🧠:[<think>Okay, so I need to find the cosine of the angle between the vectors (a + b) and (a - b), given that the angle between vectors a and b is π/3, the magnitude of a is 1, and the magnitude of b is 2. Let me think about how to approach this.First, I remember that the cosine of the angle between two vectors can be found using the dot product formula. Specifically, for two vectors u and v, the cosine of the angle θ between them is given by:cosθ = (u · v) / (|u| |v|)So, I need to compute the dot product of (a + b) and (a - b), and then divide that by the product of their magnitudes. Let me break this down step by step.First, let's compute the dot product (a + b) · (a - b). Expanding this out using the distributive property of the dot product:(a + b) · (a - b) = a · a - a · b + b · a - b · bBut I know that the dot product is commutative, so a · b = b · a. Therefore, this simplifies to:= a · a - a · b + a · b - b · bWait, the -a · b and +a · b terms cancel each other out, right? So that leaves us with:= a · a - b · bBut a · a is the magnitude of a squared, which is |a|², and similarly, b · b is |b|². Given that |a| = 1 and |b| = 2, this becomes:= |a|² - |b|² = 1² - 2² = 1 - 4 = -3So the dot product of (a + b) and (a - b) is -3. Hmm, interesting. So that's the numerator in our cosine formula.Now, I need to find the magnitudes of (a + b) and (a - b) to get the denominator. Let's start with |a + b|. The magnitude of a vector is given by the square root of the dot product with itself. So:|a + b|² = (a + b) · (a + b) = a · a + 2a · b + b · bAgain, a · a is |a|² = 1, and b · b is |b|² = 4. The middle term is 2a · b. We know the angle between a and b is π/3, so the dot product a · b is |a||b|cos(π/3) = 1*2*(1/2) = 1. Therefore:|a + b|² = 1 + 2*1 + 4 = 1 + 2 + 4 = 7So |a + b| = sqrt(7).Similarly, let's compute |a - b|. Using the same method:|a - b|² = (a - b) · (a - b) = a · a - 2a · b + b · bPlugging in the known values:= 1 - 2*1 + 4 = 1 - 2 + 4 = 3So |a - b| = sqrt(3).Therefore, the denominator in the cosine formula is sqrt(7) * sqrt(3) = sqrt(21).Putting it all together, the cosine of the angle θ between (a + b) and (a - b) is:cosθ = ( -3 ) / ( sqrt(21) )But simplifying this, we can write it as -3 / sqrt(21). However, rationalizing the denominator, we multiply numerator and denominator by sqrt(21):= (-3 * sqrt(21)) / (sqrt(21) * sqrt(21)) = (-3 sqrt(21)) / 21 = (- sqrt(21)) / 7So the cosine value is -sqrt(21)/7. Looking at the options given:A: sqrt(21)/7B: 1/7C: -1/7D: -sqrt(21)/7Therefore, the correct answer is option D.Wait, just to double-check, maybe I made a mistake in the calculation. Let me verify each step again.First, the dot product (a + b)·(a - b) = |a|² - |b|² = 1 - 4 = -3. That seems correct.Magnitude of (a + b): sqrt(1 + 4 + 2*1*2*cos(π/3)) = sqrt(1 + 4 + 2*1) = sqrt(7). Wait, actually, hold on. Wait, when you compute |a + b|², it's |a|² + |b|² + 2|a||b|cosθ. The angle θ between a and b is π/3, so cosθ is 1/2. Therefore:|a + b|² = 1² + 2² + 2*1*2*(1/2) = 1 + 4 + 2 = 7. Yes, that's correct. So sqrt(7) is correct.Similarly, |a - b|² is |a|² + |b|² - 2|a||b|cosθ. So that's 1 + 4 - 2*1*2*(1/2) = 5 - 2 = 3. Therefore sqrt(3) is correct.So the denominator is sqrt(7)*sqrt(3) = sqrt(21). Then the cosine is -3/sqrt(21). Rationalized, that's -sqrt(21)/7, which is option D. So yes, seems correct.Alternatively, maybe the answer is C? Wait, but -1/7 is not equal to -sqrt(21)/7. Let's see. sqrt(21) is approximately 4.583, so sqrt(21)/7 is roughly 0.654, so -sqrt(21)/7 is approximately -0.654. But -1/7 is about -0.142. So clearly different.Therefore, the answer is D.But just to be thorough, let me check once more. Maybe there's a mistake in the dot product expansion?Wait, (a + b)·(a - b) = a·a - a·b + b·a - b·b. Then, a·a is |a|², -a·b + b·a is -a·b + a·b (since dot product is commutative), so they cancel, leaving |a|² - |b|². That's correct. Therefore, the dot product is indeed 1 - 4 = -3.Magnitudes calculated correctly. So yes, everything checks out. The answer is D: -sqrt(21)/7.Final Answerboxed{D}</think>Given the angle between the plane vectors (overrightarrow{a}) and (overrightarrow{b}) is (frac{pi}{3}), with (|overrightarrow{a}| = 1) and (|overrightarrow{b}| = 2), we need to find the cosine value of the angle between (overrightarrow{a} + overrightarrow{b}) and (overrightarrow{a} - overrightarrow{b}).First, we compute the dot product of (overrightarrow{a} + overrightarrow{b}) and (overrightarrow{a} - overrightarrow{b}):[(overrightarrow{a} + overrightarrow{b}) cdot (overrightarrow{a} - overrightarrow{b}) = overrightarrow{a} cdot overrightarrow{a} - overrightarrow{a} cdot overrightarrow{b} + overrightarrow{b} cdot overrightarrow{a} - overrightarrow{b} cdot overrightarrow{b}]Since the dot product is commutative, this simplifies to:[|overrightarrow{a}|^2 - |overrightarrow{b}|^2 = 1^2 - 2^2 = 1 - 4 = -3]Next, we find the magnitudes of (overrightarrow{a} + overrightarrow{b}) and (overrightarrow{a} - overrightarrow{b}). The magnitude of (overrightarrow{a} + overrightarrow{b}) is calculated as:[|overrightarrow{a} + overrightarrow{b}|^2 = |overrightarrow{a}|^2 + |overrightarrow{b}|^2 + 2|overrightarrow{a}||overrightarrow{b}|cosleft(frac{pi}{3}right) = 1 + 4 + 2 cdot 1 cdot 2 cdot frac{1}{2} = 7]Thus, (|overrightarrow{a} + overrightarrow{b}| = sqrt{7}).Similarly, the magnitude of (overrightarrow{a} - overrightarrow{b}) is:[|overrightarrow{a} - overrightarrow{b}|^2 = |overrightarrow{a}|^2 + |overrightarrow{b}|^2 - 2|overrightarrow{a}||overrightarrow{b}|cosleft(frac{pi}{3}right) = 1 + 4 - 2 cdot 1 cdot 2 cdot frac{1}{2} = 3]Thus, (|overrightarrow{a} - overrightarrow{b}| = sqrt{3}).The cosine of the angle between (overrightarrow{a} + overrightarrow{b}) and (overrightarrow{a} - overrightarrow{b}) is:[costheta = frac{(overrightarrow{a} + overrightarrow{b}) cdot (overrightarrow{a} - overrightarrow{b})}{|overrightarrow{a} + overrightarrow{b}| |overrightarrow{a} - overrightarrow{b}|} = frac{-3}{sqrt{7} cdot sqrt{3}} = frac{-3}{sqrt{21}} = frac{-sqrt{21}}{7}]Thus, the answer is (boxed{D}).,<think>Okay, let's see. The problem is about finding the cosine of the angle between two vectors, which are (a + b) and (a - b). The given information is that the angle between vectors a and b is π/3, the magnitude of vector a is 1, and the magnitude of vector b is 2. The answer choices are A to D. Hmm, okay.First, I remember that to find the cosine of the angle between two vectors, we can use the dot product formula. The formula is: cosθ = (u · v) / (|u| |v|)So, the cosine of the angle between vectors u and v is equal to their dot product divided by the product of their magnitudes. That makes sense. So, in this case, u is (a + b) and v is (a - b). Therefore, I need to compute the dot product of (a + b) and (a - b), and then divide that by the product of their magnitudes.Let me write that down:cosθ = [(a + b) · (a - b)] / [|a + b| |a - b|]Okay, now I need to compute both the numerator and the denominator.Starting with the numerator: (a + b) · (a - b). I recall that the dot product distributes over addition and subtraction, so this should expand similarly to algebra. Let me expand it:(a + b) · (a - b) = a · a - a · b + b · a - b · bWait, but dot product is commutative, so a · b = b · a. So, simplifying this:= a · a - a · b + a · b - b · bThe middle terms (-a · b and +a · b) cancel each other out. So that leaves:= a · a - b · bWhich is the same as |a|² - |b|². Because the dot product of a vector with itself is the square of its magnitude.Given that |a| = 1 and |b| = 2, so |a|² = 1² = 1, and |b|² = 2² = 4. Therefore, the numerator is 1 - 4 = -3.Okay, so numerator is -3. That's straightforward.Now, the denominator is the product of |a + b| and |a - b|. Hmm, so I need to compute the magnitudes of (a + b) and (a - b) first.To find the magnitude of a vector, we can use |v| = sqrt(v · v). So, let's compute |a + b| and |a - b|.Starting with |a + b|:|a + b|² = (a + b) · (a + b) = a · a + 2a · b + b · bAgain, using the fact that the dot product is distributive and commutative. Then:= |a|² + 2(a · b) + |b|²We know |a| = 1, |b| = 2, and the angle between a and b is π/3, so the dot product a · b can be calculated as |a||b|cos(π/3).So, a · b = (1)(2)(cos π/3) = 2*(1/2) = 1.Therefore, substituting back into |a + b|²:= 1² + 2*(1) + 2² = 1 + 2 + 4 = 7So, |a + b| = sqrt(7)Similarly, compute |a - b|:|a - b|² = (a - b) · (a - b) = a · a - 2a · b + b · b= |a|² - 2(a · b) + |b|²Again, substituting the known values:= 1² - 2*(1) + 2² = 1 - 2 + 4 = 3Therefore, |a - b| = sqrt(3)So, the denominator is sqrt(7) * sqrt(3) = sqrt(21)Putting numerator and denominator together, the cosine of the angle is:cosθ = -3 / sqrt(21)Hmm, but the answer choices are given as fractions with denominators 7 or sqrt(21)/7. Let me rationalize the denominator here.Multiplying numerator and denominator by sqrt(21):cosθ = (-3 * sqrt(21)) / (sqrt(21) * sqrt(21)) = (-3 sqrt(21)) / 21 = (- sqrt(21)) / 7So, simplifying, that's -sqrt(21)/7. Looking at the options, D is -sqrt(21)/7. So the answer should be D.Wait, but let me double-check my calculations because that's a negative cosine value. The angle between vectors is between 0 and π, so the cosine can be negative if the angle is greater than π/2. Let me confirm if my steps are correct.Starting with the dot product of (a + b) and (a - b) leading to |a|² - |b|². Since |a| is 1 and |b| is 2, that gives 1 - 4 = -3. That's correct.For |a + b|, sqrt(7) is correct, as 1 + 2*1 + 4 = 7, and sqrt(7). For |a - b|, sqrt(3) is correct, as 1 - 2 + 4 = 3, sqrt(3). Therefore, denominator is sqrt(21). So the cosine is -3/sqrt(21), which simplifies to -sqrt(21)/7. Yes, that's correct.Looking back at the answer choices:A: sqrt(21)/7B: 1/7C: -1/7D: -sqrt(21)/7Therefore, D is the correct answer. However, let me check if there's any mistake in the sign.Wait, when we calculated the dot product (a + b) · (a - b) = |a|² - |b|² = 1 - 4 = -3, which is negative. Therefore, the cosine of the angle is negative, so the angle is obtuse. But the answer options include both C and D as negative. However, according to the calculation, it's -sqrt(21)/7, which is option D. So that's correct.Wait, but let me check if when computing the magnitude of (a - b), perhaps I made a mistake. Let me recompute |a - b|²:|a - b|² = |a|² - 2a·b + |b|² = 1 - 2(1) + 4 = 1 - 2 + 4 = 3. So |a - b| = sqrt(3). That's correct.Similarly, |a + b|² = 1 + 2(1) + 4 = 7, so sqrt(7). Correct.So denominator is sqrt(7)*sqrt(3)=sqrt(21), numerator is -3.Therefore, cosθ = -3 / sqrt(21) = (-3/sqrt(21)) * (sqrt(21)/sqrt(21)) = (-3 sqrt(21))/21 = (-sqrt(21))/7. So yes, D is correct.Alternatively, maybe the problem expects a different approach? Let me think.Alternatively, if I consider the angle between (a + b) and (a - b), maybe there's another way to compute it. But no, the method using dot product is standard. Let's see if there's any alternative formula.Alternatively, if I can compute the angle between the two vectors (a + b) and (a - b), perhaps using the law of cosines in the parallelogram formed by a and b. But probably the method we did before is the most straightforward.Alternatively, check the answer options. Since we have a negative cosine, the answer must be either C or D. The numerator was -3, the denominator sqrt(21). So -3/sqrt(21) is equivalent to -sqrt(21)/7. Since sqrt(21)/7 is equal to sqrt(21)/7, and 3/sqrt(21) simplifies to sqrt(21)/7. Let me verify:sqrt(21)/7 = sqrt(21)/7. Let's compute 3/sqrt(21). Multiply numerator and denominator by sqrt(21):3/sqrt(21) = 3 sqrt(21)/21 = sqrt(21)/7. Therefore, 3/sqrt(21) is sqrt(21)/7, so -3/sqrt(21) is -sqrt(21)/7. Therefore, yes, that's correct.So, the answer is D. So I think that is correct. Therefore, I would select option D.Final Answerboxed{D}</think>To find the cosine value of the angle between the vectors (overrightarrow{a} + overrightarrow{b}) and (overrightarrow{a} - overrightarrow{b}), we use the dot product formula:[costheta = frac{(overrightarrow{a} + overrightarrow{b}) cdot (overrightarrow{a} - overrightarrow{b})}{|overrightarrow{a} + overrightarrow{b}| |overrightarrow{a} - overrightarrow{b}|}]First, we calculate the numerator:[(overrightarrow{a} + overrightarrow{b}) cdot (overrightarrow{a} - overrightarrow{b}) = overrightarrow{a} cdot overrightarrow{a} - overrightarrow{a} cdot overrightarrow{b} + overrightarrow{b} cdot overrightarrow{a} - overrightarrow{b} cdot overrightarrow{b}]Since (overrightarrow{a} cdot overrightarrow{b} = overrightarrow{b} cdot overrightarrow{a}), this simplifies to:[|overrightarrow{a}|^2 - |overrightarrow{b}|^2]Given (|overrightarrow{a}| = 1) and (|overrightarrow{b}| = 2), we have:[1^2 - 2^2 = 1 - 4 = -3]Next, we calculate the magnitudes of (overrightarrow{a} + overrightarrow{b}) and (overrightarrow{a} - overrightarrow{b}). For (|overrightarrow{a} + overrightarrow{b}|):[|overrightarrow{a} + overrightarrow{b}|^2 = |overrightarrow{a}|^2 + 2overrightarrow{a} cdot overrightarrow{b} + |overrightarrow{b}|^2]Using (overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}||overrightarrow{b}|cosfrac{pi}{3} = 1 cdot 2 cdot frac{1}{2} = 1), we get:[1 + 2 cdot 1 + 4 = 7 implies |overrightarrow{a} + overrightarrow{b}| = sqrt{7}]For (|overrightarrow{a} - overrightarrow{b}|):[|overrightarrow{a} - overrightarrow{b}|^2 = |overrightarrow{a}|^2 - 2overrightarrow{a} cdot overrightarrow{b} + |overrightarrow{b}|^2][1 - 2 cdot 1 + 4 = 3 implies |overrightarrow{a} - overrightarrow{b}| = sqrt{3}]Thus, the denominator is:[sqrt{7} cdot sqrt{3} = sqrt{21}]Putting it all together, the cosine value is:[costheta = frac{-3}{sqrt{21}} = frac{-3sqrt{21}}{21} = frac{-sqrt{21}}{7}]The answer is (boxed{D}).]

🔑:Given that overrightarrow{a}cdot overrightarrow{b}=1×2×cos dfrac {π}{3}=1,We know that ( overrightarrow{a}+ overrightarrow{b})^{2}= overrightarrow{a}^{2}+2 overrightarrow{a}cdot overrightarrow{b}+ overrightarrow{b}^{2}=1+2+4=7, and ( overrightarrow{a}- overrightarrow{b})^{2}= overrightarrow{a}^{2}-2 overrightarrow{a}cdot overrightarrow{b}+ overrightarrow{b}^{2}=1-2+4=3,Hence, | overrightarrow{a}+ overrightarrow{b}|= sqrt {7}, and | overrightarrow{a}- overrightarrow{b}|= sqrt {3},Also, ( overrightarrow{a}+ overrightarrow{b})( overrightarrow{a}- overrightarrow{b})= overrightarrow{a}^{2}- overrightarrow{b}^{2}=-3. Let the angle between overrightarrow{a}+ overrightarrow{b} and overrightarrow{a}- overrightarrow{b} be θ,Thus, cos θ= dfrac {( overrightarrow{a}+ overrightarrow{b})( overrightarrow{a}- overrightarrow{b})}{| overrightarrow{a}+ overrightarrow{b}|| overrightarrow{a}- overrightarrow{b}|}= dfrac {-3}{ sqrt {3}cdot sqrt {7}}=- dfrac { sqrt {21}}{7}.Therefore, the answer is boxed{D}.To solve this problem, we need to calculate ( overrightarrow{a}+ overrightarrow{b})( overrightarrow{a}- overrightarrow{b}), | overrightarrow{a}+ overrightarrow{b}|, and | overrightarrow{a}- overrightarrow{b}|, and then substitute these values into the formula for the cosine of an angle. This problem tests your understanding of the dot product operation for plane vectors and is of moderate difficulty.

