🧠:[<think>Okay, so I need to figure out for which values of 'a' the equation |x² - 5x| = a has exactly two real roots. Hmm, let's start by understanding what this equation represents. The absolute value function |x² - 5x| means that the expression inside can be either positive or negative, but the result is always non-negative. So, the equation |x² - 5x| = a is equivalent to two separate equations: 1. x² - 5x = a2. x² - 5x = -aEach of these is a quadratic equation, and the original equation will have real roots based on the solutions of these two quadratics. The problem states that we need exactly two real roots in total. Therefore, we need to find the values of 'a' such that the combined number of real roots from both equations is exactly two.First, let's analyze each quadratic equation separately.Starting with the first equation: x² - 5x - a = 0The discriminant of a quadratic equation ax² + bx + c = 0 is D = b² - 4ac. For the equation to have real roots, the discriminant must be non-negative (D ≥ 0). Here, a (the coefficient of x²) is 1, b is -5, and c is -a. So, discriminant D1 = (-5)² - 4*1*(-a) = 25 + 4a. For real roots, D1 ≥ 0 ⇒ 25 + 4a ≥ 0 ⇒ 4a ≥ -25 ⇒ a ≥ -25/4. However, since the original equation involves an absolute value, |x² - 5x| = a, 'a' must be non-negative (because absolute value can't be negative). So, even though the discriminant allows a ≥ -25/4, 'a' must be ≥ 0. Therefore, the first equation x² - 5x = a will have real roots when a ≥ 0, but we need to consider how many roots it has.Similarly, the second equation: x² - 5x + a = 0Here, the discriminant D2 = (-5)² - 4*1*a = 25 - 4a. For real roots, D2 ≥ 0 ⇒ 25 - 4a ≥ 0 ⇒ 4a ≤ 25 ⇒ a ≤ 25/4. Again, since 'a' must be non-negative, the second equation will have real roots when 0 ≤ a ≤ 25/4.Now, the original equation |x² - 5x| = a will have solutions when either of the two quadratics have real solutions. However, the number of real roots depends on how many solutions each quadratic contributes. But the problem states that the total number of real roots should be exactly two. So, we need to find the values of 'a' where the combined real roots from both equations are two.But we have to be careful here. The two quadratics could have overlapping roots, so we need to check if there's any intersection between the solutions of x² -5x = a and x² -5x = -a. Let me see. Suppose a root satisfies both equations, then x² -5x = a and x² -5x = -a would imply that a = -a ⇒ a = 0. So, when a = 0, both equations reduce to x² -5x = 0, which has roots at x = 0 and x = 5. So, in this case, the total number of roots is two, but both equations contribute the same roots. Therefore, for a = 0, even though both equations have two roots, they are the same, so total roots are two. But wait, is that the case? Let me check.If a = 0, then |x² -5x| = 0 implies x² -5x = 0, which factors to x(x -5) = 0, so roots at x = 0 and x = 5. So that's two roots. So, in this case, the equation has two real roots.But if a is greater than 0, then we need to see how many roots each quadratic contributes. Let's consider different ranges of 'a':Case 1: a < 0. But since 'a' is the right-hand side of an absolute value equation, this is impossible. So, a must be ≥ 0.Case 2: a = 0. As above, equation has two real roots.Case 3: 0 < a < 25/4. Then, both quadratics have two real roots each (since D1 and D2 are positive). So, the first equation x² -5x -a = 0 has two roots, and the second equation x² -5x +a = 0 also has two roots. However, we need to check if these roots are distinct. If the roots of the two quadratics are different, then the total number of roots would be 4. But if there's overlap, then the total could be less. However, let's verify this.Suppose a is such that 0 < a < 25/4. Let's take a specific example, say a = 1. Then the equations are x² -5x -1 = 0 and x² -5x +1 = 0. Let's compute their roots.For x² -5x -1 = 0: roots are [5 ± sqrt(25 + 4)]/2 = [5 ± sqrt(29)]/2.For x² -5x +1 = 0: roots are [5 ± sqrt(25 - 4)]/2 = [5 ± sqrt(21)]/2.These roots are all distinct. Therefore, total roots are 4. Hence, for a = 1, which is in 0 < a < 25/4, there are four real roots.But the problem asks for exactly two real roots. So, this case (0 < a < 25/4) gives four roots, which is more than desired. Therefore, these values of 'a' are not in the desired range.Case 4: a = 25/4. Then, let's check the discriminants. For the first equation, D1 = 25 + 4*(25/4) = 25 +25 = 50 > 0, so two distinct roots. For the second equation, D2 = 25 -4*(25/4) = 25 -25 = 0, so one repeated root. Thus, total roots: 2 (from first equation) + 1 (from second equation) = 3. But since roots might overlap, but in this case, the roots of the second equation would be x = [5 ± 0]/2 = 5/2. Let's check if 5/2 is a root of the first equation: plug into x² -5x -25/4. Compute (25/4) -25/2 -25/4 = (25/4 -50/4 -25/4) = (-50/4) = -12.5 ≠ 0. So, no overlap. Therefore, total roots are 3. So, a =25/4 gives three roots. Not two.Case 5: a >25/4. Then, D2 = 25 -4a <0, so the second equation x² -5x +a =0 has no real roots. The first equation x² -5x -a =0 has discriminant D1=25 +4a, which is positive because a >25/4 implies 4a >25, so 25 +4a >50, so positive. Hence, the first equation has two real roots, and the second equation has none. Therefore, total roots are two. Wait a second. So, if a >25/4, then only the first equation contributes two real roots, and the second equation has none. So, total roots are two. Therefore, for a >25/4, the equation |x² -5x| =a has two real roots. But then, when a =0, we have two roots. When 0 <a <25/4, four roots. When a =25/4, three roots. When a >25/4, two roots. Therefore, the equation has exactly two real roots when a =0 or a >25/4. But wait, let's check a=0. When a=0, |x² -5x|=0 implies x² -5x=0, which gives x=0 or x=5. So two real roots. Correct. Then, when a>25/4, the equation |x² -5x|=a would correspond to x² -5x =a or x² -5x=-a. But since a>25/4, x² -5x=-a would require x² -5x to be negative, but x² -5x can attain a minimum value. Let's check the graph of y =x² -5x. It's a parabola opening upwards with vertex at x=5/2. The vertex value is (5/2)^2 -5*(5/2)=25/4 -25/2=25/4 -50/4= -25/4. So, the minimum value of x² -5x is -25/4. Therefore, the absolute value |x² -5x| has a minimum value of 0 and a maximum that goes to infinity as |x| increases. The graph of |x² -5x| will be V-shaped at the vertex (5/2, 25/4). So, when a >25/4, the equation |x² -5x|=a would correspond to two points on either side of the parabola where the absolute value is equal to a. But since the original parabola x² -5x opens upwards, the absolute value graph will have a "V" at the vertex (5/2, -25/4), flipping the part of the parabola below the x-axis to above. So, the graph of |x² -5x| will touch the y=0 line at x=0 and x=5, and then it will have a minimum point at (5/2, 25/4). Wait, no. Wait, the minimum value of |x² -5x| is 0, achieved at x=0 and x=5, and the graph actually has a local maximum at the vertex? Wait, no. Let's think again.Wait, the original function x² -5x is a parabola opening upwards, vertex at x=2.5, y= -6.25. So, |x² -5x| will take the negative part (between x=0 and x=5) and reflect it above the x-axis. So, the graph of |x² -5x| will be a W-shaped graph? Wait, no. Let me sketch it mentally. For x <0 and x >5, x² -5x is positive because x² dominates. Between x=0 and x=5, x² -5x is negative. Taking absolute value would flip that part. So, the graph of |x² -5x| will be:- For x ≤0 and x ≥5: same as x² -5x (positive).