term is even larger, so inequality holds. If a<1/2, then at x=1, we have (1/2 -1 +a) = (-1/2 +a), which is negative if a<1/2, which would violate the inequality. Therefore, the minimal a is 1/2.Therefore, the range of a is [1/2, ∞). So answer is a ≥1/2.Problem (3): When a=0, the equation m f(x) = x² has a unique real solution. Find the value of the positive number m.First, when a=0, the function f(x) is ln x -0 +x = ln x +x. So the equation is m(ln x +x) = x². So m(ln x + x) =x². We need to find m >0 such that this equation has exactly one real solution.So, m = x² / (ln x +x) for x >0 (since ln x is defined for x>0). So the equation m = x² / (ln x +x) has exactly one real solution. Therefore, we need to find m such that the horizontal line y=m intersects the graph of h(x)=x²/(ln x +x) exactly once.Therefore, we need to find m where h(x)=x²/(ln x +x) has a unique solution. This would occur either at a extremum (where h(x) has a horizontal tangent, so m is equal to the extremum value) or at a point where the function is tangent to the line y=m.Alternatively, perhaps the function h(x) is monotonic, but since h(x) =x²/(x + ln x), let's check its behavior.First, analyze h(x):h(x) = x²/(x + ln x). Let's check the domain. x + ln x must not be zero, and x >0.First, when x approaches 0+:ln x approaches -infinity, so denominator x + ln x approaches -infinity. x² approaches 0. So h(x) approaches 0 / (-infty) = 0.But need to verify the limit:lim_{x→0+} x²/(x + ln x). Let's compute this. The numerator tends to 0, denominator tends to -infty. So overall, the limit is 0.At x=1: h(1)=1/(1 +0)=1.As x increases beyond 1, let's see the behavior. For x>1, ln x is positive, but x is dominant. Let's see:As x approaches infinity, h(x) ~ x² /x =x, so h(x) tends to infinity.But let's check the derivative of h(x) to see if it's increasing or has a critical point.Compute h'(x):h(x) = x²/(x + ln x). Let's use the quotient rule.h'(x) = [2x(x + ln x) - x²(1 + 1/x)] / (x + ln x)^2.Simplify numerator:2x(x + ln x) - x²(1 +1/x) = 2x^2 + 2x ln x -x² -x.= (2x² -x²) + 2x ln x -x= x² + 2x ln x -x.So h'(x) = [x² + 2x ln x -x] / (x + ln x)^2.To find critical points, set numerator equal to zero:x² + 2x ln x -x =0.Factor x:x(x + 2 ln x -1) =0.Since x>0, the critical points occur when x + 2 ln x -1=0.Let’s solve x + 2 ln x -1=0.Let’s define k(x) =x + 2 ln x -1. Find x such that k(x)=0.We can attempt to solve this equation.First, check x=1: 1 +0 -1=0. So x=1 is a solution.Is there another solution?Check behavior of k(x):As x approaches 0+, 2 ln x approaches -infty, so k(x) approaches -infty.At x=1, k(1)=0.Compute derivative of k(x): k’(x)=1 + 2/x. Since x>0, k’(x)=1 +2/x >0 always. Therefore, k(x) is strictly increasing from -infty to +infty as x increases from 0 to infinity. Since it crosses zero at x=1, that is the only solution. Therefore, the only critical point is at x=1.Therefore, h(x) has critical point only at x=1.Now, check the value at x=1: h(1)=1/(1 +0)=1.Now, let's check if it's a maximum or minimum.Looking at h'(x) around x=1:Since k(x) =x + 2 ln x -1, and k'(x)=1 +2/x. Since k(x) is increasing, before x=1, k(x) <0 and after x=1, k(x) >0. But since the numerator of h'(x) is x(x + 2 ln x -1). When x <1, x +2 ln x -1 <0, so numerator is positive (x>0, x +2 ln x -1 <0, so product x*(negative) is negative. Wait, but h'(x) numerator is x² +2x ln x -x. Wait, perhaps I made a mistake here.Wait, actually, the critical points occur when x +2 ln x -1=0. Since x=1 is the only solution, so when x <1, x +2 ln x -1 <0, and when x >1, x +2 ln x -1 >0. Therefore, the numerator x² +2x ln x -x can be written as x(x +2 ln x -1). So for x <1, x +2 ln x -1 <0, so the numerator is positive (since x>0 and x +2 ln x -1 <0, so x*(negative) is negative). Wait, but the numerator is x² +2x ln x -x. Let me compute the numerator at x approaching 0+:x² approaches 0, 2x ln x approaches 0 (since x ln x approaches 0 as x→0+), and -x approaches 0. So the numerator approaches 0. But near x=0, maybe the numerator is negative? Let's take x=0.1:Numerator = 0.01 + 2*0.1*ln(0.1) -0.1 ≈ 0.01 + 0.2*(-2.3026) -0.1 ≈ 0.01 -0.4605 -0.1 ≈ -0.5505 <0.At x=1: numerator=1 +0 -1=0.At x=2:Numerator=4 + 2*2*ln2 -2 ≈4 +4*0.6931 -2≈4 +2.7724 -2≈4.7724>0.Therefore, the numerator is negative for x in (0,1) and positive for x>1. Therefore, h'(x) is negative for x in (0,1) and positive for x>1. Therefore, h(x) is decreasing on (0,1) and increasing on (1, ∞). Therefore, h(x) has a minimum at x=1.Therefore, the graph of h(x) decreases from x=0 to x=1, reaching a minimum at x=1, then increases to infinity as x→∞.Thus, the function h(x)=x²/(x + ln x) has a minimum value at x=1, which is h(1)=1.Therefore, the equation m =x²/(x + ln x) will have:- No solution if m <1.