🧠:[<think>Okay, so I need to solve this problem about an ellipse. There are two parts. Let me start with part (1). The problem says: Ellipse E has the equation x²/a² + y²/b² = 1, where a > b > 0. It passes through the point A(0, -1), and its eccentricity is √2/2. We need to find the equation of the ellipse E.Alright, so first, since the ellipse passes through point A(0, -1), we can substitute x=0 and y=-1 into the ellipse equation to find a relation between a and b. Let me do that.Substituting (0, -1) into the equation:0²/a² + (-1)²/b² = 1Which simplifies to 0 + 1/b² = 1, so 1/b² = 1. Therefore, b² = 1, so b = 1. Since b > 0, we take the positive root. So, b = 1.Now, we also know the eccentricity e = √2/2. For an ellipse, the eccentricity is given by e = c/a, where c is the distance from the center to each focus, and c = √(a² - b²). So, we can write:e = √(a² - b²)/aWe know e = √2/2 and b² = 1. Let's substitute those values in:√(a² - 1)/a = √2/2Let me square both sides to eliminate the square roots:(a² - 1)/a² = ( (√2)/2 )² = (2)/4 = 1/2So:(a² - 1)/a² = 1/2Multiply both sides by a²:a² - 1 = (1/2)a²Subtract (1/2)a² from both sides:a² - 1 - (1/2)a² = 0 => (1/2)a² -1 = 0So, (1/2)a² = 1 => a² = 2. Hence, a = √2, since a > 0.Therefore, the equation of the ellipse is x²/(√2)² + y²/1² = 1, which simplifies to x²/2 + y² = 1.Wait, let me check that again. If a² = 2, then yes, so the equation is x²/2 + y² = 1. That seems right. Let me verify with the given eccentricity.Given a² = 2 and b² = 1, then c² = a² - b² = 2 - 1 = 1, so c = 1. Then e = c/a = 1/√2 = √2/2, which matches the given eccentricity. Good. Also, since a > b, √2 > 1, which is true. So part (1) is solved. The equation is x²/2 + y² = 1.Now, moving on to part (2). The problem states: A line passing through point (1, 1) with slope k intersects ellipse E at two distinct points P and Q (both different from point A). Determine whether the sum of the slopes of lines AP and AQ is a constant. If it is a constant, find its value; if not, explain why.Hmm. Let's parse this. We have a line with slope k passing through (1,1). This line intersects the ellipse E (which we found to be x²/2 + y² = 1) at points P and Q. Both P and Q are different from point A, which is (0, -1). Then, we need to find the slopes of lines AP and AQ and see if their sum is a constant, regardless of k, or if it depends on k.First, let's think about how to approach this. The line through (1,1) with slope k can be written as y - 1 = k(x - 1). So, the equation of the line is y = kx - k + 1.We can find the points of intersection between this line and the ellipse by substituting y from the line equation into the ellipse equation. Then, solve for x. The solutions will be the x-coordinates of points P and Q. Then, we can find the corresponding y-coordinates. Then, for each point P and Q, compute the slopes of AP and AQ, then sum them and see if the result is a constant.Alternatively, perhaps there is a more elegant way using properties of conic sections or Vieta's formulas, without having to compute the coordinates explicitly. Let me consider both approaches.First, let's try substituting the line into the ellipse equation.Given ellipse: x²/2 + y² = 1Substitute y = kx - k + 1 into the ellipse equation:x²/2 + (kx - k + 1)^2 = 1Let me expand this:x²/2 + [k^2x^2 - 2k(k - 1)x + (k - 1)^2] = 1Multiply out the terms:x²/2 + k²x² - 2k(k - 1)x + (k - 1)^2 = 1Combine like terms:(1/2 + k²)x² - 2k(k - 1)x + (k - 1)^2 - 1 = 0Let me compute each term step by step.First coefficient of x²: 1/2 + k²Coefficient of x: -2k(k - 1) = -2k² + 2kConstant term: (k - 1)^2 - 1 = k² - 2k + 1 - 1 = k² - 2kSo the quadratic equation in x is:(1/2 + k²)x² + (-2k² + 2k)x + (k² - 2k) = 0Let me double-check that.Original substitution:y = kx - k + 1, so y = k(x - 1) + 1.Substitute into x²/2 + y² = 1:x²/2 + [k(x - 1) + 1]^2 = 1Expanding [k(x - 1) + 1]^2:= [k(x - 1)]^2 + 2*k(x - 1)*1 + 1^2= k²(x - 1)^2 + 2k(x - 1) + 1= k²(x² - 2x + 1) + 2k(x - 1) + 1= k²x² - 2k²x + k² + 2kx - 2k + 1So adding x²/2:x²/2 + k²x² - 2k²x + k² + 2kx - 2k + 1 = 1Combine terms:x²/2 + k²x² becomes (1/2 + k²)x²Then, the x terms: -2k²x + 2kx = (-2k² + 2k)xConstants: k² - 2k + 1So the equation is:(1/2 + k²)x² + (-2k² + 2k)x + (k² - 2k + 1) = 1Subtract 1 from both sides:(1/2 + k²)x² + (-2k² + 2k)x + (k² - 2k + 1 - 1) = 0Simplify the constant term:k² - 2k + 1 - 1 = k² - 2kHence, same as before: (1/2 + k²)x² + (-2k² + 2k)x + (k² - 2k) = 0So this quadratic in x will have two solutions, corresponding to points P and Q. Let's denote the roots as x1 and x2. Then, for each root, the corresponding y-coordinate is y1 = kx1 - k + 1 and y2 = kx2 - k + 1.Then, points P and Q are (x1, y1) and (x2, y2). We need to find the slopes of lines AP and AQ. Point A is (0, -1). So, the slope of line AP is (y1 - (-1))/(x1 - 0) = (y1 + 1)/x1. Similarly, slope of AQ is (y2 + 1)/x2.Therefore, the sum of the slopes is (y1 + 1)/x1 + (y2 + 1)/x2.We need to compute this sum and see if it's a constant.Alternatively, maybe we can use Vieta's formulas here. Let's note that x1 and x2 are roots of the quadratic equation we derived. Let's recall that for a quadratic equation ax² + bx + c = 0, the sum of roots is -b/a and the product is c/a.So, let's compute sum x1 + x2 and product x1 x2.In our equation:a = (1/2 + k²)b = (-2k² + 2k)c = (k² - 2k)Therefore,Sum of roots x1 + x2 = -b/a = [2k² - 2k]/(1/2 + k²)Product of roots x1 x2 = c/a = (k² - 2k)/(1/2 + k²)Now, let's compute y1 + 1 and y2 + 1. Since y1 = kx1 - k + 1, then y1 + 1 = kx1 - k + 1 + 1 = kx1 - k + 2 = k(x1 - 1) + 2. Similarly, y2 + 1 = k(x2 - 1) + 2.Therefore, the slopes are (y1 + 1)/x1 = [k(x1 - 1) + 2]/x1 = k - k/x1 + 2/x1. Similarly, (y2 + 1)/x2 = k - k/x2 + 2/x2.Therefore, the sum of the slopes is:[k - k/x1 + 2/x1] + [k - k/x2 + 2/x2] = 2k - k(1/x1 + 1/x2) + 2(1/x1 + 1/x2)Factor out (1/x1 + 1/x2):= 2k + [ -k + 2 ]*(1/x1 + 1/x2 )Now, let's compute (1/x1 + 1/x2). This is (x1 + x2)/(x1 x2). From Vieta's, we have x1 + x2 = [2k² - 2k]/(1/2 + k²) and x1 x2 = (k² - 2k)/(1/2 + k²). Therefore, (1/x1 + 1/x2) = [ (2k² - 2k)/(1/2 + k²) ] / [ (k² - 2k)/(1/2 + k²) ] = (2k² - 2k)/(k² - 2k) ) = [2k(k - 1)]/[k(k - 2)] = [2(k - 1)]/(k - 2)Wait, let's check that:Numerator: 2k² - 2k = 2k(k - 1)Denominator: k² - 2k = k(k - 2)So (2k(k - 1))/(k(k - 2)) ) = 2(k - 1)/(k - 2)Yes, simplifies to 2(k - 1)/(k - 2)Therefore, (1/x1 + 1/x2 ) = 2(k - 1)/(k - 2)Therefore, going back to the sum of slopes:Sum = 2k + [ -k + 2 ] * [ 2(k - 1)/(k - 2) ]Let me compute [ -k + 2 ] as (2 - k)So Sum = 2k + (2 - k) * [ 2(k - 1)/(k - 2) ]Note that (2 - k) = -(k - 2). So we can write:Sum = 2k - (k - 2) * [ 2(k - 1)/(k - 2) ]Simplify:(k - 2) cancels with denominator (k - 2):Sum = 2k - 2(k - 1) = 2k - 2k + 2 = 2Wow, that's nice. So regardless of k, the sum is 2. Therefore, the sum of the slopes of lines AP and AQ is a constant, which is 2.Wait, let me check that step again where we canceled (k - 2). Let me do that more carefully:Sum = 2k + (2 - k) * [ 2(k - 1)/(k - 2) ]Factor (2 - k) as -(k - 2):Sum = 2k - (k - 2) * [ 2(k - 1)/(k - 2) ]Now, the (k - 2) in the numerator cancels with the denominator:Sum = 2k - 2(k - 1)Then, distribute the 2:Sum = 2k - 2k + 2 = 0 + 2 = 2Yes, that's correct. The k terms cancel, leaving 2. Therefore, the sum of the slopes is indeed 2, which is a constant. Therefore, the answer is yes, it's a constant, and the value is 2.But let me verify this with a concrete example to make sure. Let's pick a specific value of k and check.Let me choose k = 0. Then the line is horizontal: y = 0*(x - 1) + 1 = 1. So y = 1.Find intersection points with the ellipse x²/2 + y² = 1. Substitute y = 1:x²/2 + 1 = 1 => x²/2 = 0 => x = 0. But that's a single point (0,1). However, the problem states that the line intersects at two distinct points P and Q different from A(0, -1). So k=0 gives a line that is tangent to the ellipse? Wait, but x²/2 + 1 = 1 implies x²=0, so x=0. So the line y=1 touches the ellipse at (0,1). So it's a tangent, hence only one point. But the problem statement says "two distinct points P and Q". Therefore, k=0 might not be allowed. Let's pick another k.Let me choose k = 1. Then the line is y = 1*(x - 1) + 1 = x - 1 + 1 = x. So y = x.Find intersections with ellipse x²/2 + y² = 1. Substitute y = x:x²/2 + x² = 1 => (3/2)x² = 1 => x² = 2/3 => x = ±√(2/3). Therefore, points P and Q are (√(2/3), √(2/3)) and (-√(2/3), -√(2/3)).Now, compute slopes of AP and AQ. Point A is (0, -1).First, point P (√(2/3), √(2/3)):Slope AP = (√(2/3) - (-1)) / (√(2/3) - 0) = (√(2/3) + 1)/√(2/3) = 1 + 1/√(2/3) = 1 + √(3/2)Similarly, point Q (-√(2/3), -√(2/3)):Slope AQ = (-√(2/3) - (-1))/(-√(2/3) - 0) = (-√(2/3) + 1)/(-√(2/3)) = [1 - √(2/3)]/(-√(2/3)) = - [1 - √(2/3)]/√(2/3) = -1/√(2/3) + 1Which is -√(3/2) + 1Therefore, sum of slopes:[1 + √(3/2)] + [ -√(3/2) + 1 ] = 1 + 1 + √(3/2) - √(3/2) = 2. So yes, the sum is 2. That checks out.Let's try another k, say k = 1/2. Then the line is y = (1/2)(x - 1) + 1 = (1/2)x - 1/2 + 1 = (1/2)x + 1/2.Find intersections with ellipse x²/2 + y² = 1.Substitute y = (1/2)x + 1/2 into the ellipse:x²/2 + [(1/2)x + 1/2]^2 = 1Compute [(1/2)x + 1/2]^2 = (1/4)x² + (1/2)x + 1/4So, x²/2 + (1/4)x² + (1/2)x + 1/4 = 1Combine like terms:(1/2 + 1/4)x² + (1/2)x + (1/4 - 1) = 0Which is (3/4)x² + (1/2)x - 3/4 = 0Multiply all terms by 4 to eliminate denominators:3x² + 2x - 3 = 0Solve using quadratic formula:x = [-2 ± √(4 + 36)]/(2*3) = [-2 ± √40]/6 = [-2 ± 2√10]/6 = [-1 ± √10]/3Therefore, x-coordinates are x1 = (-1 + √10)/3 and x2 = (-1 - √10)/3Corresponding y-coordinates:y1 = (1/2)x1 + 1/2 = (1/2)[(-1 + √10)/3] + 1/2 = (-1 + √10)/6 + 3/6 = (2 + √10)/6Similarly, y2 = (1/2)x2 + 1/2 = (1/2)[(-1 - √10)/3] + 1/2 = (-1 - √10)/6 + 3/6 = (2 - √10)/6Now, compute slopes of AP and AQ.Slope AP: (y1 - (-1))/(x1 - 0) = [ (2 + √10)/6 + 1 ] / [ (-1 + √10)/3 ]Convert 1 to 6/6:= [ (2 + √10 + 6)/6 ] / [ (-1 + √10)/3 ] = [ (8 + √10)/6 ] / [ (-1 + √10)/3 ] = [ (8 + √10)/6 ] * [ 3/(-1 + √10) ] = [ (8 + √10)/2 ] * [ 1/(-1 + √10) ]Multiply numerator and denominator by (1 + √10) to rationalize denominator:= [ (8 + √10)(1 + √10) ] / [ 2*(-1 + √10)(1 + √10) ] = [ (8(1) + 8√10 + √10*1 + √10*√10) ] / [ 2*(-1 - √10 + √10 + 10) ]Compute numerator:8 + 8√10 + √10 + 10 = 18 + 9√10Denominator:2*(-1 + 10) = 2*9 = 18Therefore, slope AP = (18 + 9√10)/18 = (2 + √10)/2Similarly, slope AQ: (y2 + 1)/x2 = [ (2 - √10)/6 + 1 ] / [ (-1 - √10)/3 ]Convert 1 to 6/6:= [ (2 - √10 + 6)/6 ] / [ (-1 - √10)/3 ] = [ (8 - √10)/6 ] / [ (-1 - √10)/3 ] = [ (8 - √10)/6 ] * [ 3/(-1 - √10) ] = [ (8 - √10)/2 ] * [ 1/(-1 - √10) ]Multiply numerator and denominator by (-1 + √10) to rationalize denominator:= [ (8 - √10)(-1 + √10) ] / [ 2*(-1 - √10)(-1 + √10) ]Compute numerator:8*(-1) + 8*√10 - √10*(-1) + (-√10)*√10 = -8 + 8√10 + √10 -10 = (-8 -10) + (8√10 + √10) = -18 + 9√10Denominator:2*(-1)^2 - (√10)^2 = 2*(1 -10) = 2*(-9) = -18So slope AQ = (-18 + 9√10)/(-18) = (-18)/(-18) + (9√10)/(-18) = 1 - (√10)/2Therefore, sum of slopes:Slope AP + slope AQ = (2 + √10)/2 + 1 - √10/2 = 2/2 + √10/2 + 1 - √10/2 = 1 + 1 = 2Yes, again, the sum is 2. So even with k = 1/2, it works. Hence, seems like the sum is indeed always 2, regardless of k (as long as the line intersects the ellipse at two distinct points).Therefore, the answer to part (2) is that the sum is a constant, and its value is 2.Final Answer(1) The equation of ellipse ( E ) is (boxed{dfrac{x^2}{2} + y^2 = 1}).(2) The sum of the slopes is a constant, and its value is (boxed{2}).</think>Given the problem of finding the equation of an ellipse ( E ) and determining the sum of the slopes of lines ( AP ) and ( AQ ):1. Equation of Ellipse ( E ): - The ellipse passes through the point ( A(0, -1) ). Substituting ( x = 0 ) and ( y = -1 ) into the ellipse equation ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), we get ( frac{1}{b^2} = 1 ), leading to ( b^2 = 1 ) and ( b = 1 ). - The eccentricity ( e = frac{sqrt{2}}{2} ). Using the relationship ( e = frac{sqrt{a^2 - b^2}}{a} ), we substitute ( e ) and ( b ) to find ( a ): [ frac{sqrt{a^2 - 1}}{a} = frac{sqrt{2}}{2} ] Squaring both sides, we get: [ frac{a^2 - 1}{a^2} = frac{1}{2} implies a^2 - 1 = frac{1}{2}a^2 implies frac{1}{2}a^2 = 1 implies a^2 = 2 implies a = sqrt{2} ] - Thus, the equation of the ellipse is: [ boxed{frac{x^2}{2} + y^2 = 1} ]2. Sum of the Slopes of Lines ( AP ) and ( AQ ): - The line passing through ( (1, 1) ) with slope ( k ) is ( y = kx - k + 1 ). Substituting this into the ellipse equation: [ frac{x^2}{2} + (kx - k + 1)^2 = 1 ] Expanding and simplifying, we get a quadratic equation in ( x ): [ left( frac{1}{2} + k^2 right)x^2 + (-2k^2 + 2k)x + (k^2 - 2k) = 0 ] - Using Vieta's formulas, the sum and product of the roots ( x_1 ) and ( x_2 ) are: [ x_1 + x_2 = frac{2k^2 - 2k}{frac{1}{2} + k^2}, quad x_1 x_2 = frac{k^2 - 2k}{frac{1}{2} + k^2} ] - The slopes of ( AP ) and ( AQ ) are ( frac{y_1 + 1}{x_1} ) and ( frac{y_2 + 1}{x_2} ). Using ( y = kx - k + 1 ), we find: [ frac{y_1 + 1}{x_1} = frac{kx_1 - k + 2}{x_1} = k - frac{k}{x_1} + frac{2}{x_1} ] Similarly for ( frac{y_2 + 1}{x_2} ). Summing these slopes and simplifying using Vieta's formulas, we find the sum simplifies to a constant: [ 2k + (2 - k) cdot frac{2(k - 1)}{k - 2} = 2 ] - Therefore, the sum of the slopes is a constant: [ boxed{2} ],<think>Okay, so I need to solve this problem about an ellipse. There are two parts. Let me start with the first one: finding the equation of ellipse E. The ellipse is given by the standard equation x²/a² + y²/b² = 1, where a > b > 0. It passes through the point A(0, -1), and its eccentricity is √2/2. First, let me recall that the eccentricity e of an ellipse is given by e = c/a, where c is the distance from the center to each focus, and c² = a² - b². Since the eccentricity is given as √2/2, that means c/a = √2/2, so c = (√2/2)a. Then, squaring both sides, c² = (2/4)a² = a²/2. But since c² = a² - b², that means a² - b² = a²/2. Let me write that down:a² - b² = (a²)/2Subtracting (a²)/2 from both sides:a²/2 - b² = 0 => b² = a²/2So, b² is half of a². Got that. Now, the ellipse passes through the point A(0, -1). Let's plug that point into the ellipse equation to find a relationship between a and b. Plugging x=0 and y=-1 into x²/a² + y²/b² = 1:0²/a² + (-1)²/b² = 1 => 0 + 1/b² = 1 => 1/b² = 1 => b² = 1So, b² = 1. But earlier, we found that b² = a²/2. Therefore:a²/2 = 1 => a² = 2So, a = √2, since a > 0. Therefore, the equation of the ellipse is:x²/(√2)² + y²/(1)² = 1 => x²/2 + y² = 1So, part (1) seems to be solved. Let me check again. The eccentricity e = c/a = √(a² - b²)/a. Plugging a² = 2 and b² = 1, we have c = √(2 - 1)/√2 = √1/√2 = 1/√2, so e = (1/√2)/√2 = 1/2. Wait, that's 1/2. But the problem says the eccentricity is √2/2. Hmm, wait, that seems conflicting. Let me check my steps again.Wait, wait, let's recast the eccentricity. Given that e = √2/2. Then e = c/a = √2/2. Then c = (√2/2)a. Then c² = (2/4)a² = a²/2. But c² = a² - b², so a² - b² = a²/2, which gives b² = a² - a²/2 = a²/2. So that's correct. Then when plugging the point A(0, -1), we get b² = 1, so a²/2 = 1 => a² = 2. So a = √2. Then e = c/a = (√(a² - b²))/a = √(2 - 1)/√2 = 1/√2 = √2/2. Ah, right, because 1/√2 is equal to √2/2. Yes, that's correct. So the eccentricity is indeed √2/2. So all checks out. So the equation is x²/2 + y² = 1. So part (1) done.Now part (2). A line passing through (1, 1) with slope k intersects the ellipse E at two distinct points P and Q (both different from point A). We need to determine if the sum of the slopes of lines AP and AQ is a constant. If yes, find the value; if no, explain why.First, let me understand the problem. We have the ellipse from part (1): x²/2 + y² = 1. There is a line with slope k passing through (1,1). This line intersects the ellipse at two points P and Q. Both P and Q are different from point A, which is (0, -1). Then, we need to find the slopes of lines AP and AQ, sum them, and check if this sum is a constant, independent of k. So, steps:1. Find the equation of the line passing through (1,1) with slope k.2. Find the intersection points P and Q of this line with the ellipse.3. Find the coordinates of P and Q.4. Compute the slopes of AP and AQ.5. Sum these slopes and check if the result is a constant.Alternatively, maybe there's a smarter way using properties of ellipses or using Vieta's formula, so we don't have to compute everything explicitly.Let me try to outline the steps algebraically.First, equation of the line: passing through (1,1) with slope k. So the equation is y - 1 = k(x - 1), which can be written as y = kx - k + 1.We need to find where this line intersects the ellipse x²/2 + y² = 1. Substitute y = kx - k + 1 into the ellipse equation.So, substitute into x²/2 + (kx - k + 1)^2 = 1.Expand this equation and solve for x. Then, the solutions will be the x-coordinates of P and Q. Then, for each x, compute y. Then, compute the slopes AP and AQ.But since point A is (0, -1), the slope of AP is (y_P - (-1))/(x_P - 0) = (y_P + 1)/x_P. Similarly for AQ: (y_Q + 1)/x_Q. Then the sum would be [(y_P + 1)/x_P + (y_Q + 1)/x_Q]. We need to check if this sum is constant.Alternatively, perhaps we can use Vieta's formula. Let me try to set up the equation.Substitute y = kx - k + 1 into x²/2 + y² = 1:x²/2 + (kx - k + 1)^2 = 1.Let me expand (kx - k + 1)^2:= [kx - (k - 1)]²= (kx)^2 - 2kx(k - 1) + (k - 1)^2= k²x² - 2k(k - 1)x + (k - 1)^2.So the equation becomes:x²/2 + k²x² - 2k(k - 1)x + (k - 1)^2 = 1.Combine like terms:x²/2 + k²x² = x²(1/2 + k²)-2k(k - 1)x is the linear term.Constant term: (k - 1)^2 - 1.So putting it all together:x²(1/2 + k²) - 2k(k - 1)x + [(k - 1)^2 - 1] = 0.Let me compute the constant term:(k - 1)^2 - 1 = k² - 2k + 1 - 1 = k² - 2k.So the equation is:(1/2 + k²)x² - 2k(k - 1)x + (k² - 2k) = 0.Multiply both sides by 2 to eliminate the fraction:(1 + 2k²)x² - 4k(k - 1)x + 2k² - 4k = 0.So:(1 + 2k²)x² - 4k(k - 1)x + (2k² - 4k) = 0.This is a quadratic in x. Let me denote this as Ax² + Bx + C = 0, where:A = 1 + 2k²B = -4k(k - 1)C = 2k² - 4kWe can use Vieta's formula here. The sum of the roots x_P + x_Q = -B/A = [4k(k - 1)] / (1 + 2k²)The product of the roots x_P x_Q = C/A = (2k² - 4k)/(1 + 2k²)Similarly, once we have x_P and x_Q, we can find y_P and y_Q using y = kx - k + 1.Then, the slopes of AP and AQ would be (y_P + 1)/x_P and (y_Q + 1)/x_Q.So let's compute these slopes.First, y_P = k x_P - k + 1. Therefore, y_P + 1 = k x_P - k + 1 + 1 = k x_P - k + 2 = k(x_P - 1) + 2.Similarly, slope of AP is (y_P + 1)/x_P = [k(x_P - 1) + 2]/x_P = k - (k - 2)/x_P.Wait, maybe it's better to write as [k x_P - k + 2]/x_P = k + (-k + 2)/x_P.Hmm, not sure. Alternatively, perhaps we can write:Slope AP: (y_P + 1)/x_P = [k x_P - k + 1 + 1]/x_P = [k x_P - k + 2]/x_P = k - (k - 2)/x_P.But this might not be helpful. Alternatively, maybe express everything in terms of x_P and x_Q.Alternatively, since y = kx - k + 1, then y + 1 = kx - k + 1 + 1 = kx - k + 2 = k(x - 1) + 2.So (y + 1)/x = [k(x - 1) + 2]/x = k(x - 1)/x + 2/x = k - k/x + 2/x.Hmm, so slope of AP is k - (k - 2)/x_P.Similarly for AQ: slope AQ is k - (k - 2)/x_Q.Therefore, the sum of the slopes would be [k - (k - 2)/x_P] + [k - (k - 2)/x_Q] = 2k - (k - 2)(1/x_P + 1/x_Q).So if we can compute 1/x_P + 1/x_Q, we can express the sum of the slopes in terms of k.But 1/x_P + 1/x_Q = (x_P + x_Q)/(x_P x_Q). From Vieta's formula, we have x_P + x_Q = 4k(k - 1)/(1 + 2k²) and x_P x_Q = (2k² - 4k)/(1 + 2k²). Therefore:1/x_P + 1/x_Q = [4k(k - 1)/(1 + 2k²)] / [(2k² - 4k)/(1 + 2k²)] = [4k(k - 1)] / (2k² - 4k) = [4k(k - 1)] / [2k(k - 2)] = (4k(k - 1)) / (2k(k - 2)) ) = [4(k - 1)] / [2(k - 2)] = 2(k - 1)/(k - 2)Simplify numerator and denominator: 2(k - 1)/(k - 2)Therefore, the sum of the slopes is 2k - (k - 2)*(2(k - 1)/(k - 2))).Wait, note that (k - 2) cancels with denominator (k - 2):So (k - 2)*(2(k - 1)/(k - 2)) = 2(k - 1)Therefore, the sum becomes 2k - 2(k - 1) = 2k - 2k + 2 = 2.Wow, so the sum is 2. Therefore, the sum of the slopes of AP and AQ is a constant 2. So the answer is yes, it's a constant, and its value is 2.But let me verify this because sometimes sign errors can occur. Let's check step by step.First, the line equation is y = kx - k + 1. Then, substituting into the ellipse:x²/2 + (kx - k + 1)^2 = 1. Expand that:x²/2 + k²x² - 2k(k - 1)x + (k - 1)^2 = 1.Then, x²/2 + k²x² is x²(1/2 + k²). The linear term is -2k(k - 1)x. The constants are (k - 1)^2 - 1 = k² - 2k. So equation becomes x²(1/2 + k²) - 2k(k - 1)x + (k² - 2k) = 0. Multiply by 2: (1 + 2k²)x² - 4k(k - 1)x + (2k² - 4k) = 0.Vieta's formulas: sum of roots x1 + x2 = [4k(k - 1)] / (1 + 2k²); product x1x2 = (2k² - 4k)/(1 + 2k²).Then, 1/x1 + 1/x2 = (x1 + x2)/(x1x2) = [4k(k - 1)/(1 + 2k²)] / [(2k² - 4k)/(1 + 2k²)] = [4k(k - 1)] / [2k(k - 2)] = [4(k - 1)] / [2(k - 2)] = 2(k - 1)/(k - 2).Then, the sum of slopes:Slope AP + Slope AQ = [ (y1 + 1)/x1 + (y2 + 1)/x2 ]But y1 = kx1 - k + 1, so y1 + 1 = kx1 - k + 1 + 1 = kx1 - k + 2 = k(x1 - 1) + 2. Similarly for y2 + 1 = k(x2 - 1) + 2.Therefore, slopes:Slope AP = [k(x1 - 1) + 2]/x1 = k - k/x1 + 2/x1.Similarly, Slope AQ = k - k/x2 + 2/x2.Therefore, sum:Slope AP + Slope AQ = 2k - k(1/x1 + 1/x2) + 2(1/x1 + 1/x2)= 2k - (k - 2)(1/x1 + 1/x2)Wait, let's compute:= 2k - k*(1/x1 + 1/x2) + 2*(1/x1 + 1/x2)= 2k + (-k + 2)(1/x1 + 1/x2)We already found that 1/x1 + 1/x2 = 2(k - 1)/(k - 2)Thus:Slope sum = 2k + (-k + 2)*[2(k - 1)/(k - 2)]Let me compute (-k + 2) = -(k - 2). So:Slope sum = 2k + [ - (k - 2) ]*[2(k - 1)/(k - 2)] = 2k - 2(k - 1)Simplify:2k - 2k + 2 = 2.Yes, that cancels out the k terms and leaves 2. Therefore, regardless of k, the sum is 2. Therefore, the answer is yes, it's a constant, and the value is 2.But let me test this with a specific example to make sure.Take k = 0. Then the line is horizontal through (1,1): y = 1. Intersect with ellipse x²/2 + y² = 1.Substitute y = 1: x²/2 + 1 = 1 => x²/2 = 0 => x = 0. So the only intersection point is (0,1). But the line y=1 would intersect the ellipse at (0,1). But wait, the problem states that the line intersects at two distinct points P and Q, both different from A(0, -1). However, if k=0, the line y=1 intersects the ellipse only at (0,1), which is a single point. So this case is invalid because the problem requires two distinct points. So k cannot be such that the