- For 0 <x <5: -(x² -5x) = -x² +5x.So, the graph will have two parts: the original parabola on the left and right sides, and an upside-down parabola in the middle. The minimum of the original parabola is at x=2.5, y=-6.25, so when we take absolute value, the minimum becomes a maximum at y=6.25 (i.e., 25/4). Therefore, the graph of |x² -5x| has a "valley" at x=0 and x=5 (touching the x-axis) and a peak at x=2.5, y=25/4. Therefore, the equation |x² -5x|=a will have:- Two solutions when a >25/4 (intersecting the two outer arms of the W-shaped graph)- Four solutions when 0 <a <25/4 (intersecting both the outer arms and the middle inverted parabola)- Three solutions when a=25/4 (touching the peak and intersecting the outer arms)- Two solutions when a=0 (touching at x=0 and x=5)Therefore, the equation |x² -5x|=a has exactly two real roots when a=0 or a >25/4. But the problem states "the range of values of a" for which the equation has exactly two real roots. So, this would be a=0 and a >25/4. But since a=0 is a single value, not a range, maybe the problem considers a=0 as part of a range? Wait, the problem says "range of values", so perhaps including a=0 and the interval a >25/4. But in the problem statement, the answer is supposed to be a range. If we write it as a ≥0, but that would include a=0 and 0 <a <25/4 and a ≥25/4. But we know that for 0 <a <25/4, there are four roots, which is not desired. So, the correct range is a=0 and a >25/4. But how to write that as a range?In interval notation, we can write a=0 union with (25/4, ∞). So, {0} ∪ (25/4, ∞). But depending on the problem's expectation, maybe they consider a=0 as part of the lower bound. But according to the graph explanation, when a=0, it's just two roots, and when a>25/4, two roots. So, the answer is a=0 or a>25/4. But since the question says "range of values", perhaps writing two intervals: a=0 and a>25/4. But in some contexts, they might consider 0 ≤ a ≤25/4, but that's not the case here.Wait, no. Wait, when a=0, you have two roots, when 0 <a <25/4, four roots, when a=25/4, three roots, and when a >25/4, two roots. Therefore, the values of a where there are exactly two real roots are a=0 and a>25/4. But in the problem statement, it says "range of values". So, perhaps combining them as two separate intervals: a=0 and a>25/4. But in the answer, they might want to write it as a=0 or a>25/4. But since a=0 is a single point, it's technically not a range. However, in some cases, a single point can be considered part of the answer. Alternatively, maybe there's a mistake in my reasoning. Let me check again.Wait, when a=0: two roots. For a>25/4: two roots. So, the range is a=0 and a>25/4. So, the answer is a=0 or a>25/4. But the problem asks for the "range of values", so perhaps writing in interval notation: {0} ∪ (25/4, ∞). But how is that usually presented? Maybe in the problem's context, it's acceptable to write two separate conditions.Alternatively, maybe there's a different interpretation. Let's think again. The equation |x² -5x| =a. Let's consider the function f(x) = |x² -5x|. The number of real roots of the equation f(x) =a is determined by the intersections of the horizontal line y=a with the graph of f(x). As established earlier, f(x) has a minimum value of 0 at x=0 and x=5, reaches a local maximum of 25/4 at x=5/2. The graph of f(x) is V-shaped at x=0 and x=5, with a peak at x=5/2. Therefore, the intersections:- If a <0: no solution.- If a=0: two solutions (x=0 and x=5).- If 0 <a <25/4: four solutions (two from the left side of the V, two from the right side, and two from the inverted parabola in the middle? Wait, maybe not. Wait, when you graph f(x)=|x² -5x|, between x=0 and x=5, it's the reflection of the original parabola, which opens downward. So, the graph would have a "W" shape? Wait, no.Wait, let's clarify. The original function x² -5x is a parabola opening upwards, vertex at x=2.5, y=-6.25. Taking absolute value would reflect the part of the parabola below the x-axis (between x=0 and x=5) to above. So, the graph of f(x) would be:- For x ≤0: x² -5x (which is positive because x is negative, x² positive, -5x positive).- For 0 <x <5: -(x² -5x) = -x² +5x (a downward-opening parabola with vertex at x=2.5, y=6.25).- For x ≥5: x² -5x (positive again, since x² dominates).Therefore, the graph of f(x) is a combination of the right arm of the original parabola (x≥5), the left arm (x≤0), and an upside-down parabola between 0 and 5. Therefore, when we draw horizontal lines y=a:- For a <0: no intersection.- For a=0: intersects at x=0 and x=5.- For 0 <a <25/4: intersects the left arm (x≤0) once, the right arm (x≥5) once, and the middle inverted parabola twice. So, total four intersections.- For a=25/4: intersects the middle inverted parabola exactly at its vertex (x=2.5), and also intersects the left and right arms. So, three intersections.- For a>25/4: the horizontal line is above the peak of the middle inverted parabola, so it only intersects the left and right arms once each. So, two intersections.Therefore, the equation f(x)=a has exactly two real roots when a=0 (two intersections) or a>25/4 (two intersections). Thus, the range of values of a is a=0 or a>25/4. But how to write this as a range? In interval notation, it would be {0} ∪ (25/4, ∞). However, sometimes in exams, they might accept writing it as a ≥0, but we know that's incorrect because between 0 and 25/4, there are four roots. So, the correct answer is a=0 or a>25/4. But the problem says "range of values". Since a=0 is a single value and a>25/4 is an interval, we can write the solution as a=0 or a>25/4. In boxed notation, probably expressed as a ∈ {0} ∪ (25/4, ∞). But maybe converting 25/4 to 6.25, but it's better to use fractions. So, 25/4 is 6.25.But let me check with specific examples. When a=0: two roots. When a=7 (which is >25/4=6.25): solve |x² -5x|=7. So, x² -5x=7 or x² -5x=-7. First equation: x² -5x -7=0. Discriminant 25 +28=53, two real roots. Second equation: x² -5x +7=0. Discriminant 25 -28=-3, no real roots. So, total two roots. Correct. When a=6 (which is less than 25/4=6.25): |x² -5x|=6. Then, x² -5x=6 or x² -5x=-6. First equation: x² -5x -6=0. Discriminant 25 +24=49, roots (5 ±7)/2: 12/2=6 and -2/2=-1. Second equation: x² -5x +6=0. Discriminant 25 -24=1, roots (5 ±1)/2: 3 and 2. So, four roots: x=6, -1, 3, 2. Thus, four roots. Correct. When a=25/4=6.25: |x² -5x|=6.25. So, x² -5x=6.25 or x² -5x=-6.25. First equation: x² -5x -6.25=0. Discriminant 25 +25=50, roots (5 ±√50)/2 = (5 ±5√2)/2. Second equation: x² -5x +6.25=0. This factors as (x -2.5)^2=0, so double root at x=2.5. Therefore, total roots: two from the first equation and one from the second, but 2.5 is a single root. So, three roots total. Correct. Thus, confirming that for a=0 and a>25/4, there are exactly two real roots, and otherwise more or less. Therefore, the answer is a=0 or a>25/4. But in terms of a range, since a=0 is a single value, the answer is all real numbers a such that a=0 or a>25/4. But the problem asks for the range of values. So, in boxed form, how should this be written? The problem may expect the answer in interval notation, combining both the point and the interval. So, {0} ∪ (25/4, ∞). But if the problem expects a union, then writing that. Alternatively, since in some contexts, writing a=0 or a>25/4 is acceptable. But since 25/4 is 6.25, but fractional form is preferred. But let me check if there's another way. Suppose I consider that when a=0, it's a boundary case where the equation touches the x-axis at two points. Then, for a>25/4, the equation |x² -5x|=a is equivalent to x² -5x=±a, but since a>25/4, x² -5x=-a has no real roots (as the left-hand side x² -5x can't be less than -25/4, and -a < -25/4). Therefore, only x² -5x=a has two real roots, giving total two roots. Hence, combining with a=0, we have two cases where the equation has two roots. So, the answer is a=0 or a>25/4. In mathematical notation, a ∈ {0} ∪ (25/4, ∞). Therefore, boxed, it would be boxed{[0] cup left( dfrac{25}{4}, infty right)}. But sometimes, they might accept a=0 and a>25/4 written as separate conditions. But since the question says "range of values", interval notation is more appropriate.But in some textbooks or sources, they might write the answer as a=0 or a>25/4, so in boxed form, written as boxed{a = 0 text{ or } a > dfrac{25}{4}}.Alternatively, considering that a=0 is a single point and the rest is an interval, but typically in such problems, if the only valid ranges are a=0 and a>25/4, you need to mention both. So, the correct answer is a=0 or a>25/4. Hence, the boxed answer should reflect that. However, in some cases, the problem might consider a=0 as part of a continuous range, but according to our analysis, between a=0 and a=25/4, there are four roots. So, a=0 is isolated.Therefore, the final answer is boxed{a = 0 text{ or } a > dfrac{25}{4}}.But let me check again. If I graph the function f(x) = |x² -5x|, the minimum value is 0 at x=0 and x=5, and there's a local maximum at x=2.5, which is 25/4. So, the graph is above the x-axis, touching it at 0 and 5, peaking at 2.5 with 25/4. Therefore, for a=0, it's two roots. For a between 0 and 25/4, four roots. At a=25/4, three roots. For a above 25/4, two roots. So, indeed, exactly two roots when a=0 or a>25/4. Hence, the answer is correct.Final AnswerThe range of values of ( a ) is (boxed{a = 0 text{ or } a > dfrac{25}{4}}).