- Exactly one solution when m=1.- Two solutions when m>1 (one in (0,1) and one in (1, ∞)).But wait, hold on. Wait, at x approaching 0+, h(x) approaches 0. At x=1, h(x)=1, and as x→∞, h(x)→∞. Since h(x) is decreasing on (0,1) from 0 to 1, and increasing from 1 to ∞ on (1, ∞). Therefore:- For m between 0 and 1, the equation m =h(x) has two solutions: one in (0,1) and one in (1, ∞).- For m=1, it has exactly one solution at x=1.- For m ≤0, no solution, but since m is a positive number, we don't consider m ≤0.Wait, but the problem states that when a=0, the equation m f(x)=x² has a unique real solution. So m must be such that the equation has exactly one solution. From the above, the only m where this is true is m=1, since for m>1, there are two solutions (but wait, no: when m>1, as x approaches infinity, h(x) tends to infinity, so for m>1, the equation h(x)=m would have two solutions? Wait, wait, let's check again.Wait, when m >1, since h(x) approaches infinity as x→∞ and h(x) has a minimum at x=1 of 1. Therefore, for m>1, there is exactly one solution in (1, ∞), since h(x) is increasing there, and since h(x) is decreasing on (0,1) from 0 to 1, so for m>1, the equation m=h(x) would have only one solution in (1, ∞). Wait, but at x=1, h(x)=1, so for m>1, we can have only one solution in (1, ∞) because h(x) is increasing from 1 to infinity as x goes from 1 to infinity.Wait, but for m>1, there is only one solution. For m=1, one solution at x=1. For m<1, two solutions. Wait, but the problem states that the equation has a unique real solution. So the answer should be m=1. However, the problem says "has a unique real solution", so if for m=1, it's unique, and for m>1, is there a unique solution?Wait, wait, no: when m>1, the equation h(x)=m would have only one solution in (1, ∞), since h(x) is increasing there and goes to infinity. But on (0,1), h(x) is decreasing from 0 to 1, so for m>1, h(x) never reaches m on (0,1). Therefore, for m≥1, there is exactly one solution (for m=1 at x=1, for m>1 at x>1). For m between 0 and1, two solutions.But the problem states that the equation has a unique real solution. Therefore, the values of m where there's exactly one solution are m≥1. Wait, but the problem says "the equation has a unique real solution, find the value of the positive number m". So it's possible that m can be any value greater than or equal to 1, but the problem says "find the value of the positive number m", implying there's a unique m. But this contradicts my analysis. Wait, maybe I made a mistake.Wait, let's think again. The problem says "the equation m f(x) =x² has a unique real solution". When a=0, f(x)=ln x +x. So m(ln x +x)=x². So m =x²/(ln x +x). So to have this equation hold for exactly one x>0, we need that the horizontal line y=m intersects the graph of h(x)=x²/(ln x +x) exactly once.From the previous analysis, h(x) is decreasing from (0,1) to 1, then increasing to infinity. So h(x) has a minimum at x=1 with h(1)=1. Therefore, if m=1, the line touches the graph at the minimum point x=1. For m>1, the line y=m intersects the graph once on (1, ∞). For m=1, once. For m<1, twice. But the problem says "has a unique real solution", so m≥1. However, the problem specifies "find the value of the positive number m", using "the value", which might suggest a single value. But maybe the answer is m=1, because if m>1, the equation still has a unique solution, but the problem might consider that m is such that the equation is tangent at x=1, which is the case only when m=1. Alternatively, perhaps my analysis is missing something.Wait, but let's check for m>1. If m>1, then h(x)=m has exactly one solution in (1, ∞), since h(x) is increasing there. So the equation m f(x)=x² would have one solution. However, the problem states "has a unique real solution". So technically, for all m≥1, the equation has exactly one real solution. But the problem says "find the value of the positive number m", implying perhaps only m=1? But that doesn't make sense because for m>1, the equation still has one solution. Maybe the problem is referring to the case where the equation has exactly one real solution in the entire domain, regardless of where it is. Wait, but in the case of m>1, the equation has one real solution. In the case m=1, it also has one real solution. So the range of m is [1, ∞). But the problem says "find the value of the positive number m". Maybe there's a misunderstanding here.Wait, perhaps I made a mistake. Let me re-examine the