line is tangent. Hence, discriminant must be positive. Let me check discriminant.The quadratic in x after substitution is (1 + 2k²)x² - 4k(k - 1)x + (2k² - 4k) = 0.Discriminant D = B² - 4AC = [ -4k(k - 1) ]² - 4*(1 + 2k²)*(2k² - 4k)Compute D:= 16k²(k - 1)^2 - 4*(1 + 2k²)*(2k² - 4k)First term: 16k²(k² - 2k + 1)Second term: 4*(1 + 2k²)*(4k - 2k²) [Wait, 2k² - 4k = -4k + 2k², so (2k² - 4k) = -2(2k - k²))Wait, let me compute it step by step.Second term:4*(1 + 2k²)*(2k² - 4k) = 4*(2k² - 4k + 4k³ - 8k²) = 4*(-6k² -4k + 4k³) = 16k³ -24k² -16kWait, perhaps better:Let me compute (1 + 2k²)(2k² - 4k):= 1*(2k² -4k) + 2k²*(2k² -4k)= 2k² -4k +4k^4 -8k³So, multiply by 4: 4*(2k² -4k +4k^4 -8k³) = 8k² -16k +16k^4 -32k³First term: 16k²(k² - 2k +1) =16k^4 -32k³ +16k²So discriminant D = (16k^4 -32k³ +16k²) - (8k² -16k +16k^4 -32k³)=16k^4 -32k³ +16k² -8k² +16k -16k^4 +32k³Simplify term by term:16k^4 -16k^4 = 0-32k³ +32k³=016k² -8k²=8k²+16kSo D=8k² +16k=8k(k +2)For the quadratic to have two distinct real roots, discriminant must be positive. So 8k(k +2) >0. So 8k(k +2) >0. Since 8 is positive, this reduces to k(k +2) >0.Therefore, k >0 or k < -2. So when k is in (-infty, -2) or (0, +infty), the discriminant is positive, hence two distinct points.Therefore, when k is between -2 and 0, discriminant is negative, so no real intersection. At k=0, discriminant is zero (as we saw earlier). Similarly, at k=-2, discriminant is zero. So valid k's are k >0 or k < -2.But regardless, for valid k's (k>0 or k<-2), the sum of the slopes is 2. Let me pick a k where the line intersects the ellipse at two points. Let's take k=1, which is in the valid range (k>0).For k=1, line equation is y = x -1 +1 = x. So y=x. Intersect with ellipse x²/2 + y² =1. Substitute y=x: x²/2 + x² = 1 => (3/2)x² =1 => x²=2/3 => x=±√(2/3). Therefore, points P and Q are (√(2/3), √(2/3)) and (-√(2/3), -√(2/3)).Compute slopes of AP and AQ. Point A is (0, -1).Slope AP: (√(2/3) - (-1)) / (√(2/3) -0) = (√(2/3) +1)/√(2/3) = 1 + 1/√(2/3) = 1 + √(3/2).Similarly, slope AQ: (-√(2/3) - (-1))/(-√(2/3) -0) = (-√(2/3)+1)/(-√(2/3)) = [1 -√(2/3)]/(-√(2/3)) = - [1 -√(2/3)]/√(2/3) = -1/√(2/3) +1.Compute these numerically:√(2/3) ≈ 0.8165, so √(3/2) ≈1.2247.Slope AP ≈1 + 1.2247 ≈2.2247Slope AQ ≈-1/0.8165 +1 ≈-1.2247 +1≈-0.2247Sum ≈2.2247 -0.2247≈2. So yes, sum is 2. Perfect.Another test: take k=2. Then line equation is y=2x -2 +1=2x -1. Find intersection with ellipse x²/2 + y²=1.Substitute y=2x -1:x²/2 + (2x -1)^2 =1 =>x²/2 +4x² -4x +1=1 =>x²/2 +4x² -4x =0 =>(9/2)x² -4x=0 =>x(9/2 x -4)=0.Solutions: x=0 and x=8/9. But x=0 gives y=2*0 -1=-1, which is point A. But the problem states that intersections are two distinct points different from A. So x=8/9, y=2*(8/9)-1=16/9 -1=7/9. So the other point is (8/9,7/9). Wait, but discriminant when k=2: D=8*2*(2 +2)=8*2*4=64, which is positive, but the quadratic equation gives only one non-A point. Wait, this seems conflicting. Wait, when we substituted k=2 into the quadratic equation:Equation: (1 + 2*(4))x² -4*2*(2 -1)x + (2*4 -4*2)= (9x²) -8x + (8 -8)=9x² -8x=0. So x(9x -8)=0, so x=0 or x=8/9. So one of the points is A(0, -1), the other is (8/9,7/9). But the problem states that both points are different from A, so this case is invalid. Hence, k=2 would not be allowed, as one of the intersection points is A. Wait, so need to check if (8/9,7/9) is different from A. Yes, it is. But also, when k=2, the line passes through (1,1) with slope 2: y=2x -1. When x=0, y=-1, so point A is on the line. Therefore, the line intersects the ellipse at A and another point. But the problem states that both points P and Q are different from A, so in this case, k=2 is invalid because one of the intersection points is A. Therefore, k=2 is not allowed. So in the quadratic equation, when x=0 is a solution, that corresponds to point A. So we need to exclude such cases. Therefore, in order for both points P and Q to be different from A, the line must not pass through A. So how can we ensure that? Let's check.Point A is (0, -1). The line passes through (1,1) and has slope k. The line is y = kx -k +1. To check if A is on the line, substitute x=0, y=-1:-1 = k*0 -k +1 => -1 = -k +1 => -k = -2 => k=2.Therefore, when k=2, the line passes through A. Hence, to ensure that both points P and Q are different from A, k must not be 2. Therefore, k≠2. However, from discriminant condition, k>0 or k<-2. So k=2 is allowed for discriminant but causes one point to be A, so must exclude k=2. So in general, when k≠2, the line does not pass through A, hence both intersection points are different from A.So in the earlier example, with k=1, it's valid. With k=2, it's invalid. So when testing, need to pick k≠2.Take k=3. Then line is y=3x -3 +1=3x -2. Substitute into ellipse: x²/2 + (3x -2)^2 =1.Compute: x²/2 +9x² -12x +4 =1 =>x²/2 +9x² -12x +3=0 => (1/2 +9)x² -12x +3=0 => (19/2)x² -12x +3=0.Multiply by 2:19x² -24x +6=0. Discriminant D=576 -4*19*6=576 -456=120>0, so two distinct points. Find their coordinates.But instead of calculating, let's trust our previous derivation. According to our general result, the sum of slopes should be 2. Therefore, regardless of k (as long as valid), the sum is 2. Therefore, answer is 2.Therefore, part (2)'s answer is that the sum is a constant 2.Final Answer(1) boxed{dfrac{x^2}{2} + y^2 = 1}(2) The sum is a constant boxed{2}</think>Given the problem of finding the equation of an ellipse and determining the sum of the slopes of lines from a specific point to the intersection points of a line with the ellipse, we proceed as follows:1. Finding the equation of ellipse (E): - The ellipse passes through the point (A(0, -1)), so substituting (x = 0) and (y = -1) into the ellipse equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) gives (b^2 = 1). - The eccentricity (e = frac{sqrt{2}}{2}) gives (c = frac{sqrt{2}}{2}a). Using (c^2 = a^2 - b^2), we find (a^2 = 2). - Therefore, the equation of the ellipse is (frac{x^2}{2} + y^2 = 1).2. Determining the sum of the slopes of lines (AP) and (AQ): - The line passing through ((1, 1)) with slope (k) is (y = kx - k + 1). - Substituting this into the ellipse equation gives a quadratic equation in (x). Using Vieta's formulas, we find the sum and product of the roots (x_P) and (x_Q). - The slopes of lines (AP) and (AQ) are calculated as (frac{y_P + 1}{x_P}) and (frac{y_Q + 1}{x_Q}). - Using algebraic manipulation and Vieta's formulas, we find that the sum of these slopes simplifies to a constant value of 2. Final Answer(1) The equation of ellipse (E) is (boxed{dfrac{x^2}{2} + y^2 = 1}).(2) The sum of the slopes of lines (AP) and (AQ) is a constant (boxed{2}).]

🔑:(1) Solution: Given that dfrac{c}{a} = dfrac{sqrt{2}}{2}, b=1, Combining a^{2}=b^{2}+c^{2}, we get a= sqrt{2}, Thus, the equation is dfrac{x^{2}}{2} +y^{2}=1; (2) Proof: Let the equation of line PQ be y=k(x-1)+1(kneq 0), Substituting into the ellipse equation dfrac{x^{2}}{2} +y^{2}=1, We get (1+2k^{2})x^{2}-4k(k-1)x+2k(k-2)=0, Given that (1,1) is outside the ellipse, Let P(x_{1},y_{1}), Q(x_{2},y_{2}), x_{1}x_{2}neq 0, Then x_{1}+x_{2}= dfrac{4k(k-1)}{1+2k^{2}}, x_{1}x_{2}= dfrac{2k(k-2)}{1+2k^{2}}, And triangle =16k^{2}(k-1)^{2}-8k(k-2)(1+2k^{2}) > 0, solving gives k > 0 or k < -2. Thus, the sum of the slopes of lines AP, AQ is k_{AP}+k_{AQ}= dfrac{y_{1}+1}{x_{1}}+ dfrac{y_{2}+1}{x_{2}} = dfrac{kx_{1}+2-k}{x_{1}}+ dfrac{kx_{2}+2-k}{x_{2}}=2k+(2-k)( dfrac{1}{x_{1}}+ dfrac{1}{x_{2}})=2k+(2-k)⋅ dfrac{x_{1}+x_{2}}{x_{1}x_{2}} =2k+(2-k)⋅ dfrac{4k(k-1)}{2k(k-2)}=2k-2(k-1)=boxed{2}. Therefore, the sum of the slopes of lines AP and AQ is boxed{2}.

🧠:[<think>Okay, let's see. The problem is about three people—A, B, and C—passing a ball. The first pass is made by A, and each time a person has the ball, they have an equal chance of passing it to either of the other two. We need to find the probability that after three passes, the ball is back with A. Hmm, so three passes mean the ball is with someone after the third pass. Let me try to figure this out step by step.First, maybe I should visualize this as a sequence of passes. The first pass is from A. So, starting at A, the first pass is either to B or C. Then, whoever gets the ball next will pass it to one of the other two, and so on for three passes. The question is asking for the probability that after three passes, the ball is back with A. Let me break it down pass by pass.Starting with A. So, initial state: A has the ball. Pass 1: A passes to either B or C. Each has a 50% chance. Then Pass 2: whoever received the ball (B or C) will pass to one of the other two. Then Pass 3: same thing. After three passes, we want to know the probability that it's back with A.Maybe I can model this as a Markov chain or use some probability trees. Let me think. Each time, the person with the ball has two choices, each with probability 1/2. So, starting from A, after the first pass, the ball is at B or C, each with probability 1/2. Then, from B, the next pass is to A or C, each with 1/2 chance. Similarly, from C, the next pass is to A or B. Then, after the second pass, we can be at A, B, or C? Wait, but after two passes, could we be back at A? Let's see.Wait, starting at A, pass 1: to B or C. Then pass 2: from B to A or C, or from C to A or B. So, after two passes, possible positions are A, C (if starting from B and passing to A or C) or A, B (if starting from C and passing to A or B). So, after two passes, there's a chance to be back at A. Let me compute the probabilities step by step.First, pass 1: A passes to B or C. Each with probability 1/2. So after pass 1, the ball is at B with 1/2 and at C with 1/2.Pass 2: From B, the ball can go to A or C. Similarly, from C, the ball can go to A or B. So, let's compute the probability distribution after pass 2.If after pass 1 it's at B (prob 1/2), then after pass 2: 1/2 chance to go to A and 1/2 chance to go to C. Similarly, if it's at C (prob 1/2), then after pass 2: 1/2 chance to go to A and 1/2 chance to go to B. So, the total probability at A after pass 2 is (1/2 * 1/2) + (1/2 * 1/2) = 1/4 + 1/4 = 1/2. Then, the probability at B after pass 2 is (1/2 * 1/2) = 1/4, and similarly at C is 1/4. Wait, that's interesting. After pass 2, the probability is 1/2 at A and 1/4 each at B and C.But let me confirm that. Starting from pass 1: 1/2 B and 1/2 C. Then, from B, 1/2 A and 1/2 C; so the contribution to A is 1/2 * 1/2 = 1/4, and to C is 1/2 * 1/2 = 1/4. From C, 1/2 A and 1/2 B; so contribution to A is 1/2 * 1/2 = 1/4, and to B is 1/2 * 1/2 = 1/4. So, adding those contributions: A gets 1/4 + 1/4 = 1/2, B gets 1/4, and C gets 1/4. Yep, that's correct.So after two passes, the probability is 1/2 at A, 1/4 at B, 1/4 at C.Now, pass 3: Starting from A, B, or C. Wait, no—after two passes, the ball is at A with 1/2, B with 1/4, C with 1/4. Then, each of these will pass to the other two with equal probability. So, from A, which has 1/2 probability, the next pass is to B or C, each with 1/2. From B, which has 1/4 probability, the next pass is to A or C, each with 1/2. From C, same as B: 1/4 probability, pass to A or B with 1/2 each.So, let's compute the probabilities after pass 3.First, contributions from A: 1/2 chance to be at A after pass 2, then passing to B or C. So, from A: the probability of going to B is 1/2 * 1/2 = 1/4, and same for C. Wait, no: Wait, the total probability contributed to B and C from A would be 1/2 (prob at A) times 1/2 (prob to pass to B or C). Wait, but actually, if you have a state where the ball is at A, then passing it to B or C would split the probability from A into B and C. Similarly for the other states.So, in general, the probability at each node after pass 3 is:Probability at B: contributions from A passing to B, C passing to B, and B passing to B? Wait, no. Wait, when you have the ball at a person, they pass it to someone else. So, if the ball is at A, they can't keep it; they pass to B or C. Similarly, if the ball is at B, they pass to A or C. So, the only way to end up at A after pass 3 is if the ball was at B or C after pass 2, and they passed it to A. Because if the ball is at A after pass 2, they have to pass it to B or C, so they can't stay at A. Therefore, the probability to be at A after pass 3 is the probability that the ball was at B after pass 2 times the probability they pass to A, plus the probability the ball was at C after pass 2 times the probability they pass to A.Similarly, the probability at B after pass 3 is the probability the ball was at A after pass 2 times the probability they pass to B, plus the probability the ball was at C after pass 2 times the probability they pass to B. Same for C.So, let's compute each:Probability at A after pass 3:(Probability at B after pass 2) * (prob passing to A) + (Probability at C after pass 2) * (prob passing to A)Which is (1/4 * 1/2) + (1/4 * 1/2) = (1/8 + 1/8) = 1/4.Probability at B after pass 3:(Probability at A after pass 2) * (prob passing to B) + (Probability at C after pass 2) * (prob passing to B)Which is (1/2 * 1/2) + (1/4 * 1/2) = (1/4) + (1/8) = 3/8.Similarly, probability at C after pass 3:(Probability at A after pass 2) * (prob passing to C) + (Probability at B after pass 2) * (prob passing to C)Which is (1/2 * 1/2) + (1/4 * 1/2) = 1/4 + 1/8 = 3/8.So, after three passes, the probability at A is 1/4, at B 3/8, and at C 3/8. Therefore, the answer should be 1/4. But let me check this again to make sure.Wait, let's walk through it step by step again. Maybe there's a mistake in the calculation.After pass 1: B (1/2) and C (1/2).After pass 2:From B, 1/2 to A and 1/2 to C.From C, 1/2 to A and 1/2 to B.So, total at A: 1/2 * 1/2 (from B) + 1/2 * 1/2 (from C) = 1/4 + 1/4 = 1/2.At B: 1/2 * 1/2 (from C) = 1/4.At C: 1/2 * 1/2 (from B) = 1/4.So after two passes, A has 1/2, B and C each 1/4. Then pass 3:From A, which has 1/2, passes to B and C each with 1/2. So that's 1/2 * 1/2 = 1/4 to B and 1/4 to C.From B, which has 1/4, passes to A and C each with 1/2. So that's 1/4 * 1/2 = 1/8 to A and 1/8 to C.From C, which has 1/4, passes to A and B each with 1/2. So that's 1/4 * 1/2 = 1/8 to A and 1/8 to B.Therefore, total at A after pass 3: 1/8 (from B) + 1/8 (from C) = 2/8 = 1/4.Total at B: 1/4 (from A) + 1/8 (from C) = 3/8.Total at C: 1/4 (from A) + 1/8 (from B) = 3/8.Yep, that checks out. So the probability at A after three passes is 1/4. So the answer should be 1/4.Alternatively, maybe another way to think about it is using recursion or state transitions. Let me try that approach to confirm.Let’s denote P(n, X) as the probability that after n passes, the ball is at person X. We need to find P(3, A).We know that initially, P(0, A) = 1, and P(0, B) = P(0, C) = 0.Then, for each pass, the probability at a person is the sum of the probabilities from the other two people passing to them. Since each pass from a person is equally likely to go to the other two, the transition is:P(n+1, A) = (P(n, B) + P(n, C)) * 1/2Similarly,P(n+1, B) = (P(n, A) + P(n, C)) * 1/2P(n+1, C) = (P(n, A) + P(n, B)) * 1/2So, let's compute step by step.n=0:P(0, A) = 1P(0, B) = 0P(0, C) = 0n=1:P(1, A) = (P(0, B) + P(0, C)) * 1/2 = (0 + 0) * 1/2 = 0P(1, B) = (P(0, A) + P(0, C)) * 1/2 = (1 + 0) * 1/2 = 1/2P(1, C) = (P(0, A) + P(0, B)) * 1/2 = (1 + 0) * 1/2 = 1/2n=2:P(2, A) = (P(1, B) + P(1, C)) * 1/2 = (1/2 + 1/2) * 1/2 = 1 * 1/2 = 1/2P(2, B) = (P(1, A) + P(1, C)) * 1/2 = (0 + 1/2) * 1/2 = 1/4P(2, C) = (P(1, A) + P(1, B)) * 1/2 = (0 + 1/2) * 1/2 = 1/4n=3:P(3, A) = (P(2, B) + P(2, C)) * 1/2 = (1/4 + 1/4) * 1/2 = 1/2 * 1/2 = 1/4P(3, B) = (P(2, A) + P(2, C)) * 1/2 = (1/2 + 1/4) * 1/2 = (3/4) * 1/2 = 3/8P(3, C) = (P(2, A) + P(2, B)) * 1/2 = (1/2 + 1/4) * 1/2 = 3/8Same result. So that confirms it. The probability is 1/4.Alternatively, another approach: since each time the ball is passed, it moves from one person to another, and since there are three people, maybe we can model this as a state machine with transitions.But it seems that both the step-by-step calculation and the recursive formula approach lead to the same answer. So, I think 1/4 is correct.Wait, but let me also think about enumerating all possible paths. Since each pass has two choices, after three passes, there are 2^3 = 8 possible paths. But since the first pass is from A, and each subsequent pass depends on where the ball is. Wait, maybe the number of possible paths is more than 8? Because each pass is dependent on the previous receiver.Wait, starting from A:Pass 1: A -> B or A -> C.If A -> B:Pass 2: B -> A or B -> C.If B -> A:Pass 3: A -> B or A -> C.If B -> C:Pass 3: C -> A or C -> B.Similarly, if A -> C:Pass 2: C -> A or C -> B.If C -> A:Pass 3: A -> B or A -> C.If C -> B:Pass 3: B -> A or B -> C.So, total number of paths:First branch: A -> B or A -> C (2 choices).Second pass: From B or C, each has 2 choices.Third pass: From A, B, or C, each has 2 choices.Wait, but after three passes, the number of paths is 2 * 2 * 2 = 8. Let's list them.Starting with A:1. A -> B -> A -> B2. A -> B -> A -> C3. A -> B -> C -> A4. A -> B -> C -> B5. A -> C -> A -> B6. A -> C -> A -> C7. A -> C -> B -> A8. A -> C -> B -> CSo, eight possible paths. Each path has a probability of (1/2)^3 = 1/8.Now, let's count how many of these paths end at A.Looking at the list:1. A -> B -> A -> B: ends at B2. A -> B -> A -> C: ends at C3. A -> B -> C -> A: ends at A4. A -> B -> C -> B: ends at B5. A -> C -> A -> B: ends at B6. A -> C -> A -> C: ends at C7. A -> C -> B -> A: ends at A8. A -> C -> B -> C: ends at CSo, paths 3 and 7 end at A. That's 2 out of 8. So the probability would be 2/8 = 1/4. Which matches our previous calculation. So that confirms it again. Therefore, the answer is 1/4.So, whether we use recursive probability calculations, state transitions, or enumerating all possible paths (and since the number is small enough), we get the same answer. Therefore, the probability that the ball is back with A after three passes is 1/4.Final AnswerThe probability that the ball is still with A after three passes is boxed{dfrac{1}{4}}.</think>Starting with A, the first pass is made by A to either B or C with equal probability. We need to determine the probability that the ball returns to A after three passes.1. After the first pass (Pass 1): - A passes to B or C with probability ( frac{1}{2} ) each. - Probability distribution: ( P(1, A) = 0 ), ( P(1, B) = frac{1}{2} ), ( P(1, C) = frac{1}{2} ).2. After the second pass (Pass 2): - From B: passes to A or C with probability ( frac{1}{2} ) each. - From C: passes to A or B with probability ( frac{1}{2} ) each. - Probability distribution: - ( P(2, A) = frac{1}{2} times frac{1}{2} + frac{1}{2} times frac{1}{2} = frac{1}{4} + frac{1}{4} = frac{1}{2} ) - ( P(2, B) = frac{1}{2} times frac{1}{2} = frac{1}{4} ) - ( P(2, C) = frac{1}{2} times frac{1}{2} = frac{1}{4} )3. After the third pass (Pass 3): - From A: passes to B or C with probability ( frac{1}{2} ) each. - From B: passes to A or C with probability ( frac{1}{2} ) each. - From C: passes to A or B with probability ( frac{1}{2} ) each. - Probability distribution: - ( P(3, A) = frac{1}{4} times frac{1}{2} + frac{1}{4} times frac{1}{2} = frac{1}{8} + frac{1}{8} = frac{1}{4} ) - ( P(3, B) = frac{1}{2} times frac{1}{2} + frac{1}{4} times frac{1}{2} = frac{1}{4} + frac{1}{8} = frac{3}{8} ) - ( P(3, C) = frac{1}{2} times frac{1}{2} + frac{1}{4} times frac{1}{2} = frac{1}{4} + frac{1}{8} = frac{3}{8} )Enumerating all possible paths confirms that 2 out of 8 paths end at A, resulting in a probability of ( frac{1}{4} ).Thus, the probability that the ball is still with A after three passes is (boxed{dfrac{1}{4}}).,<think>Okay, let's see. So the problem is about three people—A, B, and C—passing a ball. The first pass is made by A, and each time someone has the ball, they have an equal chance of passing it to either of the other two. After three passes, we need to find the probability that the ball is back with A. Hmm.Alright, let me break this down step by step. Since there are three passes, starting from A, each pass transfers the ball to someone else. So after the first pass, the ball is either with B or C. Then each subsequent pass is made by whoever has the ball, choosing randomly between the other two. After three passes, we want the probability that it ends up back at A.Let me think about how to model this. Maybe using a tree diagram? But with three passes, that might get a bit complicated. Alternatively, maybe using states and probabilities. Let's consider each step and the possible holders of the ball.Starting with A (pass 0). Then pass 1: A can pass to B or C with equal probability, so 50% each. So after pass 1, the ball is at B or C, each with probability 1/2.Then pass 2: whoever has the ball (either B or C) will pass it to one of the other two. So if it's B, he can pass to A or C, each with 1/2 chance. Similarly, if it's C, she can pass to A or B with 1/2 chance each. So after pass 2, where can the ball be?If after pass 1 it was B, then after pass 2, it's either A or C. Similarly, if after pass 1 it was C, then after pass 2, it's either A or B. So after pass 2, the possible holders are A, B, or C. Wait, but in each case, passing from B or C can lead back to A or to the other non-A person. So let's compute the probabilities.After pass 1: P(B) = 1/2, P(C) = 1/2.Pass 2:If at B: P(A) = 1/2, P(C) = 1/2.If at C: P(A) = 1/2, P(B) = 1/2.So total after pass 2:P(A) = P(pass from B to A) + P(pass from C to A) = (1/2 * 1/2) + (1/2 * 1/2) = 1/4 + 1/4 = 1/2.P(B) = P(pass from C to B) = 1/2 * 1/2 = 1/4.Similarly, P(C) = P(pass from B to C) = 1/2 * 1/2 = 1/4.Wait, is that right? So after pass 2, the probability the ball is with A is 1/2, and with B or C is 1/4 each. Hmm. Interesting.Then pass 3. Now, if after pass 2, the ball is at A, then A will pass to B or C with 1/2 chance each. If the ball is at B, he passes to A or C. Similarly for C.So pass 3:If at A: P(B) = 1/2, P(C) = 1/2.If at B: P(A) = 1/2, P(C) = 1/2.If at C: P(A) = 1/2, P(B) = 1/2.So after pass 3, the probability of being at A is:P(A at pass 3) = P(pass to A from B or C)Which would be:P(A at pass 3) = [P(B at pass 2) * P(B passes to A)] + [P(C at pass 2) * P(C passes to A)]From pass 2, P(B) = 1/4, P(C) = 1/4. Each of those passing to A is 1/2.So 1/4 * 1/2 + 1/4 * 1/2 = 1/8 + 1/8 = 1/4.Alternatively, since from pass 2, P(A) = 1/2. If the ball is at A, then pass 3 cannot stay at A; it must go to B or C. So the only way to end at A after pass 3 is if the ball was at B or C after pass 2, and then passed back to A.Therefore, the total probability is 1/4.Wait, so the answer would be 1/4? Let me check that again.Alternatively, maybe I should model this as a Markov chain with states A, B, C and transitions.But let's verify step by step:Pass 0: A (prob 1)Pass 1: B (1/2), C (1/2)Pass 2: From B: A (1/2), C (1/2); from C: A (1/2), B (1/2)So pass 2:A: 1/2 * 1/2 (from B) + 1/2 * 1/2 (from C) = 1/4 + 1/4 = 1/2B: 1/2 * 1/2 (from C) = 1/4C: 1/2 * 1/2 (from B) = 1/4So pass 2: A (1/2), B (1/4), C (1/4)Pass 3:From A: B (1/2), C (1/2)From B: A (1/2), C (1/2)From C: A (1/2), B (1/2)So pass 3:Probability at A: (Probability at A in pass 2)*0 + (Probability at B in pass 2)*(1/2) + (Probability at C in pass 2)*(1/2)Which is 0 * 1/2 + 1/4 * 1/2 + 1/4 * 1/2 = 0 + 1/8 + 1/8 = 1/4Similarly, probability at B: (Probability at A in pass 2)*(1/2) + (Probability at B in pass 2)*0 + (Probability at C in pass 2)*(1/2)= 1/2 * 1/2 + 0 + 1/4 * 1/2 = 1/4 + 0 + 1/8 = 3/8Similarly for C: same as B, so 3/8.So after pass 3, the probabilities are A: 1/4, B: 3/8, C: 3/8.So indeed, the probability that the ball is back with A after three passes is 1/4.Wait, but let me make sure I didn't make a mistake here. Let's check the steps again.Starting with A. First pass: B or C, each 1/2.Second pass: If at B, can go to A or C. Similarly for C. So after the second pass, A has 1/2, and B and C each have 1/4. Then third pass: if at A, can't stay at A, so only contributions from B and C. Each of them has 1/4 chance, and each has a 1/2 chance to pass back to A. So 1/4 * 1/2 + 1/4 * 1/2 = 1/8 + 1/8 = 1/4. That seems correct.Alternatively, let's list all possible sequences of passes and count how many end at A.First pass: A -> B or A -> C. Let's consider each case.Case 1: A -> B.Second pass: B can go to A or C.Case 1a: B -> A.Third pass: A must pass to B or C. So from A, can't end at A. So these sequences (A->B->A->B or A->B->A->C) don't end at A.Case 1b: B -> C.Third pass: C can go to A or B.So sequence A->B->C->A or A->B->C->B.So here, A->B->C->A ends at A.Similarly, for the other initial branch.Case 2: A->C.Second pass: C can go to A or B.Case 2a: C->A.Third pass: A passes to B or C. Neither ends at A.Case 2b: C->B.Third pass: B passes to A or C. So sequence A->C->B->A or A->C->B->C.Again, A->C->B->A ends at A.So total sequences:Each initial pass has two choices. Each subsequent pass has two choices. So total sequences: 2 * 2 * 2 = 8.From the above, sequences ending at A are:A->B->C->AA->C->B->ASo two sequences. Each sequence has probability (1/2)^3 = 1/8.So total probability: 2 * 1/8 = 1/4. Which matches the previous result.So the answer is 1/4.Therefore, the probability is 1/4.Final AnswerThe probability that the ball is still with A after three passes is boxed{dfrac{1}{4}}.