</think>To determine the range of values of ( a ) for which the equation ( |x^2 - 5x| = a ) has exactly two real roots, we analyze the equation by considering the absolute value function and solving the resulting quadratic equations.1. Breakdown of the Equation: - The equation ( |x^2 - 5x| = a ) can be split into two separate quadratic equations: 1. ( x^2 - 5x = a ) 2. ( x^2 - 5x = -a )2. Discriminant Analysis: - For ( x^2 - 5x - a = 0 ), the discriminant is ( 25 + 4a ). This equation has real roots if ( 25 + 4a geq 0 ), which is always true for ( a geq 0 ). - For ( x^2 - 5x + a = 0 ), the discriminant is ( 25 - 4a ). This equation has real roots if ( 25 - 4a geq 0 ), which gives ( a leq frac{25}{4} ).3. Number of Real Roots: - When ( a = 0 ): The equation reduces to ( x^2 - 5x = 0 ), which has two real roots (0 and 5). - When ( 0 < a < frac{25}{4} ): Both quadratic equations have real roots, resulting in four distinct roots. - When ( a = frac{25}{4} ): The equation ( x^2 - 5x + frac{25}{4} = 0 ) has a repeated root, leading to three real roots. - When ( a > frac{25}{4} ): Only the equation ( x^2 - 5x - a = 0 ) has real roots, resulting in two real roots.4. Conclusion: - The equation ( |x^2 - 5x| = a ) has exactly two real roots when ( a = 0 ) or ( a > frac{25}{4} ).Thus, the range of values of ( a ) is (boxed{a = 0 text{ or } a > dfrac{25}{4}}).,<think>Okay, so I need to find the range of values for ( a ) such that the equation ( |x^2 - 5x| = a ) has exactly two real roots. Hmm, let me think about how to approach this. First, I remember that absolute value equations can be split into two cases: one where the expression inside is non-negative and equals ( a ), and another where it's negative and equals ( -a ). So, maybe I can start by breaking this equation into two separate equations:1. ( x^2 - 5x = a )2. ( x^2 - 5x = -a )Each of these is a quadratic equation. The original equation ( |x^2 - 5x| = a ) will have solutions that are the union of the solutions from both equations. However, the number of real roots will depend on how many solutions each quadratic has and whether they overlap or not.But the question is asking for exactly two real roots. So I need to figure out for which values of ( a ) the combination of these two equations gives exactly two solutions. Let me take this step by step.First, let's analyze the graph of the function ( f(x) = |x^2 - 5x| ). The original function without the absolute value is ( f(x) = x^2 - 5x ), which is a parabola opening upwards with vertex at ( x = frac{5}{2} ). The vertex value is ( fleft(frac{5}{2}right) = left(frac{5}{2}right)^2 - 5 times frac{5}{2} = frac{25}{4} - frac{25}{2} = -frac{25}{4} ). So the vertex is at ( left(frac{5}{2}, -frac{25}{4}right) ).Taking the absolute value of this function reflects the part of the parabola that is below the x-axis (where ( f(x) ) is negative) to above the x-axis. So the graph of ( |x^2 - 5x| ) will have a "W" shape, with the vertex of the original parabola now pointing upwards at ( left(frac{5}{2}, frac{25}{4}right) ).To find the number of real roots of ( |x^2 - 5x| = a ), we can think of it as the intersection of the horizontal line ( y = a ) with the graph of ( y = |x^2 - 5x| ). The number of intersections (real roots) depends on the value of ( a ).If ( a ) is very large, say ( a > frac{25}{4} ), the line ( y = a ) will intersect the graph ( y = |x^2 - 5x| ) at two points because the graph tends to infinity as ( x ) approaches both positive and negative infinity, but since we have a "W" shape, maybe not. Wait, actually, the original parabola ( x^2 -5x ) is a standard parabola opening upwards. When we take its absolute value, the parts below the x-axis are reflected above. So the graph of ( |x^2 -5x| ) will have a minimum at ( x = frac{5}{2} ), with the value ( frac{25}{4} ). Then, as we move left or right from the vertex, the graph increases to infinity. So actually, it's more like a "V" shape on either side of the vertex. So when ( a ) is greater than ( frac{25}{4} ), the horizontal line ( y = a ) will intersect the graph at two points: one to the left of the vertex and one to the right. Wait, but that would mean two solutions. Hmm, but wait, maybe when ( a ) is greater than the maximum value of the original parabola before reflection? Wait, the original parabola goes to infinity as ( x ) approaches positive or negative infinity. So taking absolute value, after reflection, the graph will still go to infinity. So for very large ( a ), there should be two points of intersection: one on the left side and one on the right side. But when ( a = frac{25}{4} ), the horizontal line touches the vertex, so there's exactly one solution at that point. But wait, no. Let's think again.Wait, the graph of ( |x^2 -5x| ) is formed by taking ( x^2 -5x ) and flipping the parts where it is negative. Let me sketch the graph mentally. The original parabola ( x^2 -5x ) crosses the x-axis at ( x=0 ) and ( x=5 ). It has a vertex at ( x = frac{5}{2} ), which is ( y = -frac{25}{4} ). Taking absolute value, the parts between ( x=0 ) and ( x=5 ) (where the parabola is below the x-axis) are reflected upwards. So the graph of ( |x^2 -5x| ) will have a "W" shape between ( x=0 ) and ( x=5 ), with minima at ( x=0 ) and ( x=5 ) (both at ( y=0 )) and a maximum at ( x= frac{5}{2} ), ( y= frac{25}{4} ). Outside the interval ( [0,5] ), the original parabola is positive, so those parts remain the same. Wait, no: for ( x < 0 ), ( x^2 -5x ) is positive because ( x^2 ) is positive and ( -5x ) is positive (since ( x ) is negative). Similarly, for ( x >5 ), ( x^2 -5x ) is positive. So the absolute value only affects the portion between ( x=0 ) and ( x=5 ).Therefore, the graph of ( |x^2 -5x| ) is:- For ( x leq 0 ): ( x^2 -5x )- For ( 0 < x <5 ): ( -(x^2 -5x) = -x^2 +5x )- For ( x geq5 ): ( x^2 -5x )So the graph is V-shaped between 0 and 5, opening downward with vertex at ( x=2.5 ), ( y=25/4 ), and on either side (left of 0 and right of 5) it's the original upward-opening parabola.So, when we draw the horizontal line ( y=a ):- If ( a <0 ): no solution, since absolute value is always non-negative.- If ( a=0 ): solutions at x=0 and x=5 (where the original parabola touches the x-axis). So two solutions.- If ( 0 < a < 25/4 ): The horizontal line will intersect the graph four times: twice on the downward-opening parabola between 0 and 5, and twice on the upward-opening parts left of 0 and right of 5. Wait, is that right?Wait, let's check. Between 0 and 5, the graph is a downward-opening parabola with vertex at ( y=25/4 ). So if ( 0 < a <25/4 ), the horizontal line ( y=a ) will intersect this part twice. Then, on the left side (x <0) and right side (x >5), the graph is the original upward-opening parabola. So for the left side (x <0): the equation ( x^2 -5x = a ). Since ( x <0 ), ( x^2 ) is positive and ( -5x ) is positive, so ( x^2 -5x ) is positive. So solving ( x^2 -5x = a ) here. Similarly, on the right side (x >5): ( x^2 -5x = a ). For ( x >5 ), the parabola is increasing. So depending on the value of ( a ), there might be solutions here.Wait, but if ( a ) is between 0 and 25/4, does the equation ( x^2 -5x =a ) have two solutions in total? Let's solve ( x^2 -5x -a =0 ). The discriminant is ( 25 +4a ). Since ( a >0 ), discriminant is positive, so two real roots. However, the roots are ( x = [5 ± sqrt(25 +4a)]/2 ). Now, for x <0 and x >5, we need to see if these roots lie in those intervals.So the roots are ( [5 + sqrt(25 +4a)]/2 ) and ( [5 - sqrt(25 +4a)]/2 ). Let's analyze these:First root: ( [5 + sqrt(25 +4a)]/2 ). Since sqrt(25 +4a) >5 when a>0, because sqrt(25 +4a) >sqrt(25)=5. Wait, sqrt(25 +4a) is sqrt(25 +4a). If a is positive, then sqrt(25 +4a) >5, so the first root is [5 + something >5]/2 = (something >10)/2 >5. So the first root is greater than 5. The second root is [5 - sqrt(25 +4a)]/2. Since sqrt(25 +4a) >5, this becomes [5 - something >5]/2 = (negative number)/2 <0. So the second root is less than 0. Therefore, for ( x^2 -5x =a ), there are two real roots: one in x <0 and one in x >5. So for ( a >0 ), equation ( x^2 -5x =a ) has two solutions: one left of 0 and one right of 5.Meanwhile, in the interval 0 <x <5, the equation becomes ( -x^2 +5x =a ), which can be rewritten as ( x^2 -5x +a =0 ). The discriminant here is ( 25 -4a ). For real solutions, discriminant must be non-negative, so ( 25 -4a geq 0 implies a leq 25/4 ). So when ( a leq25/4 ), there are two real roots in the interval 0 <x <5.Therefore, putting this all together:For ( a <0 ): No solutions.For ( a =0 ): Two solutions (x=0 and x=5).For ( 0 < a <25/4 ):- Two solutions from ( x^2 -5x =a ): one in x <0, one in x >5.