equation. The problem states "the equation m f(x) =x² has a unique real solution". So m is fixed, and the equation has exactly one x>0 such that m f(x)=x². So, if we fix m and look for x's, but m is a parameter. So, how can this equation have exactly one solution?If m is too small, the equation might have two solutions. If m is exactly such that the curve is tangent to the line, then only one solution. But in our case, h(x)=x²/(ln x +x). So, the equation m = h(x). If the horizontal line is tangent to h(x), then there's exactly one solution. But h(x) has a minimum at x=1, so the only point where the horizontal line is tangent is at x=1, which is m=1. For m>1, the horizontal line intersects h(x) once, but not tangentially. However, in calculus, when we have a strictly monotonic function after a certain point, intersecting once doesn't count as a tangent. So, in the problem statement, does "unique real solution" mean only one solution in total, which can occur either at the minimum or for m>1. But in the problem statement, it's possible that the equation has a unique solution regardless of where it is. The problem says "has a unique real solution", so whether m=1 or m>1, but in the problem statement, it's possible that m=1 is the only value where there's exactly one real solution, because for m>1, even though there's one solution, maybe the problem is considering multiple solutions but requiring only one. Wait, no.Wait, no. If m>1, there is exactly one solution (on the right side). If m=1, exactly one solution (at x=1). If m<1, two solutions. Therefore, for uniqueness, m≥1. But the problem says "find the value of the positive number m". If it's asking for all such m, then the answer is m≥1. But the problem says "the equation ... has a unique real solution, find the value of the positive number m". So, in Chinese math problems, sometimes "the equation has a unique solution" can sometimes imply that the equation is tangent, i.e., the case when the line is tangent to the curve, which occurs at m=1. Otherwise, for m>1, it's not tangent but still only one intersection. Depending on interpretation, but maybe the answer is m=1.But let's check with an example. Suppose m=2. Then the equation is 2(ln x +x)=x². Is there exactly one solution?Let’s check x=2: 2(ln2 +2)=2(0.6931 +2)=2*2.6931≈5.3862. x²=4. So 5.3862≠4.x=3: 2(ln3 +3)=2(1.0986 +3)=2*4.0986≈8.1972, x²=9. Close. 8.1972 <9.x=4: 2(ln4 +4)=2(1.3863 +4)=2*5.3863≈10.7726, x²=16. Still less.x=5: 2(ln5 +5)=2(1.6094 +5)=2*6.6094≈13.2188, x²=25. Still less.Wait, h(x)=x²/(ln x +x). As x approaches infinity, h(x) ~x²/x =x, which tends to infinity. So for m=2, there must be some x>1 where h(x)=2. Let's solve x²/(x +ln x)=2.Multiply both sides: x²=2x +2 ln x.Rearranged: x² -2x -2 ln x=0.Let’s check x=2: 4 -4 -2*0.6931≈0 -1.3862≈-1.3862 <0.x=3: 9 -6 -2*1.0986≈3 -2.1972≈0.8028 >0.So by intermediate value theorem, there is a solution between 2 and3. Hence, for m=2, there is a solution. Since h(x) is increasing for x>1, only one solution. Therefore, m=2 is valid. But according to the problem statement, if m=2 is acceptable, then the range of m is [1, ∞). But the problem says "find the value of the positive number m", implying a specific value. However, the problem states "has a unique real solution", which occurs for all m≥1. So the answer might be m≥1, but the problem says "find the value of the positive number m", which is strange. Maybe there's a miscalculation.Wait, perhaps I messed up the analysis. Let's verify again. If m=1, then the equation is x² =1*(ln x +x), so x² -x -ln x=0. At x=1, it's 1 -1 -0=0. So x=1 is a solution. Is there another solution?Let’s check x approaching 0: x² -x -ln x ≈0 -0 -(-infty)=+infty, so no solution there.For x>1: derivative of x² -x -ln x is 2x -1 -1/x. At x=1, it's 2 -1 -1=0. For x>1, 2x -1 -1/x >0, since 2x -1 >1 and 1/x <1. Therefore, the function x² -x -ln x is increasing for x>1, starting from 0 at x=1 and going to infinity. Therefore, only x=1 is the solution. For m=1, only one solution.For m>1, the equation x² =m(ln x +x). For x>1, as x increases, h(x)=x²/(ln x +x) increases to infinity. So for any m>1, there is exactly one x>1 where h(x)=m. Also, for x in (0,1), h(x) decreases from 0 to1, so for m>1, there is no solution in (0,1). So total solutions: one.For m=1, solution at x=1.For m<1, solutions in (0,1) and (1, ∞), so two solutions.But the problem says "the equation has a unique real solution". Therefore, the values of m are m≥1. But the problem asks for "the value of the positive number m", which might suggest a single value, but based on analysis, it's m≥1. However, perhaps