</think>Starting with A, the ball is passed to either B or C with equal probability (1/2 each) after the first pass. After the first pass:- Probability at B: 1/2- Probability at C: 1/2After the second pass:- If the ball is at B, it can be passed to A or C with equal probability (1/2 each).- If the ball is at C, it can be passed to A or B with equal probability (1/2 each).Calculating the probabilities:- Probability at A: ( left(frac{1}{2} times frac{1}{2}right) + left(frac{1}{2} times frac{1}{2}right) = frac{1}{4} + frac{1}{4} = frac{1}{2} )- Probability at B: ( left(frac{1}{2} times frac{1}{2}right) = frac{1}{4} )- Probability at C: ( left(frac{1}{2} times frac{1}{2}right) = frac{1}{4} )After the third pass:- If the ball is at A, it can be passed to B or C (no chance to stay at A).- If the ball is at B, it can be passed to A or C with equal probability (1/2 each).- If the ball is at C, it can be passed to A or B with equal probability (1/2 each).Calculating the probability at A:- Contribution from B: ( frac{1}{4} times frac{1}{2} = frac{1}{8} )- Contribution from C: ( frac{1}{4} times frac{1}{2} = frac{1}{8} )Total probability at A after the third pass:- ( frac{1}{8} + frac{1}{8} = frac{1}{4} )Thus, the probability that the ball is still with A after three passes is boxed{dfrac{1}{4}}.]

🔑:We represent one passing sequence as A→B→C→A.All possible passing sequences are:A→B→A→B; A→B→A→C; A→B→C→A; A→B→C→B;A→C→A→B; A→C→A→C; A→C→B→A; A→C→B→C;Therefore, there are 8 possible passing sequences in total.To determine the number of sequences in which the ball comes back to A after the third pass, we consider the sequences:- A→B→A→C- A→C→A→BThese are the only two sequences out of the eight where the ball is passed back to A on the third pass.Hence, after three passes, the probability that the ball is still with A is given by the number of favorable outcomes divided by the total number of possible sequences:[ p = frac{2}{8} = frac{1}{4} ]Therefore, the final answer is:[ boxed{frac{1}{4}} ]By enumerating all possible passing sequences and identifying those in which A receives the ball on the third pass, we can calculate the probability that A has the ball after three passes. This problem tests the method of finding probabilities and is considered a fundamental problem. When solving, it's crucial to read the question carefully and apply enumeration judiciously.

🧠:[<think>Okay, let's try to figure out how to solve this problem step by step. So, the problem is about finding the length of a chord intercepted on a specific circle by a line that passes through the origin and is parallel to another given line. Let me parse the details first.The circle equation is given as x² + (y - √3)² = 7. Hmm, let's recall that the standard form of a circle is (x - h)² + (y - k)² = r², where (h,k) is the center and r is the radius. So in this case, the center of the circle is at (0, √3) and the radius is √7. Okay, that's straightforward.Next, the line l that passes through the origin (0,0) and is parallel to the line √6x - √3y + 1 = 0. We need to find the equation of line l first, right? Because once we have that, we can find the points where this line intersects the circle, and then compute the distance between those two points to get the chord length.First, let's find the slope of the given line √6x - √3y + 1 = 0. To find the slope, I can rearrange this equation into slope-intercept form y = mx + b. So:Starting with √6 x - √3 y + 1 = 0Let's solve for y:-√3 y = -√6 x - 1Divide both sides by -√3:y = (√6 / √3) x + 1/√3Simplify √6 / √3. Since √6 is √(2*3) = √2 * √3, so √6 / √3 = √2. So the equation becomes:y = √2 x + 1/√3So the slope (m) of this line is √2. Therefore, the line l that is parallel to this will have the same slope, √2. Since line l passes through the origin, its equation is simply y = √2 x. Okay, so that's our line l.Now, we need to find the points where this line intersects the circle x² + (y - √3)² = 7. To do this, we can substitute y from the line equation into the circle equation.Substituting y = √2 x into the circle equation:x² + (√2 x - √3)² = 7Let's expand the second term:(√2 x - √3)² = (√2 x)² - 2*(√2 x)*(√3) + (√3)² = 2x² - 2*√6 x + 3So putting that back into the equation:x² + 2x² - 2√6 x + 3 = 7Combine like terms:(1x² + 2x²) = 3x²So:3x² - 2√6 x + 3 = 7Subtract 7 from both sides:3x² - 2√6 x + 3 - 7 = 0Simplify:3x² - 2√6 x - 4 = 0So we have a quadratic equation in terms of x: 3x² - 2√6 x - 4 = 0We can solve this quadratic equation to find the x-coordinates of the intersection points, then find the corresponding y-coordinates using y = √2 x, and then compute the distance between those two points.Alternatively, since we need the length of the chord, perhaps there's a formula that relates the distance from the center of the circle to the line, and then using that to compute the chord length without finding the actual intersection points. Let me recall that.Yes, the length of the chord can be calculated using the formula: 2√(r² - d²), where r is the radius of the circle, and d is the distance from the center to the line. That might be quicker. Let me verify.If we have a line and a circle, the distance from the center to the line is d, then the length of the chord intercepted by the line on the circle is 2 times the square root of (radius squared minus distance squared). That's a standard formula. So maybe that's a better approach here, avoiding solving the quadratic.But let's check if that works here. First, we need to compute the distance from the center of the circle to the line l. The center is (0, √3), and the line l is y = √2 x. Let's compute the distance d from (0, √3) to this line.The formula for the distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / sqrt(a² + b²). So first, we need to write the line equation in the form ax + by + c = 0.Our line is y = √2 x. Let's rearrange this to standard form:√2 x - y = 0So a = √2, b = -1, c = 0.Now, plugging the center (0, √3) into the formula:d = |√2*0 + (-1)*√3 + 0| / sqrt((√2)² + (-1)²) = | -√3 | / sqrt(2 + 1) = √3 / √3 = √3 / √3? Wait, sqrt(2 + 1) is sqrt(3). So numerator is | -√3 | = √3, denominator is sqrt(3). So d = √3 / sqrt(3) = (√3)/(√3) = 1.Wait, so the distance from the center (0, √3) to the line l is 1. Then, using the chord length formula: 2√(r² - d²) where r is √7. So:Chord length = 2 * sqrt( (√7)^2 - (1)^2 ) = 2*sqrt(7 - 1) = 2*sqrt(6) = 2√6. So the chord length is 2√6. Is that correct? Let me verify.Alternatively, let's solve the quadratic equation to check.Earlier, we had the quadratic equation 3x² - 2√6 x - 4 = 0.We can solve for x using the quadratic formula:x = [2√6 ± sqrt( (2√6)^2 - 4*3*(-4) )]/(2*3)Compute discriminant D:(2√6)^2 = 4*6 = 244*3*(-4) = -48So D = 24 - (-48) = 24 + 48 = 72Therefore, x = [2√6 ± sqrt(72)]/(6)sqrt(72) = sqrt(36*2) = 6√2So x = [2√6 ± 6√2]/6 = factor out 2 from numerator:x = [2(√6 ± 3√2)] /6 = (√6 ± 3√2)/3So x coordinates are (√6 + 3√2)/3 and (√6 - 3√2)/3.Then y coordinates are √2 x, so:y = √2*(√6 + 3√2)/3 and y = √2*(√6 - 3√2)/3Simplify:First y coordinate: √2*√6 = √12 = 2√3, and √2*3√2 = 3*(√2)^2 = 3*2 = 6. So:(2√3 + 6)/3 = (6 + 2√3)/3 = 2 + (2√3)/3Second y coordinate: Similarly, √2*√6 = 2√3, and √2*(-3√2) = -3*2 = -6. So:(2√3 - 6)/3 = (-6 + 2√3)/3 = -2 + (2√3)/3Therefore, the two points of intersection are:Point A: ( (√6 + 3√2)/3 , 2 + (2√3)/3 )Point B: ( (√6 - 3√2)/3 , -2 + (2√3)/3 )Now, let's compute the distance between these two points. Let's denote x1, y1 as coordinates of A and x2, y2 as coordinates of B.First, compute the differences:Δx = x1 - x2 = [ (√6 + 3√2)/3 - (√6 - 3√2)/3 ] = [ (√6 + 3√2 - √6 + 3√2)/3 ] = (6√2)/3 = 2√2Δy = y1 - y2 = [2 + (2√3)/3 - (-2 + (2√3)/3 ) ] = [2 + (2√3)/3 + 2 - (2√3)/3 ] = 4So the distance between A and B is sqrt( (Δx)^2 + (Δy)^2 ) = sqrt( (2√2)^2 + 4^2 ) = sqrt( 8 + 16 ) = sqrt(24) = 2√6Which matches the result from the chord length formula. So both methods confirm that the chord length is 2√6. Therefore, the answer is 2√6.Wait, but just to make sure I didn't make any mistakes in the calculations. Let me verify once again.First method: distance from center to line is 1, radius is √7, so chord length is 2*sqrt(7 -1) = 2*sqrt(6). That seems straightforward.Second method: solving quadratic, found the points, computed the distance, also got 2√6. So that's consistent. So seems correct.But let's check the distance formula again. The center is (0, √3). The line is y = √2 x. The formula for distance is |√2*0 -1*√3 +0| / sqrt( (√2)^2 + (-1)^2 ) = | -√3 | / sqrt(2 +1) = √3 / √3 = 1. Yes, that's correct. So that part was right.So chord length is 2*sqrt(r² - d²) = 2*sqrt(7 -1)=2*sqrt(6). Perfect.Therefore, the length of the chord is 2√6. I think that's the answer. Let me just check once more for any possible miscalculations.Alternatively, maybe the problem is designed to test the knowledge of chord length formula, so using that is the right approach. But regardless, both methods led to the same answer. So confident that 2√6 is the correct answer.Final AnswerThe length of the chord is boxed{2sqrt{6}}.</think>The given circle equation is (x^2 + (y - sqrt{3})^2 = 7). The center of the circle is ((0, sqrt{3})) and the radius is (sqrt{7}).The line (l) passes through the origin and is parallel to the line (sqrt{6}x - sqrt{3}y + 1 = 0). First, we find the slope of the given line by converting it to slope-intercept form:[sqrt{6}x - sqrt{3}y + 1 = 0 implies y = sqrt{2}x + frac{1}{sqrt{3}}]The slope of this line is (sqrt{2}). Therefore, the line (l) parallel to this line and passing through the origin has the equation (y = sqrt{2}x).Next, we calculate the distance from the center of the circle ((0, sqrt{3})) to the line (y = sqrt{2}x). Rewriting the line equation in standard form, we get (sqrt{2}x - y = 0). Using the distance formula:[d = frac{|sqrt{2} cdot 0 - 1 cdot sqrt{3} + 0|}{sqrt{(sqrt{2})^2 + (-1)^2}} = frac{|sqrt{3}|}{sqrt{2 + 1}} = frac{sqrt{3}}{sqrt{3}} = 1]Using the chord length formula (2sqrt{r^2 - d^2}) where (r = sqrt{7}) and (d = 1):[text{Chord length} = 2sqrt{(sqrt{7})^2 - 1^2} = 2sqrt{7 - 1} = 2sqrt{6}]Thus, the length of the chord is (boxed{2sqrt{6}}).,<think>Okay, so I need to find the length of a chord on the circle defined by the equation x² + (y - √3)² = 7. The chord is created by a line l that passes through the origin and is parallel to the line √6x - √3y + 1 = 0. Hmm, let me think through this step by step.First, let me recall that the general equation of a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. Comparing that to the given equation x² + (y - √3)² = 7, the center of the circle must be at (0, √3), right? And the radius squared is 7, so the radius r is √7. That seems straightforward.Now, the chord is intercepted by a line l that passes through the origin (0,0) and is parallel to another line given by √6x - √3y + 1 = 0. So, first, I need to find the equation of line l. To do that, since it's parallel to the given line, they must have the same slope. Let me find the slope of the given line.The given line is √6 x - √3 y + 1 = 0. Let me rearrange this into slope-intercept form (y = mx + b) to find the slope. So, moving the terms around:√6 x + 1 = √3 yDivide both sides by √3:y = (√6 / √3) x + 1/√3Simplify √6 / √3. Well, √6 is √(2*3) = √2 * √3, so √6 / √3 = √2. Therefore, the equation becomes y = √2 x + 1/√3. So the slope (m) of this line is √2. Since line l is parallel to this line, it must also have a slope of √2. But line l passes through the origin, so its equation will be y = √2 x + 0, or simply y = √2 x. That's the equation of line l.Now, the next step is to find where this line intersects the circle. The chord in question is the segment of line l that lies inside the circle, so finding the two intersection points between line l and the circle will give me the endpoints of the chord. Then, I can compute the distance between these two points to get the length of the chord.So, let's set up the equations:Circle: x² + (y - √3)² = 7Line: y = √2 xSubstitute the expression for y from the line equation into the circle equation:x² + (√2 x - √3)² = 7Let me expand the second term:(√2 x - √3)² = (√2 x)² - 2*(√2 x)*(√3) + (√3)²Compute each part:(√2 x)² = (2)x²-2*(√2 x)*(√3) = -2*√6 x(√3)² = 3So putting that back into the equation:x² + [2x² - 2√6 x + 3] = 7Combine like terms:x² + 2x² - 2√6 x + 3 = 7So 3x² - 2√6 x + 3 - 7 = 0Simplify:3x² - 2√6 x - 4 = 0Now, this is a quadratic equation in x. Let's write it as:3x² - 2√6 x - 4 = 0To solve for x, we can use the quadratic formula. For ax² + bx + c = 0, x = [-b ± √(b² - 4ac)] / (2a)Here, a = 3, b = -2√6, c = -4. Plugging these into the formula:x = [2√6 ± √( (-2√6)^2 - 4*3*(-4) ) ] / (2*3)First, compute the discriminant:D = (-2√6)^2 - 4*3*(-4)Calculate each term:(-2√6)^2 = 4*6 = 244*3*(-4) = -48So D = 24 - (-48) = 24 + 48 = 72So sqrt(D) = sqrt(72) = 6*sqrt(2)Therefore, the solutions for x are:x = [2√6 ± 6√2] / 6Simplify numerator and denominator:We can factor out 2 in the numerator:x = [2(√6 ± 3√2)] / 6 = (√6 ± 3√2)/3So x = (√6 / 3) ± (3√2 / 3) = (√6 / 3) ± √2Simplify √6 / 3. Since √6 = √2*√3, √6 / 3 = (√2*√3)/3 = √2/√3 (because √3/3 = 1/√3). Wait, maybe it's better to leave it as √6 / 3 for now.So x = (√6 ± 3√2)/3Wait, perhaps I made a miscalculation here. Let me check again.Original calculation:x = [2√6 ± 6√2] / 6Divide numerator and denominator by 2:x = [√6 ± 3√2] / 3Yes, that's correct. So x = (√6 + 3√2)/3 and x = (√6 - 3√2)/3.Hmm. Let me verify if these values make sense. Let me compute approximate numerical values to check if they are real.First, approximate √6 ≈ 2.449 and √2 ≈ 1.414.So for x1 = (√6 + 3√2)/3 ≈ (2.449 + 3*1.414)/3 ≈ (2.449 + 4.242)/3 ≈ 6.691/3 ≈ 2.230x2 = (√6 - 3√2)/3 ≈ (2.449 - 4.242)/3 ≈ (-1.793)/3 ≈ -0.598So x is approximately 2.23 and -0.598. Let me check if plugging these back into the line equation y = √2 x gives real y-values. Yes, since √2 is a real number. So y1 ≈ 1.414*2.23 ≈ 3.154, and y2 ≈ 1.414*(-0.598) ≈ -0.846. Then plugging these into the circle equation:For (2.23, 3.154):x² + (y - √3)^2 ≈ 2.23² + (3.154 - 1.732)^2 ≈ 4.97 + (1.422)^2 ≈ 4.97 + 2.02 ≈ 6.99 ≈ 7, which is close.For (-0.598, -0.846):x² + (y - √3)^2 ≈ (-0.598)^2 + (-0.846 - 1.732)^2 ≈ 0.358 + (-2.578)^2 ≈ 0.358 + 6.645 ≈ 6.996 ≈ 7. So that checks out. Good, so the solutions seem correct.Now, so we have two points of intersection: ( (√6 + 3√2)/3 , √2*(√6 + 3√2)/3 ) and ( (√6 - 3√2)/3 , √2*(√6 - 3√2)/3 )To find the length of the chord, we can compute the distance between these two points.Alternatively, since we know the line intersects the circle at two points, there is a formula for the length of the chord given a line at distance d from the center: length = 2√(r² - d²). Maybe that's a faster way.Wait, maybe using that formula would be better. Let me recall that the length of the chord is 2 times the square root of (radius squared minus the distance from the center to the line squared). So if I can find the distance from the center of the circle to the line l, then compute the chord length as 2√(r² - d²).That might be simpler than computing the distance between the two points. Let me try that approach.First, the center of the circle is (0, √3), and the line l is y = √2 x. The distance from a point (h, k) to the line ax + by + c = 0 is |ah + bk + c| / sqrt(a² + b²). But in this case, the line is y = √2 x, which can be written as √2 x - y = 0. So a = √2, b = -1, c = 0.So the distance d from center (0, √3) to the line √2 x - y = 0 is:|√2*0 - 1*√3 + 0| / sqrt( (√2)² + (-1)² ) = | -√3 | / sqrt(2 + 1) = √3 / √3 = 1. Wait, that's interesting. The distance is 1?Wait, let me check the calculation again.Numerator: |√2*0 - 1*√3 + 0| = | -√3 | = √3Denominator: sqrt( (√2)^2 + (-1)^2 ) = sqrt(2 + 1) = sqrt(3)Therefore, distance d = √3 / √3 = 1. Yes, that's correct.So the distance from the center (0, √3) to the line l is 1.The radius of the circle is √7. Therefore, the length of the chord is 2√(r² - d²) = 2√( (√7)^2 - 1^2 ) = 2√(7 - 1) = 2√6. So the chord length is 2√6.Wait, but let me confirm this with the other method, computing the distance between the two points we found earlier. Let's see.First, let's denote the two points as:Point A: ( (√6 + 3√2)/3 , √2*(√6 + 3√2)/3 )Point B: ( (√6 - 3√2)/3 , √2*(√6 - 3√2)/3 )Let me compute the difference in x-coordinates and y-coordinates.Δx = [ (√6 + 3√2)/3 - (√6 - 3√2)/3 ] = [ (√6 + 3√2 - √6 + 3√2)/3 ] = (6√2)/3 = 2√2Δy = [ √2*(√6 + 3√2)/3 - √2*(√6 - 3√2)/3 ] = √2/3 [ (√6 + 3√2) - (√6 - 3√2) ] = √2/3 [6√2] = √2 * 6√2 / 3 = (6*2)/3 = 12/3 = 4Therefore, the distance between A and B is sqrt( (Δx)^2 + (Δy)^2 ) = sqrt( (2√2)^2 + 4^2 ) = sqrt( 8 + 16 ) = sqrt(24) = 2√6. Which matches the previous result. So that's reassuring.Therefore, the length of the chord is 2√6. That should be the answer.But just to ensure that I didn't skip any steps where I might have made a mistake, let me recap:1. Found the center and radius of the circle: center at (0, √3), radius √7.2. Converted the given line √6x - √3y +1 = 0 into slope-intercept form and found its slope to be √2. Thus, the line l passing through the origin has the same slope, giving equation y = √2x.3. Calculated the distance from the center (0, √3) to the line y = √2x, which came out to 1.4. Applied the chord length formula: 2√(r² - d²) = 2√(7 - 1) = 2√6.5. Verified by directly solving the intersection points and computing the distance between them, which also gave 2√6.Therefore, the answer should be 2√6. Let me check if there's any possible mistake here.Wait, when I used the distance formula, I had to ensure that the line was in the correct form. The line y = √2x can be written as √2x - y = 0, which is correct. Then the distance formula |√2*0 -1*√3 +0| / sqrt( (√2)^2 + (-1)^2 ) = √3 / sqrt(3) = 1. That seems correct. Then the chord length is 2*sqrt(7 -1) = 2*sqrt(6). So that's correct.Alternatively, if I had miscalculated the distance, that could be an issue, but both methods give the same answer, so that's solid.Therefore, the length of the chord is 2√6.Final AnswerThe length of the chord is boxed{2sqrt{6}}.</think>To determine the length of the chord intercepted on the circle (x^2 + (y - sqrt{3})^2 = 7) by a line (l) that passes through the origin and is parallel to the line (sqrt{6}x - sqrt{3}y + 1 = 0), we follow these steps:1. Identify the circle's center and radius: - The circle's equation is (x^2 + (y - sqrt{3})^2 = 7). - Center: ((0, sqrt{3})) - Radius: (sqrt{7})2. Determine the slope of the given line: - The line (sqrt{6}x - sqrt{3}y + 1 = 0) can be rewritten in slope-intercept form: [ y = sqrt{2}x + frac{1}{sqrt{3}} ] - Slope: (sqrt{2})3. Equation of line (l): - Since line (l) passes through the origin and is parallel to the given line, its equation is (y = sqrt{2}x).4. Calculate the distance from the circle's center to the line (l): - The line (y = sqrt{2}x) can be written as (sqrt{2}x - y = 0). - Using the distance formula from a point ((h, k)) to the line (ax + by + c = 0): [ text{Distance} = frac{|sqrt{2} cdot 0 - 1 cdot sqrt{3} + 0|}{sqrt{(sqrt{2})^2 + (-1)^2}} = frac{sqrt{3}}{sqrt{3}} = 1 ]5. Apply the chord length formula: - The length of the chord is given by (2sqrt{r^2 - d^2}): [ text{Length} = 2sqrt{(sqrt{7})^2 - 1^2} = 2sqrt{7 - 1} = 2sqrt{6} ]Thus, the length of the chord is (boxed{2sqrt{6}}).]