- Two solutions from ( x^2 -5x = -a ): two in 0 <x <5.Total solutions: 4.For ( a =25/4 ):- The equation ( x^2 -5x = -a ) becomes ( x^2 -5x +25/4 =0 ), discriminant is 25 -4*(25/4)=25 -25=0. So one real root at x=5/2. So this gives one solution at x=5/2.- The equation ( x^2 -5x =25/4 ). Solving this gives discriminant 25 +4*(25/4)=25 +25=50. So roots are [5 ± sqrt(50)]/2 = [5 ±5*sqrt(2)]/2. These are approximately (5 +7.07)/2 ≈6.035 and (5 -7.07)/2≈-1.035. So one in x>5 and one in x<0. So two solutions here.Total solutions at ( a=25/4 ): 1 (from the second equation) +2 (from the first equation) =3.For ( a >25/4 ):- The equation ( x^2 -5x =a ) still has two solutions: one in x<0, one in x>5.- The equation ( x^2 -5x = -a ) becomes ( x^2 -5x +a=0 ). Discriminant is 25 -4a. If a>25/4, discriminant is negative, so no real solutions here.Thus, total solutions: 2.Therefore, putting all together:- ( a <0 ): 0 solutions- ( a =0 ): 2 solutions- ( 0 <a <25/4 ):4 solutions- ( a=25/4 ):3 solutions- ( a>25/4 ):2 solutionsBut the question asks for the range of ( a ) where the equation has exactly two real roots. From the above:- At ( a=0 ): 2 solutions- At ( a>25/4 ):2 solutionsBut wait, when ( a=0 ), the equation is ( |x^2 -5x|=0 ), which implies ( x^2 -5x=0 ), so x=0 or x=5. So two real roots. So both ( a=0 ) and ( a>25/4 ) give exactly two real roots. But wait, the question is asking for a range of values. So when ( a=0 ), it's exactly two roots, and when ( a>25/4 ), it's exactly two roots. But between 0 and 25/4, there are four roots, and at a=25/4, three roots.Therefore, the values of ( a ) for which the equation has exactly two real roots are ( a=0 ) and ( a>25/4 ). But the problem says "range of values", so likely intervals. So combining these, the answer should be ( a=0 ) or ( a>25/4 ). But expressed as a range, that would be ( a=0 ) or ( a > frac{25}{4} ). However, depending on the problem's intention, maybe they consider ( a geq0 ), but no, because between 0 and 25/4 there are four roots. So the correct ranges are when ( a=0 ) (two roots) and ( a >25/4 ) (two roots). But how to write this as a single range? Since ( a=0 ) is a single point, and then ( a >25/4 ). So the answer is ( a=0 ) or ( a > frac{25}{4} ). But the problem says "range of values", which might imply intervals. So combining these, perhaps writing ( a in {0} cup (frac{25}{4}, infty) ).But let's verify again. For ( a=0 ), exactly two roots. For ( a >25/4 ), exactly two roots. So yes, the answer is ( a=0 ) or ( a>25/4 ). However, sometimes in such problems, they might consider ( a geq0 ), but that's not the case here. So the correct answer is that the equation has exactly two real roots when ( a=0 ) or when ( a > frac{25}{4} ). But the problem is asking for the "range of values", so probably writing it in interval notation: ( a=0 ) or ( a > frac{25}{4} ). So in boxed form, maybe ( a = 0 ) or ( a > frac{25}{4} ), but the problem might expect union of intervals. However, since ( a=0 ) is a single point, it's not an interval. Wait, but maybe in some contexts, it's acceptable to write ( a geq frac{25}{4} ), but no, because at ( a=25/4 ), there are three roots. So the answer is the union of the point ( a=0 ) and the interval ( a >25/4 ).Alternatively, if the problem allows for writing in terms of inequalities, then ( a=0 ) or ( a > frac{25}{4} ).But let me check if I made a mistake in the analysis. Let's confirm when ( a>25/4 ), the equation ( |x^2 -5x|=a ) has two real roots. Since ( |x^2 -5x| =a ) implies that ( x^2 -5x = pm a ). For ( a>25/4 ), ( x^2 -5x =a ) has two solutions (as discriminant is 25 +4a >0) and both roots lie outside the interval [0,5], as we saw. Then, ( x^2 -5x = -a ) becomes ( x^2 -5x +a=0 ), which has discriminant 25 -4a. Since ( a>25/4 ), discriminant is negative, so no real roots. Therefore, only two solutions. At ( a=25/4 ), discriminant of ( x^2 -5x +a=0 ) is zero, so one real root (x=5/2), and ( x^2 -5x =25/4 ) gives two roots, so total three roots. At ( a=0 ), as we saw, two roots (x=0 and x=5). For ( 0 <a <25/4 ), both equations ( x^2 -5x =a ) and ( x^2 -5x=-a ) contribute two solutions each, so total four. Therefore, the answer is that the equation has exactly two real roots when ( a=0 ) or ( a>25/4 ). But the problem says "range of values", which often refers to intervals. Since ( a=0 ) is a single value, maybe the answer is ( a > frac{25}{4} ) or ( a =0 ). However, in some contexts, the term "range" might require continuous intervals, but the problem might accept both. Let me check standard references.Wait, in similar problems, for example, when the equation |f(x)| = a has exactly n roots, the solution often includes isolated points and intervals. For instance, in this case, since at a=0 it's two roots, and then again two roots for a>25/4, both are acceptable. Therefore, the range is a=0 or a>25/4.Therefore, the final answer is all real numbers ( a ) such that ( a=0 ) or ( a > frac{25}{4} ). In boxed notation, since they might want interval notation, but since it's two separate regions, we can write ( a = 0 ) or ( a > frac{25}{4} ). However, in some textbooks, they might write this as ( a in {0} cup left( frac{25}{4}, infty right) ). But since the problem asks for a range of values, it's better to specify both conditions. But let me check again. If a=0, exactly two roots. If a>25/4, exactly two roots. So yes, the range is a=0 and a>25/4. Therefore, the answer is ( boxed{a = 0 text{ or } a > dfrac{25}{4}} ). But the problem might expect the answer in interval notation, combining both cases. However, since a=0 is a single point, we can't write it as an interval. Alternatively, maybe the problem allows for two separate intervals: [0,0] and (25/4, ∞). But usually, for such problems, they accept stating both conditions. Alternatively, if the problem expects the answer in terms of inequalities, writing ( a =0 ) or ( a > frac{25}{4} ). So boxed as per instruction.Final AnswerThe equation has exactly two real roots when ( a ) is zero or greater than ( dfrac{25}{4} ). Thus, the range of values is boxed{a = 0 text{ or } a > dfrac{25}{4}}.</think>To determine the range of values of ( a ) for which the equation ( |x^2 - 5x| = a ) has exactly two real roots, we analyze the graph of the function ( f(x) = |x^2 - 5x| ). The original function ( x^2 - 5x ) is a parabola opening upwards with vertex at ( left(frac{5}{2}, -frac{25}{4}right) ). Taking the absolute value reflects the part of the parabola below the x-axis to above the x-axis, resulting in a "W" shape with a minimum at ( left(frac{5}{2}, frac{25}{4}right) ).The equation ( |x^2 - 5x| = a ) can be split into two cases:1. ( x^2 - 5x = a )2. ( x^2 - 5x = -a )For ( a < 0 ), there are no solutions since the absolute value cannot be negative.For ( a = 0 ), the equation simplifies to ( x^2 - 5x = 0 ), which has two real roots: ( x = 0 ) and ( x = 5 ).For ( 0 < a < frac{25}{4} ), both equations ( x^2 - 5x = a ) and ( x^2 - 5x = -a ) have two real roots each, resulting in a total of four real roots.For ( a = frac{25}{4} ), the equation ( x^2 - 5x = -frac{25}{4} ) has one real root (at ( x = frac{5}{2} )), and the equation ( x^2 - 5x = frac{25}{4} ) has two real roots, resulting in a total of three real roots.For ( a > frac{25}{4} ), the equation ( x^2 - 5x = -a ) has no real roots (as the discriminant is negative), while the equation ( x^2 - 5x = a ) has two real roots, resulting in a total of two real roots.Thus, the equation ( |x^2 - 5x| = a ) has exactly two real roots when ( a = 0 ) or ( a > frac{25}{4} ).[boxed{a = 0 text{ or } a > dfrac{25}{4}}]]

🔑:(Tip: Let y_{1}=left|x^{2}-5 xright|, y_{2}=a. Then, the equation left|x^{2}-5 xright|=a having exactly two real roots is transformed into the graphs of the functions y_{1}=left|x^{2}-5 xright| and y_{2}=a having exactly two intersection points. It is easy to know that the range of values for a is a>frac{25}{4} or a=0.)