there is a misunderstanding. Wait, maybe the equation is required to have a unique real solution regardless of x, but the problem states "has a unique real solution", so if there exists exactly one x>0 satisfying the equation, then m≥1. But the problem says "find the value of the positive number m", which is singular. This is confusing.Wait, going back to the problem statement:"When a=0, the equation mf(x)=x² has a unique real solution, find the value of the positive number m."So, perhaps "has a unique real solution" means that there is exactly one pair (m, x) that satisfies the equation? No, that doesn't make sense. Or maybe it's a uniqueness independent of x? No, the equation is in x, so for a given m, find x. So if for a given m, there is exactly one x>0 solving the equation, then m≥1. But the problem asks to "find the value of the positive number m", which is plural if there are multiple values. The wording is confusing. But maybe the problem is translated from Chinese, and sometimes in Chinese problems, even if there's a range, they might phrase it as "the value".Alternatively, maybe there's a mistake in the analysis. Let's consider m=1. The equation is x² = ln x +x. Let's check if there are other solutions.At x=1, 1=0 +1, holds. For x>1, x² grows faster than x +ln x, so x² -x -ln x is increasing. At x=2, 4 -2 -ln2≈2 -0.693≈1.307>0. So no solution beyond x=1 except for m>1.For x between 0 and1, take x=0.5: 0.25 -0.5 -ln0.5≈-0.25 -(-0.693)≈0.443>0. Wait, x=0.5: x²=0.25, ln x +x≈-0.693+0.5≈-0.193. So 0.25= m*(-0.193). If m is positive, then right side is negative, left side is positive. No solution. So perhaps for x in (0,1), ln x +x is negative when x is small, but for x approaching1, ln x +x approaches0 +1=1.Wait, wait, let's compute ln x +x for x in (0,1). Let’s pick x=0.5: ln0.5 +0.5≈-0.693 +0.5≈-0.193. At x=0.1: ln0.1 +0.1≈-2.302 +0.1≈-2.202. At x approaching 0+, lnx +x approaches -infty. At x=1, it's 0 +1=1. So the function ln x +x increases from -infty to1 as x goes from0 to1.So in (0,1), lnx +x is negative for x < some point and positive afterwards. Let's find where lnx +x=0.Let’s solve lnx +x=0. Let’s call this x=α.At x=0.5: ln0.5 +0.5≈-0.693 +0.5≈-0.193 <0.At x=0.6: ln0.6≈-0.5108, so -0.5108 +0.6≈0.0892>0.So the solution is between 0.5 and0.6.Let’s use Newton-Raphson to approximate.Let’s define k(x)=lnx +x.Find x where k(x)=0.Starting with x0=0.55.k(0.55)=ln0.55 +0.55≈-0.5978 +0.55≈-0.0478.k’(x)=1/x +1. At x=0.55, k’=1/0.55 +1≈1.818 +1=2.818.Next approximation: x1= x0 -k(x0)/k’(x0)=0.55 - (-0.0478)/2.818≈0.55 +0.017≈0.567.k(0.567)=ln0.567 +0.567≈-0.567 +0.567≈0. So x≈0.567 is where lnx +x=0.Therefore, for x in (0,α≈0.567), lnx +x <0, and for x in (α,1), lnx +x >0.Therefore, the equation m(lnx +x)=x² in x>0.If m>0:For x in (0,α), lnx +x <0, so left side m(lnx +x) is negative, right side x² is positive: no solution.For x in (α,1), lnx +x >0, so left side positive, right side positive. Possible solutions.For x in (1, ∞), both sides positive.Therefore, in the domain x>0, the equation m(lnx +x)=x² can have solutions only in (α, ∞).But we have to analyze it more carefully.Wait, for x in (α,1):m(lnx +x)=x².Since lnx +x is positive here, m must be positive. So possible solutions.But as earlier analysis, h(x)=x²/(lnx +x) is decreasing from x=α to x=1 (since h(x) is decreasing on (0,1)), with h(α)=x²/0 which tends to infinity (as x approaches α from the right, lnx +x approaches0 from positive side, so h(x) approaches infinity). At x=1, h(x)=1. So on (α,1), h(x) decreases from +infty to1.On (1, ∞), h(x) increases from1 to+infty.Therefore, the function h(x) has a minimum at x=1 of1, but between x=α and x=1, h(x) decreases from +infty to1. Therefore, for m>1, the equation h(x)=m has two solutions: one in (α,1) and one in (1, ∞). Wait, but earlier analysis said that h(x) is decreasing on (0,1). But (α,1) is a subset of (0,1). So for m>1, in (α,1), h(x)=m has one solution, and in (1, ∞), another solution. Hence, for m>1, two solutions. But earlier analysis conflicting.Wait, this is contradictory. Let me double-check.From previous steps:h(x) =x²/(lnx +x).For x in (α,1), lnx +x >0, so h(x) is defined.Earlier, when analyzing h(x), I considered the entire domain x>0, except where lnx +x=0, which is x=α. But in terms of critical points, we found that h(x) has a critical point at x=1, with h'(x) negative on (0,1) and positive on (1, ∞). Therefore, h(x) is decreasing on (0,1) from h(α+)=+infty to h(1)=1, then increasing on (1, ∞) to +infty.Therefore, for m>1, the equation h(x)=m has two solutions: one in (α,1) and one in (1, ∞). For m=1, one solution