🔑:Let the line l that is parallel to the line sqrt {6}x- sqrt {3}y+1=0 be sqrt {6}x- sqrt {3}y+c=0.Since l passes through the origin, c=0.Thus, the line l is sqrt {6}x- sqrt {3}y=0.The center of the circle x^{2}+(y- sqrt {3})^{2}=7 is (0, sqrt {3}) and its radius is r= sqrt {7}.The distance d from the center (0, sqrt {3}) to the line l: sqrt {6}x- sqrt {3}y=0 is d= frac{| sqrt {6}times0- sqrt {3}times sqrt {3}|}{ sqrt {6+3}}=1.Therefore, the length of the chord intercepted by the line l on the circle x^{2}+(y- sqrt {3})^{2}=7 is |AB|=2 sqrt {r^{2}-d^{2}}=2 sqrt {7-1}=2 sqrt {6}.Hence, the answer is boxed{2 sqrt {6}}.To solve this problem, first determine the equation of the line l: sqrt {6}x- sqrt {3}y=0. Then find the center, radius, and the distance d from the center (0, sqrt {3}) to the line l: sqrt {6}x- sqrt {3}y=0. Using these values, you can find the length of the chord.This problem tests your understanding of chord length calculation. It is a moderate-difficulty question. When solving, be sure to carefully read the question, and make proper use of the properties of circles and the formula for the distance from a point to a line.

🧠:[<think>Okay, so I have this function f(x) = x·(ln x + a x + 1) - a x + 1, and I need to solve two parts. Let me start with part (I): finding the range of values for a such that f(x) is decreasing on [1, +∞). Hmm, okay. To determine if a function is decreasing on an interval, I remember that the derivative should be less than or equal to zero throughout that interval. So, first step is to compute the derivative of f(x).Let me write out the function again to make sure I have it right:f(x) = x·(ln x + a x + 1) - a x + 1Let me expand this first. Maybe expanding will make it easier to take the derivative. Let's distribute the x:f(x) = x·ln x + a x² + x - a x + 1Combine like terms: the x terms are x and -a x. So that's (1 - a)x. So the function simplifies to:f(x) = x ln x + a x² + (1 - a)x + 1Now, let's compute the derivative f’(x). The derivative of x ln x is ln x + 1 (using the product rule: derivative of x is 1 times ln x plus x times derivative of ln x which is 1/x). Then, derivative of a x² is 2a x. Derivative of (1 - a)x is (1 - a). The derivative of the constant 1 is 0. So putting that all together:f’(x) = ln x + 1 + 2a x + (1 - a)Simplify that:f’(x) = ln x + 1 + 1 - a + 2a xSo, f’(x) = ln x + 2 - a + 2a xWait, let me check that again:Original terms:- d/dx [x ln x] = ln x + 1- d/dx [a x²] = 2a x- d/dx [(1 - a)x] = 1 - aSo adding them up: ln x + 1 + 2a x + 1 - aSo that's ln x + 2a x + (1 + 1 - a) = ln x + 2a x + (2 - a)Yes, so f’(x) = ln x + 2a x + (2 - a)So for f(x) to be decreasing on [1, +∞), we need f’(x) ≤ 0 for all x ≥ 1.Therefore, we need ln x + 2a x + (2 - a) ≤ 0 for all x ≥ 1.Hmm. So I need to find all real numbers a such that ln x + 2a x + (2 - a) ≤ 0 for x ≥ 1.This seems a bit tricky. Let me think. Since x is in [1, ∞), maybe we can analyze this inequality. Let's try to rearrange terms to solve for a.Let me write the inequality as:2a x - a + ln x + 2 ≤ 0Factor out a from the first two terms:a(2x - 1) + ln x + 2 ≤ 0So, a(2x - 1) ≤ - (ln x + 2)Therefore, if 2x - 1 > 0, which it is for x ≥ 1, since when x = 1, 2(1) -1 =1 >0, and increasing as x increases. So since 2x -1 is positive for x ≥1, we can divide both sides by 2x -1 without changing the inequality direction. So:a ≤ [ - (ln x + 2) ] / (2x -1 )So, to satisfy this inequality for all x ≥1, a must be less than or equal to the minimum of the right-hand side expression [ - (ln x + 2) ] / (2x -1 ) for x ≥1.Wait, but [ - (ln x + 2) ] / (2x -1 ) is equal to - (ln x +2)/(2x -1). So we need a ≤ this expression for all x ≥1. Therefore, the maximum value of the right-hand side (since a has to be less than or equal to all of them) would be the minimum of the RHS. Wait, no. Wait, since a has to be ≤ [something] for all x ≥1, then the maximum lower bound for a is the supremum of those values. Wait, maybe another approach. Let's consider the inequality:a ≤ [ - (ln x + 2) ] / (2x -1 )For all x ≥1, this must hold. Therefore, a must be less than or equal to the minimum of [ - (ln x + 2) ] / (2x -1 ) over x ≥1. But since [ - (ln x + 2) ] is negative (because ln x ≥0 for x ≥1, so ln x +2 ≥2, so negative of that is ≤ -2), and denominator 2x -1 is positive. So the RHS is negative. So a must be ≤ some negative number. Therefore, the maximum possible a is the supremum of the RHS, which would be the minimum of the RHS, since the RHS is negative. Wait, let me clarify.Suppose we have a function g(x) = [ - (ln x + 2) ] / (2x -1 ). Then, to have a ≤ g(x) for all x ≥1, the maximum value a can take is the infimum of g(x) over x ≥1. Wait, no. Because if a is required to be less than or equal to g(x) for all x, then the maximum possible a is the minimum of g(x), because if a is less than the smallest g(x), then it is less than all of them. Wait, but since g(x) is negative, and we need a to be less than or equal to all those negative numbers. So the largest possible a is the maximum of the g(x). Wait, no. Let's take an example. Suppose for x1, g(x1) = -3, for x2, g(x2) = -2. Then if a must be ≤ -3 and ≤ -2, then the maximum a can be is -3, because if a is -3, then it's ≤ -3 (true) and ≤ -2 (also true). Wait, but -3 is less than -2, so the maximum a can be is the minimum of g(x). Wait, but in that case, if the minimum of g(x) is -3, then a must be ≤ -3. Hmm.Wait, perhaps I need to analyze the function g(x) = [ - (ln x + 2) ] / (2x -1 ). Let's write this as:g(x) = - (ln x + 2) / (2x -1 )We need to find the minimum value of g(x) over x ≥1, because a must be ≤ g(x) for all x ≥1, so the maximum a can be is the minimum of g(x). Let me check with an example. Suppose for some x, g(x) is -3, and for another x, g(x) is -4. Then a has to be ≤ -4, because it needs to be ≤ all g(x). Therefore, the minimum value of g(x) is the upper bound for a. Hence, the range for a is (-∞, min g(x)]. Therefore, we need to find the minimum of g(x) over x ≥1.So, let's compute g(x) and find its minimum. Let's let h(x) = - (ln x + 2) / (2x -1 ). We can find its derivative to find critical points.Compute h’(x):h(x) = - (ln x + 2) / (2x -1 )Let me denote numerator as u = - (ln x + 2), denominator as v = 2x -1.Then h’(x) = (u’ v - u v’) / v²Compute u’ = derivative of - (ln x + 2) = - (1/x)v’ = derivative of 2x -1 = 2Therefore,h’(x) = [ (-1/x)(2x -1) - (- (ln x + 2))(2) ] / (2x -1)^2Simplify numerator:First term: (-1/x)(2x -1) = -2 + 1/xSecond term: - (- (ln x +2 )) * 2 = 2(ln x +2 )So total numerator:(-2 + 1/x) + 2(ln x + 2 ) = -2 + 1/x + 2 ln x +4Simplify: (-2 +4) + 1/x + 2 ln x = 2 + 1/x + 2 ln xTherefore, h’(x) = [2 + 1/x + 2 ln x] / (2x -1)^2Since the denominator is (2x -1)^2, which is always positive for x ≥1, the sign of h’(x) depends on the numerator: 2 + 1/x + 2 ln x.So, for x ≥1, let's analyze 2 + 1/x + 2 ln x.At x=1:2 + 1/1 + 2 ln 1 = 2 +1 +0=3>0As x increases, 1/x decreases, 2 ln x increases. So 2 + 1/x + 2 ln x is increasing because 2 ln x dominates. Since at x=1, it's 3, and as x increases, it tends to infinity. Therefore, the numerator is always positive for x ≥1. Therefore, h’(x) is positive for x ≥1, which implies that h(x) is increasing on [1, +∞).Therefore, h(x) is increasing on [1, +∞). So the minimum value of h(x) occurs at x=1.Compute h(1):h(1) = - (ln 1 + 2)/(2*1 -1 ) = - (0 + 2)/1 = -2And as x approaches infinity, let's see the limit of h(x):h(x) = - (ln x + 2)/(2x -1 ) ~ - (ln x)/ (2x) as x approaches infinity, which tends to 0 (since ln x grows slower than x). So h(x) approaches 0 from below because the numerator is negative and denominator is positive. So since h(x) is increasing from -2 up to 0. Therefore, the minimum value of h(x) is -2 at x=1, and it increases towards 0.Therefore, the minimum value of g(x) = h(x) is -2, achieved at x=1. Therefore, a must be ≤ -2.But wait, let me verify this. Because h(x) is increasing, so the smallest value is at x=1, which is -2, and then it increases. Therefore, to have a ≤ h(x) for all x ≥1, a must be ≤ the minimal h(x), which is -2. Therefore, the range of a is (-∞, -2].But let me check with a = -2. If a = -2, then the derivative f’(x) becomes:f’(x) = ln x + 2*(-2)x + (2 - (-2)) = ln x -4x + 4Wait, hold on. Wait, original f’(x) was ln x + 2a x + (2 - a). So if a = -2:f’(x) = ln x + 2*(-2)x + (2 - (-2)) = ln x -4x +4We need to check if this is ≤0 for all x ≥1.At x=1: ln 1 -4*1 +4 = 0 -4 +4 =0. So derivative at x=1 is 0.For x >1, let's see. Take x=2:ln 2 -8 +4 ≈ 0.6931 -4 ≈ -3.3069 <0x=1.5:ln 1.5 ≈0.4055, -4*(1.5)= -6, +4: 0.4055 -6 +4 ≈-1.5945 <0So derivative starts at 0 and becomes negative, then tends to -infinity as x increases (since -4x dominates). So with a=-2, f’(x) ≤0 for all x ≥1. Therefore, a=-2 is acceptable.If a is less than -2, say a=-3, then f’(x) = ln x +2*(-3)x + (2 - (-3))= ln x -6x +5At x=1: ln1 -6 +5=0 -6 +5=-1 <0For x>1, term -6x dominates, so derivative is even more negative. So a=-3 is also acceptable. Hence, the range is a ≤ -2. So answer for part (I) is (-∞, -2].Wait, but let's check when a is greater than -2. Suppose a=-1. Then f’(x)= ln x +2*(-1)x + (2 - (-1))= ln x -2x +3At x=1: 0 -2 +3=1>0, which violates the decreasing condition. So indeed, a must be ≤-2. Therefore, part (I) answer is a ≤ -2.Now moving to part (II): If the maximum value of f(x) is 2, find the value of a.Hmm. So f(x) has a maximum value of 2. Since we are dealing with a function on [1, +∞), but wait, the problem doesn't specify the interval here. Wait, the first part was on [1, +∞), but part (II) just says the maximum value of f(x) is 2. Is the domain still [1, +∞)? Probably, since part (I) was on that interval, maybe part (II) is also on the same interval? The problem statement says "If the maximum value of f(x) is 2", so maybe the maximum is considered over all real numbers where f(x) is defined. Wait, but f(x) =x ln x +a x² + (1 -a)x +1. The domain of f(x) is x>0 because of the ln x term. So f(x) is defined for x>0. So maybe the maximum is over x>0. But the problem might still consider the interval [1, +∞) as in part (I), but it's not specified. Wait, let me check the original problem.Original problem:(II) If the maximum value of f(x) is 2, find the value of the real number a.It doesn't specify the interval, so likely the maximum is over the entire domain of f(x), which is x>0. But perhaps, given part (I) was on [1, ∞), maybe part (II) is also on [1, ∞). But the problem doesn't specify. Hmm. This is a bit ambiguous. Let me check the original problem again.Looking back:"(II) If the maximum value of f(x) is 2, find the value of the real number a."No interval specified. So perhaps it's over the entire domain, which is x>0. However, since in part (I) the interval was [1, +infty), maybe part (II) is also on the same interval. But maybe not. Hmm. Let me proceed, assuming that the maximum is over x>0. If I get an answer that makes sense, then it's okay. If not, maybe I need to reconsider.So, first, to find the maximum of f(x). For that, we need to find critical points by setting f’(x)=0, solve for x, and then check if that corresponds to a maximum.From part (I), we have f’(x) = ln x + 2a x + (2 - a). So set f’(x)=0:ln x + 2a x + (2 - a) = 0We need to solve this equation for x, and then ensure that the critical point is a maximum. Also, the maximum value of f(x) is 2, so f(x) at that critical point should be 2.This seems complex because the equation involves both x and ln x. Let me see if we can find a relationship between a and x from the critical point.Let me denote the critical point as x = c. Then at x=c:ln c + 2a c + (2 - a) = 0 --> equation (1)And f(c) = 2.Compute f(c):f(c) = c·(ln c + a c + 1) -a c +1Expand:= c ln c + a c² + c -a c +1= c ln c + a c² + (1 - a)c +1Set this equal to 2:c ln c + a c² + (1 - a)c +1 = 2 --> equation (2)So we have two equations:1. ln c + 2a c + (2 - a) = 02. c ln c + a c² + (1 - a)c +1 = 2We need to solve these two equations for a and c.This seems complicated, but maybe we can find a substitution or express a from equation (1) and plug into equation (2).From equation (1):ln c + 2a c + 2 - a = 0Let's solve for a:2a c - a = - ln c -2a (2c -1) = - (ln c +2 )Therefore,a = - (ln c + 2 ) / (2c -1 )Now plug this into equation (2):c ln c + a c² + (1 - a)c +1 =2Substitute a:c ln c + [ - (ln c +2 ) / (2c -1 ) ] * c² + (1 - [ - (ln c +2 ) / (2c -1 ) ]) * c +1 =2Let me compute each term step by step.First term: c ln cSecond term: [ - (ln c +2 ) / (2c -1 ) ] * c² = - c² (ln c +2 ) / (2c -1 )Third term: (1 - [ - (ln c +2 ) / (2c -1 ) ]) * c = [1 + (ln c +2 ) / (2c -1 ) ] * c = c + c (ln c +2 ) / (2c -1 )Fourth term: +1So adding all together:c ln c - c² (ln c +2 ) / (2c -1 ) + c + c (ln c +2 ) / (2c -1 ) +1 =2Let me combine the terms with fractions:The second term and the third term's second part:- c² (ln c +2 ) / (2c -1 ) + c (ln c +2 ) / (2c -1 ) = [ -c² + c ] (ln c +2 ) / (2c -1 )Factor out -c(c -1):= -c(c -1)(ln c +2 ) / (2c -1 )So now the equation becomes:c ln c + [ -c(c -1)(ln c +2 ) / (2c -1 ) ] + c +1 =2Let me write the entire equation:c ln c - [ c(c -1)(ln c +2 ) / (2c -1 ) ] + c +1 -2 =0Simplify constants: +c +1 -2 = c -1So:c ln c - [ c(c -1)(ln c +2 ) / (2c -1 ) ] + c -1 =0This is getting complicated. Let me see if we can find a common denominator or simplify terms.Alternatively, maybe substitute t = 2c -1. Let's see if that helps. Not sure. Alternatively, note that 2c -1 = 2(c - 0.5). Hmm.Alternatively, let's see if we can manipulate the terms:First, write the equation as:c ln c + c -1 = [ c(c -1)(ln c +2 ) ] / (2c -1 )Multiply both sides by (2c -1):(c ln c + c -1)(2c -1) = c(c -1)(ln c +2 )Expand the left-hand side (LHS):= c ln c * (2c -1) + c*(2c -1) -1*(2c -1)= 2c² ln c - c ln c + 2c² - c -2c +1Simplify:2c² ln c - c ln c + 2c² -3c +1Right-hand side (RHS):= c(c -1)(ln c +2 )= c(c -1) ln c + 2c(c -1 )= c² ln c - c ln c + 2c² -2cSo set LHS = RHS:2c² ln c - c ln c + 2c² -3c +1 = c² ln c - c ln c + 2c² -2cSubtract RHS from both sides:2c² ln c - c ln c + 2c² -3c +1 - c² ln c + c ln c -2c² +2c =0Simplify term by term:2c² ln c - c² ln c = c² ln c- c ln c + c ln c =02c² -2c²=0-3c +2c= -c+1 remainsThus, entire equation simplifies to:c² ln c -c +1=0So we have:c² ln c -c +1=0Hmm, need to solve this equation for c>0. Let's check if c=1 is a solution:1² ln1 -1 +1=0 -1 +1=0. Yes, c=1 is a solution.Now check if there are other solutions. Let's analyze the function k(c)=c² ln c -c +1.Find k(c)=0.At c=1, k(1)=0.Compute derivative k’(c)= 2c ln c + c²*(1/c) -1 +0= 2c ln c +c -1Simplify: k’(c)=2c ln c +c -1At c=1: 2*1*0 +1 -1=0. So c=1 is a critical point.Check behavior of k(c):For c approaching 0+:c² ln c approaches 0 (since c² goes to 0 and ln c goes to -infty, but c² dominates), so c² ln c approaches 0. Then -c +1 approaches 1. So k(c) approaches 1.For c approaching +infty:c² ln c dominates, which tends to +infty, so k(c) approaches +infty.At c=1, k(c)=0.Now check the derivative k’(c)=2c ln c +c -1At c slightly less than 1, say c=0.9:2*0.9*ln0.9 +0.9 -1 ≈2*0.9*(-0.105) +0.9 -1≈ -0.189 -0.1≈ -0.289 <0At c=1: k’(1)=0At c slightly more than 1, say c=1.1:2*1.1*ln1.1 +1.1 -1≈2.2*0.0953 +0.1≈0.209 +0.1≈0.309>0Therefore, the function k(c) decreases from c=0 to c=1, reaching 0 at c=1, then increases from c=1 onwards. Therefore, the only solution to k(c)=0 is c=1. Hence, the critical point is c=1.Therefore, c=1 is the critical point where the maximum occurs. Then, using equation (1):ln1 + 2a*1 + (2 -a)=0Which simplifies to:0 + 2a +2 -a=0 → a +2=0 → a= -2Therefore, the value of a is -2. Let's verify this.If a=-2, then the function f(x)=x ln x + (-2)x² + (1 - (-2))x +1 =x ln x -2x² +3x +1Compute f(1)=1*0 -2*1 +3*1 +1=0 -2 +3 +1=2, which matches the maximum value of 2.Also, check if this is indeed a maximum. Since we found that c=1 is the only critical point, and for a=-2, the function f(x) is decreasing on [1, +infty) (from part (I)), so x=1 is the maximum point. Therefore, the maximum value is f(1)=2. So this checks out.Hence, the answer for part (II) is a= -2.Final Answer(I) The range of values for ( a ) is (boxed{(-infty, -2]}).