🧠:[<think>Okay, let's see. I need to solve part (I) and part (II) of this problem. Let's start with part (I). When a=2, the function is f(x) = ln x - (1/2)*2*x² - x. Wait, simplifying that, the 2 in the numerator and denominator cancel out, so it becomes ln x - x² - x. Right? So, first, I need to find the intervals where this function is increasing or decreasing, and also find its extremum points. To find monotonicity and extrema, I remember I need to take the derivative of f(x) and analyze where it's positive or negative. The critical points occur where the derivative is zero or undefined. Let's compute f'(x). f(x) = ln x - x² - x. The derivative of ln x is 1/x, derivative of -x² is -2x, and derivative of -x is -1. So f'(x) = 1/x - 2x -1. So f'(x) = (1/x) - 2x - 1. Now, to find critical points, set f'(x) = 0. So,1/x - 2x - 1 = 0. Let's solve for x. Since x is in the domain of ln x, x must be greater than 0. So, multiply both sides by x to eliminate the denominator:1 - 2x² - x = 0. So, rearranged: -2x² - x +1 =0. Multiply both sides by -1 to make it easier: 2x² + x -1 =0.Now, solving quadratic equation 2x² + x -1 =0. The quadratic formula: x = [-b ± sqrt(b² -4ac)]/(2a). Here, a=2, b=1, c=-1. So discriminant is 1² -4*2*(-1) =1 +8=9. sqrt(9)=3. So,x = [-1 ±3]/4. So two solutions:x = (-1 +3)/4 =2/4=1/2x = (-1 -3)/4=-4/4=-1But x must be greater than 0, so x=-1 is rejected. So the critical point is at x=1/2.Now, we need to test intervals around x=1/2 to see where f'(x) is positive or negative. The domain is x>0.So, let's pick test points in (0, 1/2) and (1/2, ∞). Let's choose x=1/4 (which is 0.25) for the first interval and x=1 for the second.For x=1/4:f'(1/4) = 1/(1/4) -2*(1/4) -1 =4 -0.5 -1=4 -1.5=2.5>0. So f is increasing here.For x=1:f'(1)=1/1 -2*1 -1=1 -2 -1=-2<0. So f is decreasing here.Therefore, the function is increasing on (0, 1/2) and decreasing on (1/2, ∞). So the function has a maximum at x=1/2, since it goes from increasing to decreasing. There are no minima because after x=1/2 it keeps decreasing towards negative infinity as x approaches infinity, and as x approaches 0 from the right, ln x goes to negative infinity as well.Therefore, for part (I):Monotonicity: increasing on (0, 1/2), decreasing on (1/2, ∞). The extremum is a maximum at x=1/2. The value of the maximum can be calculated by plugging x=1/2 into f(x):f(1/2)= ln(1/2) - (1/2)*2*(1/2)^2 - (1/2). Let's compute that.ln(1/2) is -ln2. Then, (1/2)*2*(1/2)^2 simplifies: (1/2)*2=1, so 1*(1/4)=1/4. Then subtract (1/2). So total is -ln2 -1/4 -1/2 = -ln2 -3/4. So the maximum value is -ln2 -3/4 at x=1/2.Wait, is that right? Let me check:Wait, when a=2, the original function is ln x - (1/2)*2 x² -x. So (1/2)*2 is 1, so it's ln x -x² -x. Then at x=1/2,ln(1/2) - (1/2)^2 - (1/2) = -ln2 -1/4 -1/2 = -ln2 -3/4. Yes, correct.So part (I) is done. Now part (II): If y=f(x) has a monotonically decreasing interval, determine the range of values for a.So f(x) = ln x - (1/2) a x² -x. We need to find a such that there exists some interval where f is decreasing. But since the function's domain is x>0, we need to ensure that the derivative f'(x) is negative for some x>0. Alternatively, maybe it's required that the function is decreasing on some interval, but not necessarily everywhere? Let me think.Wait, the problem says "If y=f(x) has a monotonically decreasing interval", so there exists at least one interval where f is decreasing. So even if the function has parts where it's increasing and decreasing, as long as there is some interval where it's decreasing, that's okay. But for some values of a, maybe the function is always increasing? Wait, let's check.Wait, actually, for some a, maybe the derivative is always positive, so f is always increasing. Then, such a would not satisfy the condition. So we need to find the range of a where the derivative f'(x) is negative for some x>0. So even if f'(x) is sometimes positive and sometimes negative, as long as there exists some interval where it's negative, that's acceptable.Alternatively, maybe the problem requires that the function has a decreasing interval, but not necessarily that it's always decreasing. So even if it has increasing and decreasing parts, as long as there is at least one decreasing interval, which would be the case as long as the derivative is negative somewhere.So, to find the range of a such that f'(x) <0 for some x>0.Alternatively, maybe the problem is that the function is monotonically decreasing on its entire domain? Wait, but the question says "has a monotonically decreasing interval", not that the entire function is decreasing. So, just that there exists an interval where it's decreasing. But actually, for the function to have a decreasing interval, the derivative must be negative somewhere. However, the function's derivative as x approaches 0+ is infinity? Wait, let's compute the derivative again.f'(x) = 1/x - a x -1. As x approaches 0+ from the right, 1/x approaches infinity, so f'(x) approaches infinity, which is positive. As x approaches infinity, the term -a x dominates. If a>0, then -a x approaches -infty, so f'(x) approaches -infty. If a=0, then f'(x)=1/x -1, which approaches -1 as x approaches infinity. If a<0, then -a x is positive and dominates, so f'(x) approaches infinity.So, if a>0, then as x increases, f'(x) tends to -infty. So since f'(x) starts at +infty at x approaching 0, and tends to -infty as x approaches infinity, by the Intermediate Value Theorem, there must be some point where f'(x)=0, and the derivative goes from positive to negative. So in that case, the function would first increase, then decrease, having a maximum. So in this case, the function has increasing and decreasing intervals. So if a>0, there is a decreasing interval.If a=0, then f'(x)=1/x -1. Then f'(x)=0 when 1/x=1, so x=1. So derivative is positive for x<1, negative for x>1. So again, there is a decreasing interval (1, infinity). So a=0 is acceptable.If a<0, then as x approaches infinity, f'(x) approaches infinity (since -a x is positive and dominates). As x approaches 0+, f'(x) approaches infinity. So f'(x) is always positive? Wait, but if a is negative, let's compute f'(x)=1/x -a x -1. If a is negative, then -a x is positive. So 1/x + |a| x -1. Hmm, so for a negative, the derivative is 1/x + |a| x -1. Let's see if this can ever be negative.Let’s take a specific example. Let a=-1. Then f'(x)=1/x +x -1. Let's check at x=1: f'(1)=1 +1 -1=1>0. At x approaching 0+, it's infinity. At x approaching infinity, f'(x)=x + ... which approaches infinity. So derivative is always positive? So for a negative, the derivative is always positive? Then, the function is always increasing. Therefore, there is no decreasing interval. Wait, but wait, let me test with a=-1 and x=0.5. f'(0.5)=1/0.5 +0.5 -1=2 +0.5 -1=1.5>0. x=0.1: 10 +0.1 -1=9.1>0. So yes, if a is negative, derivative is always positive. Therefore, for a<=0, the function would have a decreasing interval only when a=0? Wait, for a=0, derivative is 1/x -1, which is positive for x<1 and negative for x>1. So for a=0, there is a decreasing interval. But if a is negative, then derivative is always positive. Therefore, the function is increasing on entire domain. Therefore, for the function to have a monotonically decreasing interval, a must be >=0? Wait, but when a=0, there is a decreasing interval. When a>0, also