at x=1. For m<1, no solution.But that contradicts the previous conclusion where for m>1, h(x)=m has one solution. Where is the mistake?Wait, no. Wait, in the interval (α,1), h(x) is decreasing from +infty to1, and in (1, ∞), increasing from1 to+infty. Therefore, for m>1, the equation h(x)=m has two solutions: one in (α,1) and one in (1, ∞). For m=1, one solution at x=1. For m between1 and +infty, two solutions. For m<1, no solution.But this contradicts the initial analysis where h(x) tends to infinity as x approaches infinity. So for m>1, there's one solution in (1, ∞). Wait, no: if h(x) is increasing on (1, ∞) from1 to+infty, then for each m>1, there's exactly one solution in (1, ∞). But in (α,1), h(x) is decreasing from +infty to1, so for each m>1, there's also exactly one solution in (α,1). Therefore, for m>1, two solutions. For m=1, one solution. For m<1, no solution.But this contradicts previous analysis. What's the correct answer?Wait, let's take m=2. Then h(x)=2.Find x in (α,1) and x in (1, ∞) where x²/(lnx +x)=2.As earlier, x in (1, ∞): h(x) increases from1 to infinity, so there's exactly one solution here.x in (α,1): h(x) decreases from +infty to1, so for m=2, there's exactly one solution here as well.Therefore, m=2 would have two solutions. But earlier, when I checked x=3, the left side was 2*(ln3 +3)≈2*(1.0986 +3)=8.394， and x²=9. So equation 8.394=9? No, wait, the equation is x²=2*(lnx +x). So for x=3, 9=2*(1.0986 +3)=8.394? Not equal. So there must be a solution x>3 where x²=2*(lnx +x). But since h(x) ~x as x approaches infinity, the solution for m=2 would be x≈2*(lnx +x). Wait, this is confusing.Alternatively, maybe my previous analysis was wrong. Let's take m=2 and see if there are two solutions.Let’s plot the function h(x)=x²/(lnx +x).At x approaching α≈0.567 from the right, h(x) approaches +infty. At x=1, h(x)=1. At x=2, h(x)=4/(ln2 +2)≈4/(0.693 +2)≈4/2.693≈1.485. At x=3, h(x)=9/(ln3 +3)≈9/(1.0986 +3)≈9/4.0986≈2.195. At x=4, h(x)=16/(ln4 +4)≈16/(1.386 +4)≈16/5.386≈2.97. At x=5, h(x)=25/(ln5 +5)≈25/(1.609 +5)=25/6.609≈3.785. So as x increases, h(x) increases. Therefore, for m=2, the equation h(x)=2 would have a solution at x≈3. So one solution. But in (α,1), if we take x=0.6, lnx +x≈-0.5108 +0.6=0.0892, h(x)=0.36/0.0892≈4.03>2. So for m=2, in (α,1), there's another solution x where h(x)=2. Let's compute x=0.8:h(0.8)=0.64/(ln0.8 +0.8)=0.64/(-0.2231 +0.8)=0.64/0.5769≈1.11. So less than 2. At x=0.7:h(x)=0.49/(ln0.7 +0.7)=0.49/(-0.3567 +0.7)=0.49/0.3433≈1.43. Still less than2.x=0.55:h(x)=0.3025/(ln0.55 +0.55)=0.3025/(-0.5978 +0.55)=0.3025/(-0.0478)≈-6.33. Wait, but x=0.55 is less than α≈0.567, so lnx +x is negative, so h(x) is negative. But m=2>0, so no solution in x<α.At x=0.6:h(x)=0.36/(ln0.6 +0.6)=0.36/(-0.5108 +0.6)=0.36/0.0892≈4.03>2.At x=0.65:ln0.65≈-0.4308, ln0.65 +0.65≈-0.4308 +0.65≈0.2192.h(x)=0.4225/0.2192≈1.927<2.So between x=0.6 and x=0.65, h(x) decreases from4.03 to1.927. So there's a solution for m=2 in (0.6,0.65). Therefore, for m=2, there are two solutions: one in (0.6,0.65) and one in (3,4). Therefore, my initial analysis was wrong. So for m>1, there are two solutions, one in (α,1) and one in (1, ∞). For m=1, one solution. For m<1, no solution.Therefore, the equation m f(x)=x² has a unique real solution only when m=1. Therefore, the value of the positive number m is 1.Hence, the answer is m=1.But why the confusion earlier? Because I initially thought that for m>1, there's one solution, but in reality, because h(x) is decreasing from +infty to1 on (α,1) and increasing from1 to+infty on (1, ∞), for each m>1, there are two solutions. Hence, the only value of m where there's exactly one solution is m=1.This seems to be the correct conclusion. Therefore, the answer is m=1.Final Answer(1) boxed{0}(2) boxed{[frac{1}{2}, +infty)}(3) boxed{1}</think> Summary(1) When (a = 2), find the maximum value of the function (f(x)):Given (f(x) = ln x - x^2 + x), we find the critical points by taking the derivative:[ f'(x) = frac{1}{x} - 2x + 1 ]Setting (f'(x) = 0):[ frac{1}{x} - 2x + 1 = 0 implies 2x^2 - x - 1 = 0 ]Solving the quadratic equation, we get (x = 1) (since (x) must be positive). Using the second derivative test:[ f''(x) = -frac{1}{x^2} - 2 ]At (x = 1), (f''(1) = -3), indicating a local maximum. Evaluating (f(1)):[ f(1) = ln 1 - 1^2 + 1 = 0 ]Thus, the maximum value is (boxed{0}).