(II) The value of the real number ( a ) is (boxed{-2}).</think>Given the function ( f(x) = x cdot (ln x + ax + 1) - ax + 1 ):(I) To find the range of values for ( a ) such that ( f(x) ) is decreasing on ([1, +infty)):1. Compute the derivative of ( f(x) ): [ f'(x) = ln x + 1 + 2ax + (1 - a) ] Simplifying, we get: [ f'(x) = ln x + 2ax + (2 - a) ]2. For ( f(x) ) to be decreasing, ( f'(x) leq 0 ) for all ( x geq 1 ): [ ln x + 2ax + (2 - a) leq 0 ]3. Rearrange the inequality to solve for ( a ): [ a(2x - 1) leq -(ln x + 2) ] [ a leq frac{-(ln x + 2)}{2x - 1} ]4. Analyze the function ( g(x) = frac{-(ln x + 2)}{2x - 1} ) for ( x geq 1 ): - Compute the derivative ( g'(x) ) and find that ( g(x) ) is increasing on ([1, +infty)). - The minimum value of ( g(x) ) occurs at ( x = 1 ): [ g(1) = -2 ]5. Therefore, ( a leq -2 ). The range of ( a ) is (boxed{(-infty, -2]}).(II) To find the value of ( a ) such that the maximum value of ( f(x) ) is 2:1. Find the critical points by setting ( f'(x) = 0 ): [ ln x + 2ax + (2 - a) = 0 ]2. Solve for ( a ) using the critical point ( x = 1 ) (since the maximum value occurs at ( x = 1 )): [ ln 1 + 2a cdot 1 + (2 - a) = 0 implies 0 + 2a + 2 - a = 0 implies a + 2 = 0 implies a = -2 ]3. Verify that ( f(1) = 2 ) when ( a = -2 ): [ f(1) = 1 cdot (ln 1 - 2 cdot 1 + 1) - (-2) cdot 1 + 1 = 2 ]Thus, the value of ( a ) is (boxed{-2}).,<think>Okay, so I have this problem here with two parts, (I) and (II). Let me start by understanding what's given. The function is f(x) = x·(ln x + a x + 1) - a x + 1. Hmm, that's a bit complex. Let me first try to simplify it if possible.So, expanding the first term: x multiplied by (ln x + a x + 1). That would be x·ln x + a x² + x. Then subtract a x and add 1. So combining like terms: x·ln x + a x² + x - a x + 1. The x and -a x can combine to (1 - a)x. So overall, f(x) = a x² + x ln x + (1 - a)x + 1. Maybe that's a bit simpler? Let me check:Original function: x(ln x + a x + 1) - a x + 1Expanding: x ln x + a x² + x - a x + 1Yes, so combining the x terms: x - a x = (1 - a)x. So f(x) = a x² + x ln x + (1 - a)x + 1. Not sure if that helps yet, but maybe.Now, part (I): If f(x) is a decreasing function on [1, +∞), find the range of a. So for a function to be decreasing on an interval, its derivative should be non-positive there, right? So f'(x) ≤ 0 for all x ≥ 1. So I need to compute the derivative of f(x) and find the values of a such that this derivative is ≤ 0 for x ≥ 1.Let me compute f'(x). Starting from the original function f(x) = x(ln x + a x + 1) - a x + 1.First, let's differentiate term by term. The derivative of x(ln x + a x + 1) is done using the product rule: derivative of x times (ln x + a x + 1) plus x times derivative of (ln x + a x + 1). So:d/dx [x(ln x + a x + 1)] = 1·(ln x + a x + 1) + x·(1/x + a + 0) = ln x + a x + 1 + (1 + a x)Wait, let's compute that again:First term: derivative of x is 1, times (ln x + a x + 1). So that's (ln x + a x + 1).Second term: x times the derivative of (ln x + a x + 1). The derivative of ln x is 1/x, derivative of a x is a, and derivative of 1 is 0. So that's (1/x + a). So the second term is x*(1/x + a) = 1 + a x.So altogether, the derivative of the first part is (ln x + a x + 1) + (1 + a x) = ln x + a x + 1 + 1 + a x = ln x + 2 a x + 2.Then, the derivative of the remaining terms: -a x + 1. The derivative of -a x is -a, and derivative of 1 is 0. So overall, f'(x) = ln x + 2 a x + 2 - a.So f'(x) = ln x + 2 a x + (2 - a).We need this derivative to be ≤ 0 for all x ≥ 1.So the problem reduces to: find real numbers a such that ln x + 2 a x + (2 - a) ≤ 0 for all x ≥ 1.Hmm. So this inequality must hold for all x ≥ 1. Let's rearrange terms to solve for a.Let me write the inequality as:ln x + 2 a x + (2 - a) ≤ 0.Let me group the terms with a:2 a x - a + ln x + 2 ≤ 0.Factor out a:a(2 x - 1) + ln x + 2 ≤ 0.So solving for a:a(2 x - 1) ≤ - (ln x + 2).If 2x - 1 > 0, which it is for x ≥ 1, since 2*1 -1 =1 >0, and x is increasing, so 2x -1 is positive. Therefore, we can divide both sides by 2x -1 without changing the inequality direction:a ≤ - (ln x + 2)/(2x - 1).So for the inequality to hold for all x ≥1, the value of a must be less than or equal to the minimum of the right-hand side expression - (ln x + 2)/(2x -1) over x ≥1. Therefore, a ≤ min_{x ≥1} [ - (ln x +2)/(2x -1) ].Alternatively, since we have a ≤ [ - (ln x +2) ] / (2x -1 ) for all x ≥1, the maximum lower bound for a is the infimum of [ - (ln x +2) ] / (2x -1 ) over x ≥1. Therefore, the maximum value of a that satisfies this is the minimum of the right-hand side. Wait, actually, since a has to be less than or equal to that expression for all x, then the maximum allowable a is the infimum (greatest lower bound) of that expression over x ≥1. But perhaps it's easier to think of the minimum of the right-hand side, but since the right-hand side is - (ln x +2)/(2x -1 ), we need the minimum of that function.But let's define g(x) = - (ln x +2)/(2x -1 ). Then to find the minimum value of g(x) over x ≥1, which would give the maximum a such that a ≤ g(x) for all x ≥1. Therefore, a must be ≤ the minimum of g(x) over x ≥1. Wait, but actually, since a has to be ≤ g(x) for all x ≥1, then the maximum possible a is the infimum of g(x) over x ≥1.But perhaps there is a minimum of g(x). Let's analyze g(x) = - (ln x +2)/(2x -1 ). Let's compute its derivative to find critical points.First, write g(x) as - (ln x +2)/(2x -1 ). Let's compute g'(x):g'(x) = - [ ( (1/x)(2x -1 ) - (ln x +2)(2) ) / (2x -1 )² ]Simplify numerator:(2x -1)/x - 2(ln x +2 )= (2 - 1/x) - 2 ln x -4= 2 - 1/x -2 ln x -4= -2 ln x -1/x -2Therefore, g'(x) = - [ (-2 ln x -1/x -2 ) ] / (2x -1 )² = (2 ln x +1/x +2 ) / (2x -1 )².Now, set g'(x) =0:2 ln x +1/x +2 =0.Hmm, need to solve 2 ln x + 1/x +2=0. Let's see if this equation has any solutions for x ≥1.At x=1:2 ln1 +1/1 +2= 0 +1 +2=3>0.As x increases, let's see the behavior:As x approaches infinity, 2 ln x tends to infinity, 1/x tends to 0, so the whole expression tends to infinity. So 2 ln x +1/x +2 is always positive for x ≥1. Therefore, the derivative g'(x) is positive for all x ≥1. Therefore, g(x) is increasing on [1, +∞).Wait, if g(x) is increasing on [1, +∞), then its minimum is at x=1. Therefore, the minimum value of g(x) is g(1)= - (ln1 +2)/(2*1 -1)= - (0 +2)/1= -2.Therefore, since g(x) is increasing from -2 upwards as x increases from 1 to infinity, the minimum value of g(x) on [1, +∞) is -2. Thus, the maximum value a can take is -2. Therefore, the range of a is a ≤ -2.Wait, but let me verify this. If a ≤ -2, then for all x ≥1, a ≤ - (ln x +2)/(2x -1 ). Since the right-hand side g(x) is increasing from -2 upwards, but if a is ≤ -2, then since for x=1, the RHS is -2, and for x>1, the RHS is greater than -2 (since g(x) is increasing), then a ≤ -2 would satisfy a ≤ g(x) for all x ≥1. For example, take a = -2. Then for x=1, the derivative f'(1) = ln1 +2*(-2)*1 + (2 - (-2)) =0 -4 +4=0. Then, for x>1, since g(x) > -2, so a=-2 is allowed, and f'(x)=ln x +2*(-2)x + (2 - (-2))=ln x -4x +4. Wait, but is this ≤0 for all x ≥1?Wait, maybe I need to check whether when a=-2, the derivative is indeed ≤0 for all x ≥1.Let's take a=-2. Then f'(x)=ln x +2*(-2)x +2 - (-2)=ln x -4x +4. We need to check if ln x -4x +4 ≤0 for all x ≥1.At x=1: ln1 -4*1 +4=0 -4 +4=0.At x=2: ln2 -8 +4 ≈0.693 -4≈-3.307 <0.As x increases, -4x dominates, so ln x -4x +4 tends to -infinity, so it's decreasing. Let's check if there's any x where it could be positive. Let's check the derivative of f'(x) when a=-2. The derivative of f'(x) is d/dx (ln x -4x +4)=1/x -4. At x=1, derivative is 1 -4=-3 <0. For x>1, 1/x -4 is always negative. So f'(x) is decreasing. Since at x=1, f'(1)=0, and it's decreasing, so for x>1, f'(x) <0. Therefore, for a=-2, f'(x) ≤0 for all x ≥1. Hence, a=-2 is allowed. Now, if a is less than -2, say a=-3, then f'(x)=ln x +2*(-3)x +2 - (-3)=ln x -6x +5. At x=1, this is 0 -6 +5=-1 <0, and as x increases, the term -6x dominates, so it's even more negative. So for a < -2, f'(x) would be even smaller, so it's ≤0. Therefore, the conclusion is that a must be ≤ -2.Therefore, part (I)'s answer is a ≤ -2.Now, moving on to part (II): If the maximum value of f(x) is 2, find the value of a.So f(x) has a maximum value of 2. Since the domain is [1, +∞), we need to check if the maximum occurs somewhere in this interval.First, note that if f(x) has a maximum at some point in [1, +∞), then that maximum must occur at a critical point where f'(x)=0, or at the endpoint x=1.But for part (II), since the maximum is 2, we need to find a such that the maximum of f(x) on [1, +∞) is exactly 2.First, let's recall that in part (I), when a ≤ -2, f(x) is decreasing on [1, +∞), so its maximum would be at x=1. So if in part (II), the maximum is 2, then if a ≤ -2, the maximum is f(1). Let's compute f(1):f(1)=1·(ln1 +a·1 +1) -a·1 +1=1·(0 +a +1) -a +1= (a +1) -a +1= 2. So f(1)=2 for any a. Wait, that's interesting. So if a ≤ -2, then f(x) is decreasing on [1, +∞), so the maximum is at x=1, which is 2. So that would satisfy part (II) as well. But part (II) says "the maximum value of f(x) is 2", so if a ≤ -2, then the maximum is 2. But the problem says "find the value of the real number a". So perhaps a is unique? Wait, but according to part (I), for all a ≤ -2, f(x) is decreasing, so maximum is 2. But maybe part (II) is considering cases where the function might have a critical point inside [1, ∞) where the maximum is 2, even if the function is not necessarily decreasing everywhere. So perhaps there's another value of a where the function has a critical point at some x ≥1, and f(x) reaches 2 there, and that's the maximum. But if so, then there could be two possible a's: one from part (I) where the function is decreasing (so maximum at x=1), and another where the function first increases then decreases, reaching a maximum at some x>1, which is also 2. Let me check.So first, check the case when a > -2. In part (I), if a > -2, then the function f(x) is not decreasing everywhere on [1, ∞). So it might have a critical point where f'(x)=0. If that critical point is a maximum, then the maximum value could be 2. So we need to consider both possibilities: either the maximum is at x=1 (which is always 2, regardless of a, as f(1)=2) or at a critical point x>1 where f(x)=2. But since the problem states that the maximum value is 2, it could be that even if there is a critical point, the value there is also 2, but not higher. So perhaps only a unique a where the critical point is exactly at x=1, but since f(1)=2, and if the derivative at x=1 is zero (which would be the case when a=-2, as in part (I)), then maybe that's the only solution? Wait, but when a=-2, the function is decreasing everywhere, so maximum at x=1. For a > -2, the function might first increase then decrease, but if the maximum value is still 2, which occurs at x=1, even though there is another critical point. Wait, this is a bit confusing. Let me proceed step by step.First, compute f(1) for any a: as above, f(1)=2. So regardless of a, f(1)=2. Now, if a > -2, then the function f(x) might have a critical point somewhere in [1, ∞). Let's find critical points by setting f'(x)=0.From part (I), f'(x)=ln x +2 a x + (2 - a).Set f'(x)=0:ln x +2 a x + (2 - a)=0.If a > -2, then perhaps this equation has a solution x >1. Let's suppose that there's a solution x=c >1 where f'(c)=0. Then, if f(c)=2, that would be the maximum. So perhaps we need to solve the system:1. ln c +2 a c + (2 - a)=0.2. f(c)=2.So two equations with two variables (a and c). Let's write down f(c)=2:f(c)=a c² +c ln c + (1 - a)c +1=2.Simplify:a c² +c ln c + (1 - a)c +1 =2.Subtract 1:a c² +c ln c + (1 - a)c =1.Factor terms with a:a(c² -c) +c ln c +c =1.So:a(c² -c) =1 -c ln c -c.Thus,a= [1 -c ln c -c ] / (c² -c )= [1 -c(ln c +1)] / [c(c -1)].So equation 1 gives:ln c +2 a c + (2 -a)=0.But a is expressed in terms of c. Let's substitute a from above into equation 1.First, let me write equation 1:ln c +2 a c +2 -a=0.Substitute a:ln c +2* [ (1 -c(ln c +1)) / (c(c -1)) ] *c +2 - [ (1 -c(ln c +1)) / (c(c -1)) ]=0.Simplify step by step:First term: ln c.Second term: 2*[ (1 -c(ln c +1)) / (c(c -1)) ]*c =2*[ (1 -c(ln c +1)) / (c -1) ].Third term: +2.Fourth term: - [ (1 -c(ln c +1)) / (c(c -1)) ].So putting it all together:ln c + [2(1 -c(ln c +1))/(c -1)] +2 - [ (1 -c(ln c +1)) / (c(c -1)) ]=0.This looks complicated. Let's try to combine the terms.Let me denote A = 1 -c(ln c +1). Then:ln c + [2A/(c -1)] +2 - [A/(c(c -1))] =0.Factor out A:ln c +2 + A[ 2/(c -1) - 1/(c(c -1)) ]=0.Compute the coefficient of A:2/(c -1) -1/(c(c -1))= [2c -1]/[c(c -1)].Therefore:ln c +2 + A*(2c -1)/[c(c -1)]=0.But A=1 -c(ln c +1)=1 -c ln c -c.So substitute back:ln c +2 + [1 -c ln c -c]*(2c -1)/[c(c -1)]=0.Now, this is a complex equation in c. Let me see if I can simplify this.Let me compute [1 -c ln c -c]*(2c -1)/[c(c -1)].First, expand the numerator:[1 -c -c ln c]*(2c -1)= (1 -c)(2c -1) -c ln c*(2c -1).Compute (1 -c)(2c -1):=2c -1 -2c² +c= (2c +c) -1 -2c²=3c -1 -2c².Then, -c ln c*(2c -1)= -2c² ln c +c ln c.Therefore, the entire numerator is:3c -1 -2c² -2c² ln c +c ln c.Divide by denominator c(c -1):So:[3c -1 -2c² -2c² ln c +c ln c]/[c(c -1)].Therefore, the entire equation is:ln c +2 + [3c -1 -2c² -2c² ln c +c ln c]/[c(c -1)] =0.This seems very complicated. Maybe there's a better approach.Alternatively, perhaps when the maximum occurs at x=1, which is 2, and if a > -2, then the function might have a local maximum somewhere else greater than 2, which would contradict the maximum being 2. Therefore, the only possible a is a=-2, where the function is decreasing, so the maximum is at x=1, which is 2. However, maybe there's another a where the function has a critical point at some x>1, and f(c)=2, and also at x=1, f(1)=2, and between 1 and c, the function doesn't exceed 2. So maybe there are two possibilities.But let's test with specific values. Let me take a=-1, which is greater than -2. Then check if the function has a maximum greater than 2.Compute f(x) when a=-1:f(x)= -1 x² +x ln x + (1 - (-1))x +1= -x² +x ln x +2x +1.At x=1: -1 +0 +2 +1=2.Now, check the derivative f'(x)=ln x +2*(-1)x +2 - (-1)=ln x -2x +3.Set derivative to zero: ln x -2x +3=0.Find x where this is true. Let's try x=1: ln1 -2 +3=0 -2 +3=1≠0.x=2: ln2 -4 +3≈0.693 -1≈-0.307≠0.x=1.5: ln1.5≈0.405, -2*1.5= -3, +3=0. So 0.405 -3 +3=0.405≠0.x=3: ln3≈1.098, -6 +3≈-3. So ≈1.098 -3≈-1.902≠0.Wait, seems like the equation ln x -2x +3=0 may not have a solution. Let's check the behavior.As x approaches 1 from the right:At x=1+, ln x~0, -2x+3~1, so positive.At x=2, derivative≈ -0.307 <0.At x approaching infinity, ln x -2x +3 tends to -infinity.So the derivative starts positive at x=1, decreases, and becomes negative. Therefore, there must be some x=c in (1, ∞) where derivative=0. So the function f(x) when a=-1 first increases from x=1 to x=c, then decreases after that. So the maximum is at x=c. Now, compute f(c). If f(c)=2, then that would be the solution. But let's see.However, solving for c in ln c -2c +3=0 is difficult analytically. Let me try numerically.Let me set g(c)=ln c -2c +3. Find c where g(c)=0.At c=1: 0 -2 +3=1.At c=1.5: ln1.5≈0.405, -3 +3=0.405>0.At c=2: ln2≈0.693, -4 +3≈-0.307. So somewhere between 1.5 and 2.Use Newton-Raphson:Let’s take c0=1.5:g(1.5)=0.4055 -3 +3=0.4055.g’(c)=1/c -2.At c=1.5, g’=1/1.5 -2≈0.6667 -2≈-1.3333.Next iteration:c1=1.5 - g(c)/g’(c)=1.5 -0.4055/(-1.3333)=1.5 +0.304≈1.804.Compute g(1.804)=ln(1.804)≈0.591, -2*1.804≈-3.608, +3=0.591 -3.608 +3≈0.591 -0.608≈-0.017.Close to zero. Compute g’(1.804)=1/1.804≈0.554 -2≈-1.446.Next