there is a decreasing interval. When a<0, no. So the range is a >=0? Wait, but let's check carefully.Wait, when a=0, f'(x)=1/x -1. So when x>1, derivative is negative. So yes, decreasing interval (1, inf). When a>0, as before, the derivative tends to -infinity as x approaches infinity, so there is some interval where derivative is negative. For a negative, the derivative is always positive, so no decreasing interval. Therefore, the range of a is a >=0? Wait, but wait, let me think again.Wait, in part (II), the question says "If y=f(x) has a monotonically decreasing interval, determine the range of values for a". So if a >=0, then there is a decreasing interval. If a <0, then there is no decreasing interval. So the answer would be a >=0? Hmm, but let's check with a=0.5. Let's see. If a=0.5, then f'(x)=1/x -0.5x -1. Let's see if this derivative can be negative somewhere. Take x=2: f'(2)=0.5 -1 -1= -1.5 <0. x=1: f'(1)=1 -0.5 -1= -0.5 <0. x=0.5: f'(0.5)=2 -0.25 -1=0.75>0. So, for a=0.5, the derivative is positive near 0.5 and negative at x=1 and x=2. So, yes, there is a decreasing interval. So a=0.5 is acceptable.But wait, what if a is between 0 and some value? Let me see. Wait, as long as a>0, then as x increases, the term -a x dominates, so derivative tends to -infinity, so there must be some x where derivative is negative. Therefore, for all a >=0, the function has a decreasing interval. For a <0, derivative is always positive. So the answer is a >=0. Therefore, the range of a is [0, ∞).Wait, but let's verify for a=0. When a=0, f'(x)=1/x -1, which is negative when x>1, so yes, decreasing interval (1, ∞). For a>0, as in part (I), there's a critical point, and after that, decreasing. For a <0, no decreasing interval. So the range is a >=0. So part (II)'s answer is a ∈ [0, ∞).But let me check with another value of a>0. Let's take a=1. Then f'(x)=1/x -x -1. At x=0.5: 2 -0.5 -1=0.5>0. At x=1:1 -1 -1=-1<0. So crosses zero somewhere between 0.5 and 1. Therefore, decreasing interval exists. For a=0.1: f'(x)=1/x -0.1x -1. At x=10: 0.1 -1 -1=-1.9<0. So even for very small a>0, as x becomes large enough, derivative is negative. So indeed, for any a>0, there exists x such that f'(x)<0. Therefore, the range of a is [0, ∞). Therefore, part (II)'s answer is a ≥0. So written in interval notation, it's [0, ∞). So summarizing:(I) When a=2, the function is increasing on (0, 1/2) and decreasing on (1/2, ∞), with a maximum at x=1/2.(II) The range of a for which f(x) has a decreasing interval is a ≥0.Final Answer(I) The function is increasing on boxed{left(0, frac{1}{2}right)} and decreasing on boxed{left(frac{1}{2}, inftyright)} with a maximum at boxed{x = frac{1}{2}}.(II) The range of values for ( a ) is boxed{[0, infty)}.</think>Given the function ( f(x) = ln x - frac{1}{2} ax^2 - x ) where ( a in mathbb{R} ):(I) When ( a = 2 ), the function simplifies to ( f(x) = ln x - x^2 - x ). To find the intervals of monotonicity and the extremum, we first compute the derivative:[ f'(x) = frac{1}{x} - 2x - 1. ]Setting ( f'(x) = 0 ) gives:[ frac{1}{x} - 2x - 1 = 0 implies 1 - 2x^2 - x = 0 implies 2x^2 + x - 1 = 0. ]Solving this quadratic equation, we get:[ x = frac{-1 pm sqrt{9}}{4} implies x = frac{1}{2} text{ (since } x > 0). ]Testing intervals around ( x = frac{1}{2} ):- For ( x in (0, frac{1}{2}) ), ( f'(x) > 0 ) (increasing).- For ( x in (frac{1}{2}, infty) ), ( f'(x) < 0 ) (decreasing).Thus, the function is increasing on ( (0, frac{1}{2}) ) and decreasing on ( (frac{1}{2}, infty) ) with a maximum at ( x = frac{1}{2} ).(II) To determine the range of ( a ) for which ( f(x) ) has a monotonically decreasing interval, we analyze the derivative:[ f'(x) = frac{1}{x} - ax - 1. ]For ( f(x) ) to have a decreasing interval, ( f'(x) < 0 ) for some ( x > 0 ). As ( x to 0^+ ), ( frac{1}{x} to infty ), and as ( x to infty ), the term ( -ax ) dominates. For ( a geq 0 ), ( f'(x) ) tends to ( -infty ) as ( x to infty ), ensuring ( f'(x) < 0 ) for some ( x ). For ( a < 0 ), ( f'(x) ) remains positive for all ( x > 0 ). Therefore, the range of ( a ) is ( [0, infty) ).Final Answers:(I) The function is increasing on boxed{left(0, frac{1}{2}right)} and decreasing on boxed{left(frac{1}{2}, inftyright)} with a maximum at boxed{x = frac{1}{2}}.(II) The range of values for ( a ) is boxed{[0, infty)}.,<think>Okay, so I need to solve this problem about the function f(x) = ln x - (1/2) a x² - x. There are two parts: part I where a=2, and I have to find the intervals where the function is increasing or decreasing, and find the extremum points. Part II is about determining the range of a for which the function has a monotonically decreasing interval. Let me start with part I.First, when a=2, the function becomes f(x) = ln x - (1/2)*2*x² - x. Simplifying that, the (1/2)*2 cancels out, so it's ln x - x² - x. Now, to find the intervals of monotonicity, I need to take the derivative of f(x) with respect to x and analyze where it's positive or negative.So, f'(x) is the derivative of ln x, which is 1/x, minus the derivative of x², which is 2x, minus the derivative of x, which is 1. Therefore, f'(x) = 1/x - 2x - 1. Wait, hold on. Wait, the original function is ln x - (1/2)*a x² - x. When a=2, that term becomes (1/2)*2 x² = x². So the derivative is 1/x - 2x -1. Correct.Now, to find critical points, set f'(x) = 0. So 1/x - 2x -1 = 0. Let's solve for x. Multiply both sides by x to eliminate the denominator (assuming x ≠ 0, but since ln x is part of the original function, x must be >0 anyway). So 1 - 2x² - x = 0. Rearranged: -2x² - x +1 =0. Multiply both sides by -1 to make it 2x² +x -1 =0.Now, solving 2x² +x -1 =0. Using quadratic formula: x = [-b ± sqrt(b² -4ac)]/(2a). Here a=2, b=1, c=-1. So discriminant is 1² -4*2*(-1) = 1 +8 =9. So sqrt(9)=3. Therefore, x = [-1 ±3]/4.So two solutions: (-1 +3)/4 = 2/4 = 1/2, and (-1 -3)/4 = -4/4 = -1. But x must be >0, so x=-1 is rejected. So the critical point is at x=1/2.Now, to determine the intervals of increase and decrease, we test the sign of f'(x) in the intervals (0, 1/2) and (1/2, ∞). Let's pick test points.First interval (0, 1/2): Let's take x=1/4. Then f'(1/4) = 1/(1/4) - 2*(1/4) -1 = 4 - 0.5 -1 = 2.5. Which is positive. So the function is increasing on (0, 1/2).Second interval (1/2, ∞): Let's take x=1. Then f'(1) = 1/1 -2*1 -1 =1 -2 -1= -2. Negative. So the function is decreasing on (1/2, ∞).Therefore, the function is increasing on (0, 1/2) and decreasing on (1/2, ∞). Therefore, there is a local maximum at x=1/2. Since it's the only critical point and the function changes from increasing to decreasing there, it's indeed a maximum. The question mentions "extremum", so we need to find the value as well.So, the extremum is at x=1/2. Let's compute f(1/2):f(1/2) = ln(1/2) - (1/2)^2 - (1/2) = ln(1/2) - 1/4 -1/2. ln(1/2) is -ln2, so that's -ln2 - 3/4. So the maximum is at x=1/2 with value -ln2 - 3/4.Therefore, part I's answer: increasing on (0, 1/2), decreasing on (1/2, ∞), with a maximum at (1/2, -ln2 -3/4).Now part II: If y=f(x) has a monotonically decreasing interval, determine the range of values for a.Hmm. So the function f(x) must have some interval where it's decreasing. But since the function is defined for x>0, we need to ensure that there exists some interval (c, d) within x>0 where f'(x) <0. However, it's possible that for certain a, f'(x) is always positive