(2) Find the range of (a) such that the slope (k) of the tangent line of (F(x)) is always (leq frac{1}{2}):Given (F(x) = ln x + frac{a}{x}), we need (F'(x) leq frac{1}{2}):[ F'(x) = frac{1}{x} - frac{a}{x^2} leq frac{1}{2} ]Rearranging:[ frac{1}{x} - frac{a}{x^2} leq frac{1}{2} implies frac{1}{2}x^2 - x + a geq 0 ]We need (a geq - frac{1}{2}x^2 + x). The maximum value of (g(x) = -frac{1}{2}x^2 + x) on (0 < x leq 3) is (frac{1}{2}) at (x = 1). Thus, (a geq frac{1}{2}), so the range is (boxed{[frac{1}{2}, +infty)}).(3) When (a = 0), find (m) such that (mf(x) = x^2) has a unique solution:Given (f(x) = ln x + x), the equation becomes (m(ln x + x) = x^2). We need (m = frac{x^2}{ln x + x}) to have a unique solution. Analyzing (h(x) = frac{x^2}{ln x + x}), we find it has a minimum at (x = 1) with (h(1) = 1). For uniqueness, (m) must be the minimum value, thus (m = boxed{1}).,<think>Okay, so I need to solve three parts of this problem. Let me start with part (1). When a = 2, the function becomes f(x) = ln x - (1/2)*2*x² + x. Simplifying that, the 2 and 1/2 cancel out, so it's ln x - x² + x. I need to find the maximum value of this function. To find the maximum, I should take the derivative and set it equal to zero. The derivative of ln x is 1/x, the derivative of -x² is -2x, and the derivative of x is 1. So f'(x) = 1/x - 2x + 1. Set this equal to zero: 1/x - 2x + 1 = 0. Let me solve for x. Multiply both sides by x to eliminate the denominator: 1 - 2x² + x = 0. Rearranging terms: -2x² + x + 1 = 0. Multiply through by -1 to make it standard: 2x² - x - 1 = 0. Now, solve the quadratic equation 2x² - x - 1 = 0. Using the quadratic formula: x = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4. So the solutions are x = (1 + 3)/4 = 1 and x = (1 - 3)/4 = -0.5. But since the original function is ln x, which is only defined for x > 0, we discard the negative solution. So critical point at x = 1.Now check if this is a maximum. Take the second derivative: f''(x). The first derivative was 1/x - 2x + 1. The derivative of 1/x is -1/x², derivative of -2x is -2, derivative of 1 is 0. So f''(x) = -1/x² - 2. At x = 1, f''(1) = -1 - 2 = -3, which is negative. Therefore, the function has a maximum at x = 1.Compute f(1): ln 1 - (1)^2 + 1 = 0 - 1 + 1 = 0. So the maximum value is 0. Wait, that seems straightforward. Let me double-check. The function at x=1 is 0. Then, as x approaches 0 from the right, ln x tends to negative infinity, so f(x) tends to negative infinity. As x approaches infinity, the -x² term dominates, so f(x) tends to negative infinity. So yes, the maximum at x=1 is indeed the global maximum. So part (1) answer is 0.Moving on to part (2). Here, F(x) is given as f(x) + (1/2)a x² - x + a/x. Let me first substitute f(x). Given f(x) = ln x - (1/2)a x² + x. So F(x) = [ln x - (1/2)a x² + x] + (1/2)a x² - x + a/x. Let's simplify this.The -(1/2)a x² and +(1/2)a x² cancel out. Similarly, +x and -x cancel out. So F(x) simplifies to ln x + a/x. So F(x) = ln x + a/x, defined for 0 < x ≤ 3. We need to ensure that the slope of the tangent line at any point on F(x) is ≤ 1/2. The slope is the derivative of F(x), so F'(x) must be ≤ 1/2 for all x in (0, 3]. Let's compute F'(x). The derivative of ln x is 1/x, and the derivative of a/x is -a/x². So F'(x) = 1/x - a/x². We need 1/x - a/x² ≤ 1/2 for all x in (0, 3]. Let's rearrange this inequality. Multiply both sides by x² (since x > 0, x² is positive, so inequality direction remains): x - a ≤ (1/2)x². So, x - (1/2)x² ≤ a. Let me write this as a ≥ x - (1/2)x² for all x in (0, 3]. Therefore, the real number a must be greater than or equal to the maximum value of the function g(x) = x - (1/2)x² on the interval (0, 3]. Therefore, we need to find the maximum of g(x) on (0, 3], and set a ≥ that maximum. Find the maximum of g(x) = x - (1/2)x². Take derivative: g'(x) = 1 - x. Set to zero: 1 - x = 0 => x = 1. Check endpoints x approaches 0 and x = 3. At x=1, g(1) = 1 - 1/2 = 1/2. At x approaching 0, g(x) approaches 0. At x=3, g(3) = 3 - (1/2)(9) = 3 - 4.5 = -1.5. So maximum is 1/2 at x=1. Therefore, a must be ≥ 1/2. Wait, but the problem says "the slope k of the tangent line at any point ... satisfies k ≤ 1/2 always holds". So our condition is F'(x) ≤ 1/2 for all x in (0,3]. Which converted to a ≥ g(x) for all x in (0,3], so a must be ≥ max g(x) = 1/2. Therefore, the range of a is [1/2, ∞). Wait, but let me check again. The inequality was x - (1/2)x² ≤ a, so a has to be greater or equal to the maximum of x - (1/2)x². Since the maximum is 1/2, then a ≥ 1/2. Therefore, the range of a is [1/2, ∞). Is that right? Let me check with x=1. If a=1/2, then F'(1) = 1/1 - (1/2)/1² = 1 - 1/2 = 1/2. So at x=1, the slope is exactly 1/2, which is allowed since the condition is k ≤ 1/2. For x near 0, F'(x) = 1/x - a/x². If a=1/2, then 1/x - (1/2)/x² = (2x - 1)/ (2x²). As x approaches 0, this tends