iteration: c2=1.804 - (-0.017)/(-1.446)=1.804 -0.017/1.446≈1.804 -0.0118≈1.792.Compute g(1.792)=ln(1.792)=0.583, -2*1.792≈-3.584, +3=0.583 -3.584 +3≈0.583 -0.584≈-0.001.Almost zero. So c≈1.792.Now compute f(c)=2?Compute f(1.792):a=-1, so f(x)=-x² +x ln x +2x +1.Compute x=1.792:x²≈3.212.x ln x≈1.792*0.583≈1.044.2x≈3.584.So f(x)= -3.212 +1.044 +3.584 +1≈(-3.212 +1.044)= -2.168 +3.584=1.416 +1≈2.416>2.Therefore, when a=-1, the maximum value is approximately 2.416>2. So the maximum exceeds 2. Therefore, a=-1 is invalid.Similarly, if we take a between -2 and some higher value, perhaps there exists an a where the maximum at x=c is exactly 2. But this requires solving the system of equations:1. ln c +2 a c +2 -a=0.2. a c² +c ln c + (1 - a)c +1=2.But this seems complicated. Alternatively, maybe there's only one solution where a=-2, and for a >-2, the maximum is greater than 2, so the only possible a is -2. But let's check for a=-1.5, which is between -2 and -1.Wait, a=-2 is the boundary from part (I). If I take a=-1.5, which is greater than -2, then the function might have a maximum higher than 2.Let’s try a=-1.5.Compute f(x)= -1.5 x² +x ln x + (1 - (-1.5))x +1= -1.5x² +x ln x +2.5x +1.Compute derivative f’(x)=ln x +2*(-1.5)x +2.5 -(-1.5)=ln x -3x +2.5 +1.5=ln x -3x +4.Set derivative zero: ln x -3x +4=0.Let’s find x where this holds.At x=1: 0 -3 +4=1>0.At x=1.5: ln1.5≈0.405, -4.5 +4= -0.5 +0.405≈-0.095.So between 1 and1.5.Use Newton-Raphson:g(c)=ln c -3c +4.g’(c)=1/c -3.Starting with c=1.2:g(1.2)=ln1.2≈0.182, -3.6 +4=0.182+0.4=0.582>0.c=1.3: ln1.3≈0.262, -3.9 +4=0.262+0.1≈0.362>0.c=1.4: ln1.4≈0.336, -4.2 +4≈0.336-0.2≈0.136>0.c=1.45: ln1.45≈0.372, -4.35 +4≈-0.35+0.372≈0.022>0.c=1.46: ln1.46≈0.378, -4.38 +4≈-0.38+0.378≈-0.002.So c≈1.46.Compute f(c):x=1.46:f(x)= -1.5*(1.46)^2 +1.46*ln1.46 +2.5*1.46 +1.Compute 1.46²≈2.13.So -1.5*2.13≈-3.195.1.46*ln1.46≈1.46*0.378≈0.552.2.5*1.46≈3.65.So f(x)= -3.195 +0.552 +3.65 +1≈(-3.195 +0.552)= -2.643 +3.65=1.007 +1≈2.007≈2.01≈2. So close to 2. Maybe with more precise calculation, it's exactly 2.Wait, that's interesting. So when a=-1.5, the maximum at x≈1.46 is approximately 2.01, which is close to 2. Maybe with a precise a, the maximum is exactly 2. So perhaps there exists another a where the maximum at critical point is exactly 2, in addition to a=-2 where the maximum is at x=1.But this requires solving the system of equations. Let me try to see if a=-2 is the only solution.Alternatively, since when a=-2, the function is decreasing, so maximum at x=1, which is 2. When a >-2, the function first increases to a maximum and then decreases. If that maximum is exactly 2, then the value at the critical point is 2. But since f(1)=2 regardless of a, if the function has a critical point x>1 where f(x)=2, then between x=1 and x=c, the function would have values higher than 2, unless the function is constant. Wait, but f(x) is not a constant function. Wait, unless the critical point is exactly at x=1, but when a=-2, f’(1)=0, and for x>1, f’(x)<0. Therefore, the maximum is at x=1. If a >-2, then f’(1)=ln1 +2a*1 +2 -a=0 +2a +2 -a= a +2. For a >-2, this is positive, so the function is increasing at x=1. Therefore, the maximum would be at some x>1, which might have f(x)>2.But in the case when a=-1.5, the maximum at x≈1.46 is approximately 2.01, which is slightly above 2. If we adjust a to be slightly less than -1.5, maybe the maximum decreases to exactly 2. However, since when a decreases, the critical point x might change, but this requires solving the equations.Alternatively, perhaps the only possible value is a=-2. Because for a >-2, the function has a maximum greater than 2, and for a ≤-2, the maximum is exactly 2. But the problem says "the maximum value of f(x) is 2", which could be achieved for a ≤-2, but the question asks for "the value of the real number a", implying a unique answer. However, in part (I), a range is requested, while part (II) asks for a specific value. This suggests that perhaps there is only one a where the maximum is exactly 2, achieved at a critical point. But when a=-2, the maximum is at x=1, which is 2. For a < -2, the function is decreasing, so maximum at x=1, which is still 2. But the problem says "the maximum value of f(x) is 2". So for all a ≤-2, the maximum is 2. But why does part (II) ask for "the value" instead of "the range"? Maybe there is a mistake in my reasoning.Wait, let's check f(1) again. f(1)=2 regardless of a. So for any a, f(1)=2. But if the function is decreasing for a ≤-2, then the maximum is at x=1, which is 2. If a >-2, then the function first increases to a peak and then decreases. If that peak is higher than 2, then the maximum is greater than 2. If that peak is exactly 2, then even though the function has a critical point, the maximum value is still 2. But how can the peak be 2? Since f(1)=2, and if the function increases after x=1, reaches a peak at x=c>1 where f(c)=2, then between x=1 and x=c, the function must be higher than 2. But that contradicts the maximum being 2. Therefore, the only way for the maximum to be 2 is that the function does not exceed 2 anywhere. Therefore, if a >-2, the function increases immediately after x=1, which would make f(x) >2 near x=1+, contradicting the maximum being 2. Hence, the only possible a is a ≤-2. But the problem states "the maximum value of f(x) is 2", which holds for all a ≤-2. However, part (II) asks for "the value of a", suggesting a unique answer, so maybe the question has a mistake, or perhaps I'm missing something.Wait, let me check when a=-2. Then f(x) is decreasing on [1, ∞), so maximum at x=1, which is 2. If a < -2, same thing. But for example, take a=-3. Then f(x)= -3x² +x ln x +4x +1. Compute f(1)= -3 +0 +4 +1=2. The derivative is ln x +2*(-3)x +2 -(-3)=ln x -6x +5. At x=1: 0 -6 +5=-1 <0, so decreasing. Thus, the maximum is 2 at x=1. Similarly for any a ≤-2. So why does part (II) ask for "the value"? Maybe there's a typo, and it should be "the minimum value of f(x) is 2", but as stated, "maximum value is 2", which is achieved for all a ≤-2. However, the problem might expect considering that when a >-2, the function could have a local maximum at some point x>1, but if that local maximum is exactly 2, then a would be a specific value. But from the previous example with a=-1.5, the local maximum was approximately 2.01, so adjusting a could bring it down to 2. However, solving for this requires the system of equations:1. f'(c)=02. f(c)=2But since f(1)=2 regardless of a, maybe the function could touch 2 at x=1 and another point, making the maximum 2. But this seems unlikely unless the function is constant, which it's not.Alternatively, maybe the problem considers only the global maximum, not allowing any higher values. So the only way to ensure that the maximum is exactly 2 is if the function is non-increasing, so a ≤-2. However, the problem says "the maximum value of f(x) is 2", which could technically be achieved for any a ≤-2, but the answer might expect a=-2, the boundary case. Alternatively, perhaps there's an additional constraint.Wait, let's check when a=-2, f(x) is decreasing, so maximum at x=1. When a < -2, f(x) is also decreasing, so maximum at x=1. So in both cases, maximum is 2. Thus, the answer for part (II) would be all a ≤-2. But since the problem says "find the value of the real number a", using "the value" suggests a unique answer, which contradicts. This might indicate an error in the problem statement or in my reasoning.Wait, let me re-examine part (II). It says "If the maximum value of f(x) is 2, find the value of the real number a". So perhaps the problem is considering that when a >-2, the function could attain maximum value greater than 2, and when a ≤-2, it is 2. So the answer would be a ≤-2. But the problem says "the value", singular. So maybe there's a misunderstanding.Alternatively, maybe there is a specific a where the function has its maximum at 2, which could occur at a critical point. But as shown earlier, when a=-1.5, the maximum is slightly above 2, so there might be a specific a where the maximum at the critical point is 2, which would be a unique solution. Let's try to find that.Assume that there exists a c>1 where f(c)=2 and f'(c)=0.From f(c)=2:a c² +c ln c + (1 -a)c +1=2.Which simplifies to:a(c² -c) +c ln c +c +1=2.So:a(c² -c)=1 -c ln c -c.Thus,a=(1 -c ln c -c)/(c² -c).From f'(c)=0:ln c +2 a c +2 -a=0.Substitute a from above:ln c +2 c*(1 -c ln c -c)/(c² -c) +2 - (1 -c ln c -c)/(c² -c)=0.This is very complex. Let me denote D=c² -c.Then:ln c + [2c(1 -c ln c -c)/D] +2 - (1 -c ln c -c)/D=0.Combine terms:ln c +2 + [2c(1 -c ln c -c) - (1 -c ln c -c)]/D=0.Factor numerator:(1 -c ln c -c)(2c -1) / D + ln c +2=0.But D=c(c -1).Thus:(1 -c ln c -c)(2c -1)/[c(c -1)] + ln c +2=0.This is the same equation as before. It seems difficult to solve analytically. Let's attempt to find c such that this equation holds.Let me guess that c=1. But then denominator becomes zero, so invalid. Try c approaches 1 from above. But not helpful. Alternatively, suppose that at c=1, the equation is satisfied, but c=1 is the endpoint. However, when c=1, f(c)=2, and a can be any ≤-2. Maybe there's no solution with c>1. But earlier numeric example with a=-1.5 showed a maximum slightly above 2. If I set up the equation f(c)=2 and f'(c)=0 and solve numerically for a.Alternatively, perhaps the only solution is a=-2. Because when a=-2, the function is decreasing, so maximum at x=1. For a>-2, the function increases past x=1, making the maximum greater than 2. Therefore, the only possible a is a=-2. Therefore, part (II)'s answer is a=-2.But why then when I computed a=-1.5, the maximum was approximately 2.01? Maybe my calculations were inaccurate. Let me recompute f(c) for a=-1.5 and c≈1.46.Compute f(1.46):a=-1.5, x=1.46f(x)= -1.5*(1.46)^2 +1.46*ln(1.46) +2.5*1.46 +1First, (1.46)^2=2.1316-1.5*2.1316≈-3.19741.46*ln1.46≈1.46*0.378≈0.5522.5*1.46=3.65So f(x)= -3.1974 +0.552 +3.65 +1≈(-3.1974 +0.552)= -2.6454 +3.65=1.0046 +1≈2.0046≈2.005. So approximately 2.005, very close to 2. If we adjust a slightly more negative, say a=-1.6.Compute a=-1.6, find critical point.f’(x)=ln x +2*(-1.6)x +2 -(-1.6)=ln x -3.2x +3.6.Set to zero: ln x -3.2x +3.6=0.Find x:At x=1.1:ln1.1≈0.095, -3.52 +3.6≈0.095+0.08≈0.175>0.x=1.2:ln1.2≈0.182, -3.84 +3.6≈0.182 -0.24≈-0.058<0.So root between 1.1 and1.2.Use linear approximation:At x=1.1, g=0.175.At x=1.2, g=-0.058.Slope≈(-0.058 -0.175)/(1.2 -1.1)=(-0.233)/0.1=-2.33 per 0.1.Need to find Δx where 0.175 -2.33Δx=0 → Δx≈0.175/2.33≈0.075.Thus, c≈1.1 +0.075≈1.175.Compute f(1.175):a=-1.6, x=1.175f(x)= -1.6*(1.175)^2 +1.175*ln(1.175) +2.6*1.175 +1.Compute:(1.175)^2≈1.3806-1.6*1.3806≈-2.209.ln1.175≈0.1625.1.175*0.1625≈0.191.2.6*1.175≈3.055.So f(x)= -2.209 +0.191 +3.055 +1≈(-2.209 +0.191)= -2.018 +3.055=1.037 +1≈2.037≈2.04.Still above 2. If I set a=-1.8.Compute critical point.f’(x)=ln x +2*(-1.8)x +2 -(-1.8)=ln x -3.6x +3.8.Set to zero: ln x -3.6x +3.8=0.At x=1:0 -3.6 +3.8=0.2>0.At x=1.05:ln1.05≈0.0488, -3.78 +3.8≈0.0488+0.02≈0.0688>0.At x=1.1:ln1.1≈0.0953, -3.96 +3.8≈0.0953 -0.16≈-0.0647<0.Thus, root between1.05 and1.1.Using linear approximation:At x=1.05, g=0.0688.At x=1.1, g=-0.0647.Slope≈(-0.0647 -0.0688)/0.05≈-0.1335/0.05≈-2.67 per 0.1.Δx needed: 0.0688 /2.67≈0.0258.Thus, c≈1.05 +0.0258≈1.0758.Compute f(1.0758):a=-1.8, x≈1.0758.f(x)= -1.8x² +x ln x + (1 -(-1.8))x +1= -1.8x² +x ln x +2.8x +1.Calculate:x²≈1.0758²≈1.1574.-1.8*1.1574≈-2.0833.x ln x≈1.0758*0.0725 (ln1.0758≈0.0725)≈0.078.2.8x≈2.8*1.0758≈3.012.So f(x)= -2.0833 +0.078 +3.012 +1≈(-2.0833 +0.078)= -2.0053 +3.012≈1.0067 +1≈2.0067≈2.007.Still above 2. It seems that as a approaches -2 from above, the maximum approaches 2. For example, if a approaches -2, then the critical point approaches x=1, and f(x) approaches 2. Let's take a approaches -2.Let a=-2 + ε, where ε approaches 0+.Then, the derivative f’(x)=ln x +2(-2 + ε)x +2 -(-2 + ε)=ln x -4x +2ε x +4 -ε.Set f’(x)=0:ln x -4x +2ε x +4 -ε=0.As ε approaches 0, this becomes ln x -4x +4=0.Solve ln x -4x +4=0.At x=1: 0 -4 +4=0. So x=1 is a solution.The derivative when ε=0 is ln x -4x +4, which at x=1 is 0, and the next derivative is 1/x -4. At x=1, it's -3 <0, so x=1 is a local maximum when ε=0. But as ε increases slightly (a approaches -2 from above), the equation ln x -4x +2ε x +4 -ε=0 will have a solution near x=1.Let’s expand around x=1 + δ, where δ is small.Let x=1 + δ, δ≈0.Then ln x≈δ -δ²/2.-4x≈-4(1 +δ)= -4 -4δ.2ε x≈2ε(1 +δ).4 -ε≈4 -ε.Plug into the equation:(δ -δ²/2) -4 -4δ +2ε(1 +δ) +4 -ε=0.Simplify:δ -δ²/2 -4 -4δ +2ε +2εδ +4 -ε=0.Combine like terms:(δ -4δ) + (-δ²/2) + (-4 +4) + (2ε -ε) +2εδ=0.→ -3δ -δ²/2 + ε +2εδ=0.Neglecting higher-order terms (δ², εδ):-3δ +ε≈0 → δ≈ε/3.So x≈1 + ε/3.Now compute f(x)= a x² +x ln x + (1 -a)x +1 with a=-2 + ε and x=1 + ε/3.Plug in:f(x)= (-2 + ε)(1 + ε/3)^2 + (1 + ε/3)ln(1 + ε/3) + (1 -(-2 + ε))(1 + ε/3) +1.Expand term by term:First term: (-2 + ε)(1 + 2ε/3 + ε²/9)≈-2(1 +2ε/3) + ε(1) ≈-2 -4ε/3 +ε= -2 -ε/3.Second term: (1 + ε/3)(ε/3 - (ε/3)^2/2)≈(1 + ε/3)(ε/3)≈ε/3 + ε²/9 ≈ε/3.Third term: (3 - ε)(1 + ε/3)=3(1 + ε/3) - ε(1 + ε/3)=3 + ε - ε - ε²/3≈3.Fourth term: +1.Thus, total f(x)≈-2 -ε/3 + ε/3 +3 +1= (-2 +3 +1)=2. So up to first order, f(x)=2. Hence, as ε approaches 0 (a approaches -2 from above), the maximum value approaches 2. Therefore, the limiting case is a=-2, where the maximum is exactly 2 at x=1. For a slightly greater than -2, the maximum is slightly above 2. Therefore, the only value where the maximum is exactly 2 is a=-2. Hence, part (II)'s answer is a=-2.Final Answer(I) boxed{(-infty, -2]}(II) boxed{-2}</think>Given the function ( f(x) = x(ln x + ax + 1) - ax + 1 ):(I) To find the range of values for ( a ) such that ( f(x) ) is a decreasing function on ([1, +infty)):1. Compute the derivative ( f'(x) ): [ f'(x) = ln x + 2ax + (2 - a) ]2. For ( f(x) ) to be decreasing, ( f'(x) leq 0 ) for all ( x geq 1 ): [ ln x + 2ax + (2 - a) leq 0 ]3. Rearrange the inequality to solve for ( a ): [ a(2x - 1) leq -(ln x + 2) ]4. Since ( 2x - 1 > 0 ) for ( x geq 1 ), divide both sides by ( 2x - 1 ): [ a leq -frac{ln x + 2}{2x - 1} ]5. Analyze the right-hand side function ( g(x) = -frac{ln x + 2}{2x - 1} ). Find its minimum value on ([1, +infty)) by checking the derivative and behavior. It is found that ( g(x) ) is increasing on ([1, +infty)), hence the minimum value is at ( x = 1 ): [ g(1) = -2 ]6. Therefore, ( a leq -2 ).(II) To find the value of ( a ) such that the maximum value of ( f(x) ) is 2:1. Check the value of ( f(x) ) at ( x = 1 ): [ f(1) = 2 text{ for any } a ]2. For ( f(x) ) to have a maximum value of 2, consider the critical points where ( f'(x) = 0 ). If ( a > -2 ), the function might have a critical point ( x > 1 ), but this leads to values greater than 2. For ( a leq -2 ), the function is decreasing, so the maximum is at ( x = 1 ).3. Detailed analysis shows that if ( a = -2 ), the function is decreasing and the maximum at ( x = 1 ) is exactly 2. For ( a < -2 ), the function is still decreasing, but the maximum remains at ( x = 1 ). However, the problem implies a unique value, which is the boundary case ( a = -2 ). Final Answer(I) boxed{(-infty, -2]}(II) boxed{-2}]

🔑:Solution: (I) If f(x) is a decreasing function on [1,+infty), then f'(x)leqslant 0 holds true on [1,+infty), f'(x)=ln x+2ax+2-aleqslant 0, therefore aleqslant - dfrac {ln x+2}{2x-1}, let g(x)=- dfrac {ln x+2}{2x-1}, then g'(x)= dfrac {2+ dfrac {1}{x}+2ln x}{(2x-1)^{2}}, since xgeqslant 1, therefore g'(x)geqslant 0, g(x) is increasing, and g(1)=-2, hence aleqslant -2; (II) From f(1)=2, to make f(x)_{max}=2, the decreasing interval of f(x) is [1,+infty), and the increasing interval is (0,1), therefore f'(1)=0, that is ln 1+2a+2-a=0, therefore a=-2.Thus, for part (I), the range of values for a is a leqslant -2, and for part (II), the value of a is boxed{-2}.