or always negative. Wait, but since f'(x) is a function of x, let's analyze f'(x) in general.Given f(x) = ln x - (1/2)a x² -x. So f'(x) = 1/x -a x -1. So f'(x) = -a x -1 + 1/x. Wait, that's equivalent to f'(x) = (1/x) - a x -1.We need to find for which a there exists some interval where f'(x) <0. But actually, for a function to have a decreasing interval, it's sufficient that f'(x) <0 somewhere. However, since the function's domain is x>0, we need to check if f'(x) can be negative for some x>0.Alternatively, maybe the problem requires that the function is decreasing on some interval, not necessarily the entire domain. Wait, but even if f'(x) is sometimes negative and sometimes positive, then there are intervals where it's decreasing. But if f'(x) is always non-negative or always non-positive, then it's monotonic. Wait, but the problem says "has a monotonically decreasing interval", so even if the function is first increasing then decreasing, that still counts as having a decreasing interval. Wait, but maybe the problem is to ensure that the function is decreasing on some interval. So even if the function has both increasing and decreasing parts, as long as there is at least one interval where it's decreasing, then a is acceptable. Wait, but the question says "If y=f(x) has a monotonically decreasing interval, determine the range of values for a". So perhaps we need to find all a such that there exists some interval where f is decreasing. But since f'(x) is continuous (as f is differentiable for x>0), if f'(x) is ever negative, then there will be an interval around that point where it's negative, hence decreasing. Therefore, the problem reduces to finding all a for which there exists some x>0 where f'(x) <0.But maybe the problem is more about the function having a decreasing interval, but not necessarily being entirely decreasing. Wait, but in part II, it's possible that for some a, the function is always increasing, hence no decreasing interval. Wait, but for example, if a is very negative, maybe the function is always increasing. Wait, let's think.Wait, let's consider f'(x) = 1/x -a x -1. Let's rearrange this as f'(x) = -a x -1 + 1/x. Hmm. To find when there exists x>0 such that f'(x) <0.Alternatively, maybe the function f'(x) can be analyzed for its minimum. Since f'(x) is a function of x, perhaps we can find if f'(x) ever goes negative. If the minimal value of f'(x) is less than zero, then there exists an interval where it's decreasing.Alternatively, if the minimal value of f'(x) is below zero, then there is a decreasing interval. So perhaps we can find the critical points of f'(x) to find its minima or maxima.Wait, f'(x) = 1/x -a x -1. Let's compute its derivative to find critical points. f''(x) = -1/x² -a. So setting f''(x)=0 gives -1/x² -a =0 => -a =1/x² => a= -1/x². But since x>0, a must be negative. So only when a is negative, f''(x)=0 has solutions. So, critical points of f'(x) occur only when a is negative. Wait, but maybe this is complicating.Alternatively, let's treat f'(x) as a function: g(x) = 1/x -a x -1. We need to find when there exists x>0 such that g(x) <0. So for which a does g(x) <0 for some x>0.We can analyze the equation 1/x -a x -1 <0. Let's rearrange:1/x -1 < a x. So (1/x -1)/x <a => (1 -x)/x² <a. But not sure.Alternatively, let's consider the inequality 1/x -a x -1 <0. Let's rearrange:-a x <1/x -1. Multiplying both sides by -1 (remember to reverse the inequality if a is positive or negative, but this complicates things). Maybe it's better to analyze the function g(x) =1/x -a x -1 and see for which a it attains negative values.Alternatively, set g(x)=0 and find when there are solutions x>0, which would imply that the function f'(x) crosses zero, hence changing sign. If there are solutions, then f(x) has both increasing and decreasing intervals. If not, then f'(x) is always positive or always negative. Wait, but the problem says "has a monotonically decreasing interval", so even if the function is decreasing on part of its domain, that's okay. So the existence of x>0 where g(x)=f'(x) <0.Alternatively, the problem is equivalent to: find all real numbers a such that there exists x>0 with 1/x -a x -1 <0.So we need to find a such that the inequality 1/x -a x -1 <0 holds for some x>0. So what values of a allow this inequality to be true?Let me rearrange the inequality:1/x -a x -1 <0=> 1/x -1 <a xIf x>0, then we can divide both sides by x (if x>0, division doesn't change inequality direction):(1/x -1)/x <a=> (1 -x)/x² <aSo the inequality becomes a > (1 -x)/x² for some x>0.Therefore, to find the range of a such that there exists x>0 where a > (1 -x)/x². So the range of a is all real numbers greater than the minimum value of (1 -x)/x² over x>0.Wait, but actually, if we can find x>0 such that a > (1 -x)/x², then a must be greater than the infimum of (1 -x)/x² over x>0. However, I need to check.Wait, no. Let me think again. The original inequality is 1/x -a x -1 <0, which is equivalent to a > (1 -x)/x² for some x>0. So the set of all a for which this inequality holds is a > the minimum value of (1 -x)/x² for x>0. Wait, but maybe the function h(x) = (1 -x)/x² has a maximum or minimum. Let's compute its extremum.Let h(x) = (1 -x)/x². Find h'(x):h'(x) = [ -1 * x² - (1 -x)*2x ] / x^4Wait, quotient rule: h(x) = (1 -x)/x². Then h'(x) = [ ( -1 ) * x² - (1 -x)*2x ] / x^4.Compute numerator:-1 * x² - (1 -x)*2x = -x² -2x +2x² = (-x² +2x²) -2x = x² -2x.Therefore, h'(x) = (x² -2x)/x^4 = x(x -2)/x^4 = (x -2)/x^3.Set h'(x)=0: (x -2)/x^3=0 => x=2. Critical point at x=2. Also, check when x approaches 0+ and infinity.As x approaches 0+, h(x) = (1 -x)/x² approaches infinity (since numerator approaches 1 and denominator approaches 0). As x approaches infinity, h(x) approaches ( -x ) /x² = -1/x, which approaches 0. So the function h(x) tends to 0 from below as x approaches infinity.At x=2, h(x) = (1 -2)/4 = -1/4. Let's see the behavior around x=2. For x <2, h'(x) is negative (since x-2 negative, x^3 positive, so h'(x) negative). For x>2, h'(x) positive. Therefore, x=2 is a minimum point. Therefore, the minimum value of h(x) is -1/4 at x=2. The function h(x) goes from infinity at x approaching 0+ to -1/4 at x=2, then increases to 0 as x approaches infinity. Therefore, the range of h(x) is (-1/4, ∞). Wait, no. Wait, when x approaches 0+, h(x) approaches infinity. Between x=0 and x=1, h(x) is positive, because 1 -x is positive (x <1) so numerator positive, denominator positive. At x=1, h(1)=0. For x>1, h(x) becomes negative, since 1 -x is negative. Then, as x increases beyond 1, h(x) becomes negative, reaches a minimum at x=2 of -1/4, then increases back towards 0 as x approaches infinity.Therefore, the range of h(x) is (-1/4, ∞). Because for x>0:- When x <1, h(x) positive, going from infinity to 0.- When x between 1 and 2, h(x) negative, decreasing to -1/4.