to -infinity, but wait, if x approaches 0, numerator is 2x -1 approaches -1, denominator approaches 0, so the whole thing approaches -infinity. Which is less than 1/2. For x=3, F'(3) = 1/3 - a/9. If a=1/2, then 1/3 - (1/2)/9 = 1/3 - 1/18 = 6/18 - 1/18 = 5/18 ≈ 0.277 < 1/2. So when a=1/2, the maximum slope is 1/2 at x=1, and all others are less. If a is larger than 1/2, then F'(x) =1/x - a/x² will be even smaller. So yes, the condition holds for a ≥1/2. So part (2) answer is a ≥1/2, which in box is boxed{[frac{1}{2}, +infty)}.Now part (3). When a=0, the equation m f(x) = x² has a unique real solution. Find positive m. First, when a=0, f(x) = ln x - 0 + x = ln x + x. So the equation is m(ln x + x) = x². We need to find m > 0 such that this equation has exactly one real solution in x >0.So m(ln x + x) = x². Let me rearrange it as m = x² / (ln x + x). So for x >0, define h(x) = x² / (ln x + x). We need to find m such that h(x) = m has exactly one solution. So we need to find m where the horizontal line y=m touches the graph of h(x) exactly once. This usually happens at a extremum point (either maximum or minimum). But since h(x) is positive for x >0 (ln x +x: for x ≥1, ln x is non-negative, and x is positive; for 0 <x <1, ln x is negative, but x is positive. Let's check if ln x +x can be zero. For x approaching 0, ln x approaches -infty, x approaches 0, so ln x +x approaches -infty. At x=1, ln1 +1=1. So lnx +x increases from -infty to 1 as x goes from 0 to1, and then increases further for x>1. Wait, derivative of lnx +x is 1/x +1, which is always positive for x>0. So lnx +x is strictly increasing from -infty to +infty as x goes from 0 to +infty. Therefore, lnx +x =0 has exactly one solution, say at x=c, where 0 <c <1. Therefore, h(x) =x²/(lnx +x) is defined for x>0 except at x=c where denominator is zero. Wait, but since lnx +x is strictly increasing, crosses zero at x=c, so for x <c, lnx +x is negative, and for x>c, it's positive. So h(x) is positive for x >c, negative for x <c. Since m is a positive number, we are only concerned with x >c where h(x) is positive. So the equation h(x)=m for m>0 is equivalent to x²/(lnx +x)=m, x>c. We need this equation to have exactly one solution. So the graph of h(x) must touch the line y=m at exactly one point. This occurs when m is the minimum value of h(x) on x >c. Because h(x) tends to infinity as x approaches c from the right (denominator approaches 0+, numerator approaches c² >0), and as x approaches infinity, h(x) behaves like x² /x =x, which also approaches infinity. Therefore, h(x) has a minimum somewhere on x >c. So the minimum value is the m we need.Therefore, we need to find the minimum of h(x) =x²/(lnx +x) for x >c (which is x >c, but since lnx +x is increasing, we can consider x >0, but only x >c where h(x) is positive). However, since we need the minimum of h(x) for x >c, let's take the derivative of h(x) and set it to zero.Let h(x) =x²/(lnx +x). Let me compute h'(x). Using quotient rule:h'(x) = [2x(lnx +x) - x²(1/x +1)] / (lnx +x)^2.Simplify numerator:2x(lnx +x) -x²( (1 +x)/x )Wait, wait, (1/x +1) = (1 +x)/x. Therefore:Numerator = 2x(lnx +x) -x²*( (1 +x)/x ) = 2x(lnx +x) -x(1 +x) = 2x lnx + 2x² -x -x² = 2x lnx + x² -x.Therefore, h'(x) = [2x lnx +x² -x] / (lnx +x)^2.Set numerator equal to zero: 2x lnx +x² -x =0.Factor x: x(2 lnx +x -1)=0. Since x >0, we can divide by x: 2 lnx +x -1=0.So we need to solve 2 lnx +x -1=0.This is a transcendental equation and likely cannot be solved algebraically. So we need to find the root numerically or possibly find that the equation has only one solution. Let's analyze the function k(x) =2 lnx +x -1.Find where k(x)=0. Let's check behavior:As x approaches 0+, 2 lnx approaches -infty, x approaches 0, so k(x) approaches -infty.At x=1: k(1)=0 +1 -1=0. So x=1 is a solution.Wait, but that's interesting. At x=1, k(1)=0. So x=1 is a solution.Check if there are other solutions. Let's take derivative of k(x): k'(x)=2/x +1. Since x>0, k'(x) is always positive (2/x +1 >0). Therefore, k(x) is strictly increasing from -infty to +infty as x increases from 0 to infinity. Therefore, x=1 is the only solution. Therefore, the critical point is at x=1. So the minimum of h(x) occurs at x=1. Therefore, the minimum value is h(1)=1²/(ln1 +1)=1/(0 +1)=1. Therefore, m=1. Wait, so if m=1, the equation h(x)=1 has exactly one solution at x=1. Let me confirm. Since h(x) has a minimum at x=1, and h(x) approaches infinity as x approaches c+ and as x approaches infinity, so the horizontal line y=1 touches the graph at x=1 only. Hence, m=1 is the required value. But wait, let's check if x=1 is the