🧠:[<think>Alright, so I need to find the smallest positive period for these two functions. Let's start with problem (1): the function is f(x) = cube root of sin((3/2)x) minus the fifth root of cos((2/3)x). Hmm, okay. I remember that when you have functions combined together, the period of the whole function is the least common multiple (LCM) of the periods of the individual functions. So first, I need to find the periods of each part separately and then find their LCM.Starting with the first part: cube root of sin((3/2)x). Let's think about the period of sin((3/2)x) first. The general formula for the period of sin(kx) is 2π divided by |k|. So here, k is 3/2, so the period would be 2π divided by (3/2), which is 2π * (2/3) = 4π/3. But wait, the cube root is applied to the sine function. Does taking the cube root affect the period? Well, the cube root is an odd function, and since sine is also odd, the composition might preserve some properties. But in general, applying a function like a root or a power can sometimes affect the period, but in this case, since the cube root is a monotonic function over the real numbers (unlike even roots which can cause issues with periodicity due to non-injectivity), I think the period remains the same. Because the cube root doesn't change where the function repeats; it just scales the output values. So the period of the first term should still be 4π/3.Now moving on to the second part: fifth root of cos((2/3)x). Similarly, let's find the period of cos((2/3)x) first. The period is 2π divided by |k|, where k is 2/3. So that's 2π / (2/3) = 3π. Then, applying the fifth root. Again, the fifth root is an odd function (since 5 is odd), so it's injective over the real numbers. Therefore, similar to the cube root, taking the fifth root of cosine shouldn't change the period; it just affects the amplitude. So the period of the second term is 3π.Now, the function f(x) is the difference of these two terms. So to find the smallest period of f(x), I need to find the LCM of 4π/3 and 3π. How do you compute LCM for fractions? I recall that LCM for fractions can be found by taking LCM of the numerators divided by GCD of the denominators. So let's represent 4π/3 and 3π as fractions. The first one is (4π)/3, and the second is (3π)/1. So numerators are 4π and 3π, denominators are 3 and 1.Wait, but LCM for numbers with π? Hmm, maybe it's easier to factor out π. Let’s let T1 = 4/3 and T2 = 3, so the periods in terms of multiples of π. Then the LCM of T1 and T2 would be LCM(4/3, 3). To compute LCM of 4/3 and 3:First, convert them to fractions with the same denominator. 4/3 and 9/3. Then LCM of 4 and 9 is 36. So LCM is 36/3 = 12. Wait, but LCM of 4/3 and 3 is 12? Let me check that. Alternatively, LCM(a/b, c/d) is LCM(a, c) / GCD(b, d). So LCM(4, 3)/GCD(3,1) = 12 / 1 = 12. So LCM of 4/3 and 3 is 12. Therefore, the LCM period is 12π? Wait, 12 times π? But original periods were 4π/3 and 3π, so LCM is 12π? Let me verify.Suppose we have two periods T1 and T2. The LCM is the smallest positive T such that T = n*T1 = m*T2 for integers n and m. So T = n*(4π/3) = m*(3π). Let's divide both sides by π: n*(4/3) = m*3. So 4n/3 = 3m. Multiply both sides by 3: 4n = 9m. We need integers n and m such that 4n = 9m. The smallest n and m would be n=9 and m=4, because 4*9 = 36 and 9*4=36. Then T = n*(4π/3) = 9*(4π/3) = 12π. Similarly, T = m*3π = 4*3π = 12π. So yes, LCM is 12π. Therefore, the smallest positive period of f(x) is 12π.Okay, that seems solid. Let me check if there's any chance that the periods could interfere or if there's a smaller period. For example, if one period is a multiple of the other, then the LCM would just be the larger one. But here, 4π/3 and 3π are not multiples. 3π divided by 4π/3 is (3π)/(4π/3) = 9/4, which is 2.25, not integer. So they are incommensurate in a way that their LCM is 12π. So that should be the answer for part (1).Moving on to problem (2): f(x) = 2 sin 3x + 3|sin 4x|. Again, need to find the smallest positive period. So similar approach: find the periods of each term and then compute their LCM.First term: 2 sin 3x. The period of sin(kx) is 2π / |k|. Here, k=3, so period is 2π/3. The coefficient 2 in front doesn't affect the period, just the amplitude.Second term: 3|sin 4x|. The absolute value of sine. I remember that |sin x| has a period of π, since the absolute value reflects the negative part of the sine wave, effectively halving the period. So normally, sin x has period 2π, but |sin x| has period π. So for |sin 4x|, the period would be π / 4. Because the period of sin 4x is 2π/4 = π/2, and then taking absolute value would halve it again? Wait, wait. Wait, hold on.Wait, let me think. For |sin x|, the original period is 2π, but after absolute value, it's π. So the period is halved. Similarly, if we have |sin kx|, the period is π / k. Wait, let me verify. The function sin(kx) has period 2π / k. Then taking absolute value would reflect the negative parts, so the period becomes π / k. Because in the interval [0, π/k], the function goes from 0 up to 1, back down to 0, and then the absolute value makes the next half-wave a mirror image, but since it's already mirrored by the absolute value, the period is halved. So yes, |sin kx| has period π / k. Therefore, |sin 4x| has period π /4. Then multiplied by 3, but again, scaling by 3 doesn't affect the period. So the period of the second term is π /4.Therefore, the two periods are 2π/3 and π/4. Need to find LCM of 2π/3 and π/4. Let's do similar steps as before. Let’s factor out π, so T1 = 2/3 and T2 = 1/4. Then LCM(2/3, 1/4). Using the formula: LCM of fractions is LCM(numerators)/GCD(denominators). Numerators are 2 and 1, denominators are 3 and 4. LCM(2,1) is 2. GCD(3,4) is 1. So LCM is 2/1 = 2. Therefore, LCM period is 2π? Wait, let me check that.Wait, the formula for LCM(a/b, c/d) is LCM(a,c)/GCD(b,d). So LCM(2,1)/GCD(3,4) = 2 /1 =2. So LCM is 2. But since we had factored out π, the actual LCM period is 2π. Hmm, but let me verify using the previous method. Suppose T = LCM(2π/3, π/4). So T must satisfy T = n*(2π/3) = m*(π/4) for integers n and m. Dividing both sides by π: n*(2/3) = m*(1/4). Multiply both sides by 12 to eliminate denominators: n*8 = m*3. So 8n = 3m. The smallest integers n and m satisfying this are n=3 and m=8. Then T = 3*(2π/3) = 2π, and T =8*(π/4)= 2π. So yes, LCM is 2π.But wait, is there a smaller period? Let me check. Suppose we try π. Let's see if π is a period. Let's check for the first term: 2 sin 3x. If we replace x with x + π, then 2 sin(3(x + π)) = 2 sin(3x + 3π) = 2 sin(3x + π) = -2 sin3x. So that's not equal to 2 sin3x, so π is not a period for the first term. Therefore, π cannot be a period for the entire function. Then 2π is the LCM. But let's check the second term. The second term is 3|sin4x|, which has period π/4. So 2π is a multiple of π/4 (since 2π / (π/4) = 8, which is integer). So yes, 2π is a period for both terms, and since there's no smaller common period (since the first term requires 2π/3 and the LCM of 2π/3 and π/4 is 2π), then 2π is indeed the smallest period.Wait, but I have a feeling that sometimes when you combine functions with different periods, especially with absolute values, the period might be smaller. Let me test with a specific example. Let's take x = 0. f(0) = 2 sin0 + 3|sin0| = 0 + 0 =0. Now check f(x + T) where T=2π. f(0 + 2π) = 2 sin(6π) + 3|sin8π| =0+0=0, same as f(0). What about halfway, like T=π. Let's check f(π). 2 sin(3π) +3|sin4π| =0+0=0, same as f(0). But what about another point, like x=π/6. f(π/6) = 2 sin(3*(π/6)) +3|sin4*(π/6)| = 2 sin(π/2) +3|sin(2π/3)| =2*1 +3*(√3/2)=2 + (3√3)/2. Now f(π/6 + π) = f(7π/6) =2 sin(3*(7π/6)) +3|sin4*(7π/6)| =2 sin(7π/2) +3|sin(14π/3)|. Sin(7π/2)=sin(3π + π/2)= -1, so 2*(-1)=-2. Sin(14π/3)=sin(4π + 2π/3)=sin(2π/3)=√3/2, so absolute value is √3/2. So 3*(√3/2)= (3√3)/2. Therefore, f(7π/6) = -2 + (3√3)/2. Comparing to f(π/6)=2 + (3√3)/2, they are not equal. So T=π is not a period. Therefore, 2π is indeed the minimal period.Alternatively, maybe even a smaller multiple? Let's suppose T=π/2. Let's check f(π/6 + π/2)=f(2π/3). Compute f(2π/3)=2 sin(3*(2π/3)) +3|sin4*(2π/3)|=2 sin(2π)+3|sin(8π/3)|=0 +3|sin(8π/3 - 2π)|=3|sin(2π/3)|=3*(√3/2). Original f(π/6)=2 + (3√3)/2. These are different, so T=π/2 is not a period. Similarly, trying T=π/4. f(π/6 + π/4)=f(5π/12). 2 sin(3*(5π/12))=2 sin(5π/4)=2*(-√2/2)=-√2. 3|sin4*(5π/12)|=3|sin(5π/3)|=3|sin(-π/3)|=3*(√3/2). So f(5π/12)= -√2 + (3√3)/2. Which is different from f(π/6)=2 + (3√3)/2. So T=π/4 isn't a period either. Therefore, 2π seems correct.Therefore, the answer for part (2) is 2π.Final Answer(1) boxed{12pi}(2) boxed{2pi}</think>To find the smallest positive period of the given functions, we need to determine the periods of each component function and then compute their least common multiple (LCM). Problem (1)The function is ( f(x) = sqrt[3]{sin frac{3}{2} x} - sqrt[5]{cos frac{2}{3} x} ).1. First term: (sqrt[3]{sin frac{3}{2} x}) - The period of (sin frac{3}{2} x) is (frac{2pi}{frac{3}{2}} = frac{4pi}{3}). - Applying the cube root does not change the period, so the period remains (frac{4pi}{3}).2. Second term: (sqrt[5]{cos frac{2}{3} x}) - The period of (cos frac{2}{3} x) is (frac{2pi}{frac{2}{3}} = 3pi). - Applying the fifth root does not change the period, so the period remains (3pi).3. LCM of (frac{4pi}{3}) and (3pi) - Convert the periods to fractions of (pi): (frac{4}{3}) and (3). - LCM of (frac{4}{3}) and (3) is calculated as follows: - LCM of numerators (4 and 3) is 12. - GCD of denominators (3 and 1) is 1. - LCM is (frac{12}{1} = 12), so the LCM period is (12pi).Thus, the smallest positive period of (f(x)) is (boxed{12pi}). Problem (2)The function is ( f(x) = 2 sin 3x + 3|sin 4x| ).1. First term: (2 sin 3x) - The period of (sin 3x) is (frac{2pi}{3}).2. Second term: (3|sin 4x|) - The period of (sin 4x) is (frac{2pi}{4} = frac{pi}{2}). - Taking the absolute value halves the period, so the period becomes (frac{pi}{2} / 2 = frac{pi}{4}).3. LCM of (frac{2pi}{3}) and (frac{pi}{4}) - Convert the periods to fractions of (pi): (frac{2}{3}) and (frac{1}{4}). - LCM of (frac{2}{3}) and (frac{1}{4}) is calculated as follows: - LCM of numerators (2 and 1) is 2. - GCD of denominators (3 and 4) is 1. - LCM is (frac{2}{1} = 2), so the LCM period is (2pi).Thus, the smallest positive period of (f(x)) is (boxed{2pi}).,<think>Okay, so I need to find the smallest positive period for two different functions. Let's start with the first one: f(x) = cube root of sin(3/2 x) minus fifth root of cos(2/3 x). Hmm. Periods can be tricky when you have multiple trigonometric functions combined. I remember that the period of a function like sin(kx) is 2π divided by k. So for each part of the function, I need to find their individual periods and then determine the least common multiple (LCM) of those periods to get the overall period of the combined function.First, let's break down the function into its components. The first part is the cube root of sin(3/2 x). The cube root doesn't affect the period, right? Because the period of sin(kx) is determined by the coefficient inside the sine function. So even if you take the cube root of it, the period should remain the same. So the period of sin(3/2 x) is 2π divided by 3/2, which is (2π)*(2/3) = 4π/3. Since the cube root doesn't change the period, the first part has a period of 4π/3.Now the second part is the fifth root of cos(2/3 x). Similarly, the fifth root doesn't affect the period. The period of cos(2/3 x) is 2π divided by 2/3, which is (2π)*(3/2) = 3π. So the second part has a period of 3π.So now, the function f(x) is a combination of two functions with periods 4π/3 and 3π. To find the smallest period of the entire function, I need to find the LCM of 4π/3 and 3π. LCM for periods is a bit different because we have to consider the LCM of the numerical coefficients when the variable is scaled out.Let me convert them into fractions with the same denominator to make it easier. 4π/3 and 3π. Let's express 3π as 9π/3. Now, looking at the numerators: 4π and 9π. The LCM of 4 and 9 is 36. So LCM of 4π/3 and 9π/3 is 36π/3 = 12π. Therefore, the LCM of 4π/3 and 3π is 12π. So the smallest positive period of f(x) is 12π. Let me check that.If we have 12π, then for the first term, 12π divided by 4π/3 is (12π)*(3)/(4π) = 9, which is an integer. For the second term, 12π divided by 3π is 4, which is also an integer. So yes, 12π is a common multiple. Since 36 is LCM of 4 and 9, and we divided by 3, which is the denominator, giving 12π. That seems right.Okay, so part (1) is 12π. Now moving on to part (2): Find the smallest positive period of the function f(x) = 2 sin 3x + 3|sin 4x|. Alright, again, this is a combination of two functions: 2 sin 3x and 3|sin 4x|. Let's find their individual periods first.The period of sin(3x) is 2π/3. The absolute value of sin(4x), |sin 4x|, has a period. Normally, sin(4x) has a period of 2π/4 = π/2. But when you take the absolute value, the negative part of the sine wave is flipped to positive, which effectively makes the period half of the original. So the period of |sin 4x| would be π/2 divided by 2, which is π/4. Wait, is that right? Let me think. For example, |sin x| has a period of π instead of 2π. So yes, taking the absolute value halves the period. Therefore, |sin 4x| would have a period of π/4.So the first term, 2 sin 3x, has a period of 2π/3, and the second term, 3|sin 4x|, has a period of π/4. Now we need to find the LCM of 2π/3 and π/4. Again, let's convert them to have the same denominator. 2π/3 and π/4 can be expressed as 8π/12 and 3π/12. The LCM of 8π and 3π is 24π. Then divide by 12: 24π/12 = 2π. Wait, but that might not be the case. Wait, LCM of 2π/3 and π/4.Alternatively, think of LCM in terms of multiples. Let's express the periods as fractions:First period: 2π/3. Second period: π/4. To find the LCM, we can find the LCM of the numerators divided by the greatest common divisor (GCD) of the denominators. Wait, no. Maybe another approach.The LCM of two periods T1 and T2 is the smallest positive T such that T = m*T1 = n*T2 for integers m and n. So, LCM(T1, T2) is the smallest T where T/T1 and T/T2 are integers.So, let me write T1 = 2π/3 and T2 = π/4. So, we need T such that T/(2π/3) = 3T/(2π) is an integer, and T/(π/4) = 4T/π is an integer. Let’s denote T = kπ, where k is a rational number. Then:3(kπ)/(2π) = 3k/2 must be integer. So 3k/2 is integer, which implies k must be a multiple of 2/3.Similarly, 4(kπ)/π = 4k must be integer. So 4k must be integer. Therefore, k must be a multiple of 1/4.So we need k such that k is a multiple of 2/3 and 1/4. So LCM of 2/3 and 1/4. How do you compute LCM of fractions? LCM of fractions is LCM(numerators)/GCD(denominators). So LCM(2,1)/GCD(3,4) = 2 / 1 = 2. Wait, but 2 is LCM of 2/3 and 1/4? Wait, maybe not. Let's check.Alternatively, find the smallest k such that k is a multiple of 2/3 and 1/4. So 2/3, 1/4. The LCM is the smallest number that both 2/3 and 1/4 divide into. So multiples of 2/3: 2/3, 4/3, 6/3=2, 8/3, etc. Multiples of 1/4: 1/4, 1/2, 3/4, 1, 5/4, 3/2, etc. The smallest common multiple is 2. Because 2 is 3*(2/3)=2 and 8*(1/4)=2. So k=2. Therefore, T=2π.But wait, let's test this. If T=2π, then for the first term, 2 sin 3x, period is 2π/3. 2π divided by 2π/3 is 3, which is an integer. For the second term, |sin4x| has period π/4. 2π divided by π/4 is 8, which is an integer. So 2π is a common period. But is it the smallest?Wait, let's check with a smaller multiple. Let's see if π works. For the first term: π divided by 2π/3 is 3/2, which is 1.5, not integer. So π is not a multiple of 2π/3. How about π/2? π/2 divided by 2π/3 is (π/2)*(3/2π) = 3/4, not integer. π/1: already checked. 3π/2: 3π/2 divided by 2π/3 is (3π/2)*(3/(2π)) = 9/4, not integer. Hmm. So maybe 2π is indeed the LCM.Wait, but maybe there is a smaller period. Let me check another way. Let's think about the function 2 sin 3x + 3|sin4x|. The first term is a sine wave with frequency 3, and the second term is an absolute sine wave with frequency 4, but the absolute value makes it frequency 8? Wait, no. Wait, |sin4x| has a period of π/4. So the frequency is 8/π? Wait, perhaps thinking in terms of frequency is confusing. Let's stick with periods.So periods are 2π/3 and π/4. The LCM of 2π/3 and π/4 is 2π. But let me see another example. Suppose T1 = 2 and T2 = 3. LCM is 6. Similarly, if T1 and T2 are periods, LCM is their LCM. But when they are fractions, we can convert them to fractions over the same denominator. For 2π/3 and π/4, which can be written as 8π/12 and 3π/12. LCM of 8π and 3π is 24π. Then 24π/12 = 2π. So that method gives the same answer. So 2π is the LCM.Wait, but let me check with a smaller multiple. Suppose T = π. Then for the first term: π/(2π/3) = 3/2, not integer. For the second term: π/(π/4) = 4, integer. So π is a period of the second term but not the first. Therefore, π cannot be the period of the entire function.What about 2π/1? Already checked. 2π is the LCM.Alternatively, let's consider the periods of the components. The first term has period 2π/3, the second term has period π/4. The LCM is the smallest number that is a multiple of both 2π/3 and π/4. So 2π/3 is equal to 8π/12, and π/4 is equal to 3π/12. So LCM of 8π/12 and 3π/12 is LCM(8π, 3π)/12. LCM(8π,3π) is 24π. Therefore, 24π/12=2π.Therefore, the smallest period is 2π. Let me verify by plugging into the function. Let's check if f(x + T) = f(x) for T = 2π.For the first term: 2 sin3(x + 2π) = 2 sin(3x + 6π) = 2 sin3x, since sin has period 2π. So that's the same.For the second term: 3|sin4(x + 2π)| = 3|sin(4x + 8π)| = 3|sin4x|, since sin is 2π periodic. So both terms repeat every 2π. Therefore, 2π is indeed a period.Is there a smaller period? Let's suppose T = π. Then for the first term: 2 sin3(x + π) = 2 sin(3x + 3π) = 2 sin(3x + π) = -2 sin3x. For the second term: 3|sin4(x + π)| = 3|sin(4x + 4π)| = 3|sin4x|. So f(x + π) = -2 sin3x + 3|sin4x|, which is not equal to f(x). So π is not a period.What about T = π/2? For the first term: 2 sin3(x + π/2) = 2 sin(3x + 3π/2) = 2*(-cos3x) = -2 cos3x. The second term: 3|sin4(x + π/2)| = 3|sin(4x + 2π)| = 3|sin4x|. So f(x + π/2) = -2 cos3x + 3|sin4x|, which is different from f(x). So T=π/2 is not a period.How about T=π/3? First term: 2 sin3(x + π/3) = 2 sin(3x + π) = -2 sin3x. Second term: 3|sin4(x + π/3)| = 3|sin(4x + 4π/3)|. That's a different argument, so not sure. Let's check f(x + π/3) = -2 sin3x + 3|sin(4x + 4π/3)|. Which is not the same as f(x) = 2 sin3x + 3|sin4x|. So not equal.How about T=π/4? For the first term: 2 sin3(x + π/4) = 2 sin(3x + 3π/4). Second term: 3|sin4(x + π/4)| = 3|sin(4x + π)| = 3| -sin4x | = 3|sin4x|. So f(x + π/4) = 2 sin(3x + 3π/4) + 3|sin4x|. Which is different from the original function because the first term is phase-shifted. So unless 3π/4 is a multiple of 2π, which it's not, the first term isn't the same. Hence, T=π/4 is not a period.Trying T=2π/3. For the first term: 2 sin3(x + 2π/3) = 2 sin(3x + 2π) = 2 sin3x. For the second term: 3|sin4(x + 2π/3)| = 3|sin(4x + 8π/3)|. 8π/3 is equivalent to 8π/3 - 2π = 8π/3 - 6π/3 = 2π/3. So sin(4x + 2π/3) which is not the same as sin4x. Therefore, the second term is |sin(4x + 2π/3)|, which is different from |sin4x|. Therefore, f(x + 2π/3) = 2 sin3x + 3|sin(4x + 2π/3)|, which is not equal to f(x). Hence, 2π/3 is not a period.Similarly, trying T=3π/4. For first term: 2 sin3(x + 3π/4) = 2 sin(3x + 9π/4) = 2 sin(3x + π/4) (since 9π/4 - 2π = π/4). Second term: 3|sin4(x + 3π/4)| = 3|sin(4x + 3π)| = 3|sin4x + 3π| = 3| -sin4x | = 3|sin4x|. So f(x + 3π/4) = 2 sin(3x + π/4) + 3|sin4x|. Again, different from original f(x).Hmm, so seems like all smaller periods don't work. Therefore, 2π is indeed the smallest period. Therefore, the answer for part (2) is 2π.Final Answer(1) The smallest positive period is boxed{12pi}.(2) The smallest positive period is boxed{2pi}.</think>To find the smallest positive period of the given functions, we need to determine the periods of each component function and then find their least common multiple (LCM). Part (1)The function is ( f(x) = sqrt[3]{sin frac{3}{2} x} - sqrt[5]{cos frac{2}{3} x} ).1. Period of (sqrt[3]{sin frac{3}{2} x}): - The period of (sin frac{3}{2} x) is (frac{2pi}{frac{3}{2}} = frac{4pi}{3}). - The cube root does not affect the period, so this component has a period of (frac{4pi}{3}).2. Period of (sqrt[5]{cos frac{2}{3} x}): - The period of (cos frac{2}{3} x) is (frac{2pi}{frac{2}{3}} = 3pi). - The fifth root does not affect the period, so this component has a period of (3pi).3. LCM of (frac{4pi}{3}) and (3pi): - Convert to common denominators: (frac{4pi}{3}) and (frac{9pi}{3}). - LCM of the numerators (4pi) and (9pi) is (36pi). - Divide by the common denominator: (frac{36pi}{3} = 12pi).Thus, the smallest positive period of (f(x)) is (boxed{12pi}). Part (2)The function is ( f(x) = 2 sin 3x + 3|sin 4x| ).1. Period of (2 sin 3x): - The period of (sin 3x) is (frac{2pi}{3}).2. Period of (3|sin 4x|): - The period of (sin 4x) is (frac{2pi}{4} = frac{pi}{2}). - The absolute value halves the period, so the period becomes (frac{pi}{2} / 2 = frac{pi}{4}).3. LCM of (frac{2pi}{3}) and (frac{pi}{4}): - Convert to common denominators: (frac{8pi}{12}) and (frac{3pi}{12}). - LCM of the numerators (8pi) and (3pi) is (24pi). - Divide by the common denominator: (frac{24pi}{12} = 2pi).Thus, the smallest positive period of (f(x)) is (boxed{2pi}).]

🔑:(1) For any T(neq 0), sqrt[5]{sin frac{3}{2}(x+T)}=sqrt[5]{sin frac{3}{2} x} Leftrightarrow sin frac{3}{2}(x+T)=sin frac{3}{2} x,Therefore, the smallest positive period of the odd function f_{1}(x)=sqrt[3]{sin frac{3}{2} x} is T_{1}=frac{2 pi}{3 / 2}=frac{4}{3} pi, and the smallest positive period of the even function f_{2}(x)= -sqrt[5]{cos frac{2}{3} x} is T_{2}=frac{2 pi}{2 / 3}=3 pi, hence the smallest positive period of f(x) is the least common multiple of T_{1} and T_{2}, which is 12 pi.(2) The smallest positive period of the odd function f_{1}(x)=2 sin 3 x is T_{1}=frac{2}{3} pi, and the smallest positive period of the even function f_{2}(x)=3|sin 4 x| is T_{2}=frac{pi}{8}, hence the smallest positive period of f(x) is the least common multiple of T_{1} and T_{2}, which is 2 pi.Explanation: The above conclusion can not only be used to find the smallest positive period of the sum of odd and even functions, but also, through translation transformations, to find the smallest positive period of periodic functions that can be transformed into the sum of odd and even functions. For example:Find the smallest positive period of f(x)=(sin x)^{2008}+(cos x)^{20107}.Solution: fleft(frac{pi}{2}-xright)=(cos x)^{2000}+(sin x)^{2007}, it is easy to see that f_{1}(x)=(sin x)^{2007} is an odd function, and f_{1}(x+T)=f_{1}(x) Leftrightarrow sin (x+T)=sin x. Therefore, the smallest positive period of f_{1}(x) is the same as that of sin x, both being T_{1}= 2 pi; f_{2}(x)=(cos x)^{20018} is an even function, and f_{2}(x+T)=f_{2}(x) Leftrightarrow|cos (x+T)|=|cos x|, so the smallest positive period of f_{2}(x) is the same as that of |cos x|, both being T_{2}=pi. Hence, the smallest positive period of fleft(frac{pi}{2}-xright) is the least common multiple of T_{1} and T_{2}, which is 2 pi. Since the smallest positive period of f(x) is the same as that of fleft(frac{pi}{2}-xright), the smallest positive period of f(x) is 2 pi.In general, if f_{1}(x) represents one of the six basic trigonometric functions (sin x, cos x, tan x, cot x, sec x, csc x), and f(x)=a f_{1}^{2 n}(x)+b cos ^{2 m-1} xleft(n, m in mathbf{N}^{*}, a b neq 0right), then the smallest positive period of f(x) is 2 pi.Note: When finding T_{1} and T_{2}, it is essential to consider that the domain of f_{1}(x) and f_{2}(x) is the common domain D, otherwise the conclusion may not hold.Counterexample: Let n in mathbf{Z}, the odd function f_{1}(x)=sin pi x, x neq 2 n, and the even function f_{2}(x)=left{begin{array}{l}0, x neq 4 n 1, x=4 nend{array}right., then T_{1}=2, T_{2}=4, and the least common multiple of T_{1} and T_{2} is 4, but the smallest positive period of f(x)=f_{1}(x)+f_{2}(x)=sin pi x, x neq 2 n is T=2.Why is this so? The reason lies in the fact that when finding T_{1}, the domain of f_{1}(x) is mathbf{R}, not the common domain D={x mid x neq 2 n, n in mathbf{Z}, x in mathbf{R}} of f_{1}(x) and f_{2}(x). In fact, the smallest positive period of f_{1}(x)(x in D) is T_{1}=2.