only solution. Suppose m=1, then x²/(lnx +x)=1 => x² = lnx +x => x² -x -lnx =0. Let's check x=1: 1 -1 -0=0. So x=1 is a solution. Let me check if there's another solution. Let's consider the function p(x)=x² -x -lnx. Find its roots. At x approaching 0+, x² approaches 0, -x approaches 0, -lnx approaches +infty. So p(x) approaches +infty. At x=1, p(1)=0. For x>1, p(x)=x² -x -lnx. The derivative p’(x)=2x -1 -1/x. At x=1, p’(1)=2 -1 -1=0. For x>1, p’(x)=2x -1 -1/x. Let's check at x=2: 4 -1 -0.5=2.5>0. Therefore, p(x) is increasing for x>1. Since p(1)=0 and p(x) increasing for x>1, there are no roots for x>1. For x between 0 and1, let's check x=0.5: p(0.5)=0.25 -0.5 -ln0.5≈0.25 -0.5 -(-0.693)=0.25 -0.5 +0.693≈0.443>0. Wait, p(0.5) is positive. Wait, but as x approaches 0+, p(x) approaches +infty? Wait, x² approaches 0, -x approaches 0, -lnx approaches +infty. So p(x)=x² -x -lnx approaches +infty as x approaches 0+. Wait, no: x² and -x go to 0, -lnx goes to +infty, so overall p(x) approaches +infty. At x=0.5, p(x)=0.25 -0.5 -(-0.693)=0.25 -0.5 +0.693≈0.443>0. At x approaching 1 from the left, p(x)=x² -x -lnx. Let’s take x=0.9: p(0.9)=0.81 -0.9 -ln0.9≈0.81 -0.9 -(-0.105)=0.81 -0.9 +0.105≈0.015>0. At x=0.95: 0.9025 -0.95 -ln0.95≈0.9025 -0.95 -(-0.0513)=0.9025 -0.95 +0.0513≈0.0038>0. So p(x) approaches 0 from above as x approaches 1 from the left. Therefore, p(x) is always positive for x in (0,1) and x>1, and zero at x=1. Therefore, the equation p(x)=0 has only x=1 as solution. Therefore, m=1 is indeed the value where the equation has a unique real solution. Therefore, part (3) answer is 1.Final Answer(1) boxed{0}(2) boxed{[dfrac{1}{2}, +infty)}(3) boxed{1}</think>(1) When (a = 2), the function (f(x) = ln x - x^2 + x). To find the maximum value, we first compute the derivative:[f'(x) = frac{1}{x} - 2x + 1]Setting (f'(x) = 0) gives:[frac{1}{x} - 2x + 1 = 0 implies 1 - 2x^2 + x = 0 implies 2x^2 - x - 1 = 0]Solving the quadratic equation, we get (x = 1) (since (x = -0.5) is discarded). The second derivative (f''(x) = -frac{1}{x^2} - 2) is negative at (x = 1), confirming a maximum. Evaluating (f(1)) gives:[f(1) = ln 1 - 1 + 1 = 0]Thus, the maximum value is (boxed{0}).(2) Given (F(x) = ln x + frac{a}{x}), we need (F'(x) leq frac{1}{2}) for all (0 < x leq 3). The derivative is:[F'(x) = frac{1}{x} - frac{a}{x^2}]Rearranging the inequality:[frac{1}{x} - frac{a}{x^2} leq frac{1}{2} implies x - a leq frac{1}{2}x^2 implies a geq x - frac{1}{2}x^2]Maximizing (g(x) = x - frac{1}{2}x^2) on (0 < x leq 3), we find the maximum at (x = 1) with (g(1) = frac{1}{2}). Thus, (a geq frac{1}{2}). The range of (a) is (boxed{[dfrac{1}{2}, +infty)}).(3) When (a = 0), the equation (mf(x) = x^2) becomes (m(ln x + x) = x^2). Rearranging gives:[m = frac{x^2}{ln x + x}]We need (m) such that this equation has a unique solution. The function (h(x) = frac{x^2}{ln x + x}) has a minimum at (x = 1) (found by solving (2 ln x + x - 1 = 0)). Evaluating (h(1)) gives:[h(1) = frac{1^2}{ln 1 + 1} = 1]Thus, the value of (m) is (boxed{1}).]

🔑:Solution:(1) When a=2, f(x)=ln x+x-x^{2},Then f'(x)=frac{1}{x}+1-2x,Let f'(x)=0, solving gives x=1 or x=-frac{1}{2} (discard this solution).When xin(0,1), f'(x) > 0,which means f(x) is monotonically increasing in this interval,When xin(1,+infty), f'(x) < 0,which means f(x) is monotonically decreasing in this interval,Therefore, the maximum value of f(x) is f(1)=0.boxed{0}(2) From the problem, we have F(x)=ln x+frac{a}{x} (0 < xleqslant 3),Then k=F'(x_{0})=frac{1}{x_{0}}-frac{a}{x_{0}^{2}} (0 < x_0leqslant 3).Given kleqslant frac{1}{2} always holds,we get ageqslant x_{0}-frac{1}{2}x_{0}^{2}=-frac{1}{2}(x_{0}-1)^{2}+frac{1}{2} always holds.Since -frac{1}{2}(x_{0}-1)^{2}+frac{1}{2}leqslant frac{1}{2}, equality holds only when x_0=1,Thus, ageqslant frac{1}{2}.Hence, the range of the real number a is boxed{[frac{1}{2},+infty)}.(3) When a=0, the equation mf(x)=x^2,which is x^2-mx-mln x=0.Let g(x)=x^{2}-mx-mln x,Then g'(x)=2x-m-frac{m}{x}.Let g'(x)=0, solving gives x_{1}=frac{m-sqrt{m^{2}+8m}}{4} (discard this solution), x_{2}=frac{m+sqrt{m^{2}+8m}}{4}.It is known that g(x) is monotonically decreasing in the interval (0,x_{2}) and monotonically increasing in the interval (x_{2},+infty), with the minimum value at g(x_{2}).Since the equation mf(x)=x^{2} has a unique real solution, i.e., g(x) has a unique zero point, thus g(x_{2})=0.From begin{cases} g'(x_{2})=0, g(x_{2})=0, end{cases} we get x_{2}+2ln x_{2}-1=0.Let h(x)=x+2ln x-1,Since h(x) is monotonically increasing in the interval and h(1)=0,Thus, x_{2}=1,Therefore, m=1. The value of the positive number m is boxed{1}.