- When x >2, h(x) negative, increasing back to 0.Therefore, h(x) can take any value greater than -1/4. So the infimum of h(x) is -1/4. Therefore, for a > -1/4, there exists x>0 such that a > h(x) = (1 -x)/x². Therefore, the set of a such that there exists x>0 with a > (1 -x)/x² is a > -1/4. Hence, if a > -1/4, then the original inequality 1/x -a x -1 <0 holds for some x>0, meaning f'(x) <0 in that x, hence f(x) has a decreasing interval.Wait, but let me verify. Wait, h(x) has a minimum value of -1/4. So if a > -1/4, then there exists some x>0 such that a > h(x). Therefore, for such a, the inequality holds. However, if a ≤ -1/4, then h(x) ≥ a for all x>0. Wait, because the minimum value of h(x) is -1/4, so h(x) ≥ -1/4 for all x>0. So if a ≤ -1/4, then h(x) ≥ -1/4 ≥ a, so h(x) ≥ a for all x>0, meaning that (1 -x)/x² ≥a for all x>0, which implies that 1/x -a x -1 ≥0 for all x>0. Therefore, f'(x) ≥0 for all x>0, so the function is non-decreasing everywhere. Hence, if a ≤ -1/4, f(x) is non-decreasing on its entire domain, so there are no decreasing intervals. Therefore, the range of a for which f(x) has a decreasing interval is a > -1/4.But wait, let me check with a specific example. Let's take a=0. Then f'(x)=1/x -0 -1 =1/x -1. So f'(x)=1/x -1. Setting this less than zero: 1/x -1 <0 => x>1. So for a=0, which is greater than -1/4, the function is decreasing for x>1. So that works.If a=-1/4, then h(x) ≥ -1/4. So h(x)= (1 -x)/x² ≥ -1/4. Therefore, a=-1/4, and the inequality is 1/x - (-1/4) x -1 <0? Wait, no. Wait, for a=-1/4, the original inequality is 1/x -a x -1 <0 => 1/x -(-1/4)x -1 <0 => 1/x + (1/4)x -1 <0. So we need to see if this can be negative for some x>0.But according to the previous analysis, when a=-1/4, since a equals the minimum of h(x), then (1 -x)/x² ≥ -1/4 for all x>0, hence 1/x - (-1/4)x -1 =1/x + (1/4)x -1= [1/x -1] + (1/4)x. Hmm. Let me compute f'(x) when a=-1/4: f'(x)=1/x - (-1/4)x -1=1/x + (x/4) -1. Let's see if this is ever negative. Let's take x=2. f'(2)=1/2 + 2/4 -1=1/2 +1/2 -1=0. At x=2, f'(2)=0. At x=4, f'(4)=1/4 +4/4 -1=1/4 +1 -1=1/4>0. At x=1, f'(1)=1 +1/4 -1=1/4>0. At x approaching 0+, f'(x)=1/x +0 -1 approaches infinity. At x approaching infinity, f'(x)=0 + (x/4) -1 approaches infinity. So when a=-1/4, f'(x) is always ≥0, because at x=2, it's 0, and elsewhere it's positive. Therefore, the function is non-decreasing for a=-1/4, so no decreasing intervals. Hence, the critical value is a>-1/4.Therefore, the range of a is a > -1/4.Wait, but let me check with a=-1/2, which is less than -1/4. Then f'(x)=1/x - (-1/2)x -1=1/x + (x/2) -1. Let's see if this is ever negative. Let's take x=1: f'(1)=1 +0.5 -1=0.5>0. x=2: f'(2)=0.5 +1 -1=0.5>0. x=0.5: f'(0.5)=2 +0.25 -1=1.25>0. So seems like f'(x) is always positive. So indeed, when a ≤ -1/4, the derivative is non-negative everywhere, so function is non-decreasing. Therefore, the function only has a decreasing interval when a > -1/4.Therefore, the answer to part II is a > -1/4. So the range is (-1/4, ∞).But wait, let me verify with another value of a between -1/4 and 0. Let's take a=-1/8. Then f'(x)=1/x - (-1/8)x -1=1/x +x/8 -1. Let's check at x=2. f'(2)=0.5 + 0.25 -1= -0.25<0. So at x=2, f'(x) is negative. Therefore, there is a decreasing interval. Hence, for a=-1/8 > -1/4, there exists an interval where the function is decreasing. Hence, confirming our conclusion.Therefore, the range of a is a > -1/4.Thus, summarizing:(I) When a=2, the function is increasing on (0, 1/2) and decreasing on (1/2, ∞), with a maximum at x=1/2.(II) The range of a is a > -1/4.Final Answer(I) The function is increasing on boxed{left(0, frac{1}{2}right)} and decreasing on boxed{left(frac{1}{2}, inftyright)} with a local maximum at boxed{left(frac{1}{2}, -ln 2 - frac{3}{4}right)}.(II) The range of values for ( a ) is boxed{(-frac{1}{4}, infty)}.</think>Given the function ( f(x) = ln x - frac{1}{2} ax^2 - x ) where ( a in mathbb{R} ):(I) When ( a = 2 ), the function simplifies to ( f(x) = ln x - x^2 - x ). To find the intervals of monotonicity and the extremum, we first compute the derivative:[ f'(x) = frac{1}{x} - 2x - 1. ]Setting ( f'(x) = 0 ) gives:[ frac{1}{x} - 2x - 1 = 0 implies 1 - 2x^2 - x = 0 implies 2x^2 + x - 1 = 0. ]Solving the quadratic equation, we get:[ x = frac{-1 pm sqrt{1 + 8}}{4} = frac{-1 pm 3}{4}. ]The critical points are ( x = frac{1}{2} ) (since ( x = -1 ) is not in the domain ( x > 0 )). Testing intervals around ( x = frac{1}{2} ):- For ( 0 < x < frac{1}{2} ), ( f'(x) > 0 ) (increasing).- For ( x > frac{1}{2} ), ( f'(x) < 0 ) (decreasing).Thus, the function is increasing on ( (0, frac{1}{2}) ) and decreasing on ( (frac{1}{2}, infty) ) with a local maximum at ( x = frac{1}{2} ). Evaluating ( f left( frac{1}{2} right) ):[ fleft( frac{1}{2} right) = ln left( frac{1}{2} right) - left( frac{1}{2} right)^2 - frac{1}{2} = -ln 2 - frac{3}{4}. ](II) To determine the range of ( a ) for which ( f(x) ) has a monotonically decreasing interval, we analyze the derivative:[ f'(x) = frac{1}{x} - ax - 1. ]We need ( f'(x) < 0 ) for some ( x > 0 ). Rewriting the inequality:[ frac{1}{x} - ax - 1 < 0 implies frac{1 - x}{x^2} < a. ]Let ( h(x) = frac{1 - x}{x^2} ). The minimum value of ( h(x) ) occurs at ( x = 2 ):[ h(2) = frac{1 - 2}{2^2} = -frac{1}{4}. ]Thus, ( h(x) geq -frac{1}{4} ) for all ( x > 0 ). For ( f'(x) < 0 ) to hold, ( a ) must be greater than the minimum value of ( h(x) ):[ a > -frac{1}{4}. ] Final Answer(I) The function is increasing on boxed{left(0, frac{1}{2}right)} and decreasing on boxed{left(frac{1}{2}, inftyright)} with a local maximum at boxed{left(frac{1}{2}, -ln 2 - frac{3}{4}right)}.(II) The range of values for ( a ) is boxed{(-frac{1}{4}, infty)}.]

🔑:(I) When a=2, the function becomes f(x) = ln x - x^2 - x, with its domain being (0, +infty).To find the monotonic intervals and extremum, we first find the derivative of f(x):f'(x) = frac{1}{x} - 2x - 1 = - frac{(x+1)(2x-1)}{x}.For f'(x) frac{1}{2}.For f'(x)>0, we get 0 0 for x>0.When a=0, x-1>0 is satisfied for all x>1.When a>0, the inequality ax^2 + x - 1 > 0 is satisfied for all x>0 because the quadratic curve opens upwards and has a negative y-intercept.When a1, the inequality holds if the quadratic ax^2 + x - 1 = 0 has two positive roots. This requires the discriminant to be positive:Delta=1+4a>0,which gives a > -frac{1}{4}.Therefore, combining all cases, the range of a for which the function is monotonically decreasing is boxed{(-frac{1}{4},+infty)}.

🧠:[<think>Okay, so I need to figure out which of the propositions A, B, C, D are correct. Let's take them one by one.Starting with proposition A: If there are exactly two distinct points on the circle C: x² + y² = 4 that are at a distance of 1 from the line l: x + y + m = 0, then the range of m is √2 < m < 3√2.Hmm. So the problem is about finding the values of m such that the line x + y + m = 0 is at a certain distance from the circle, such that exactly two points on the circle are at distance 1 from the line. Let me recall that the distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²). Here, the line is x + y + m = 0, so the distance from any point (x, y) on the circle to this line would be |x + y + m| / sqrt(2). Now, the problem states there are exactly two distinct points on the circle where this distance is 1. That would mean that the line x + y + m = 0 is such that when we move it parallel to itself by a distance of 1 on either side, those two lines intersect the circle in exactly two points each, or maybe in some way that there's only two points where the distance is exactly 1? Wait, actually, the set of points at distance 1 from the line l would be two lines parallel to l, each at a distance of 1 from l. So, if the original line l is such that these two parallel lines intersect the circle in four points (two on each line), but if the original line is too far from the circle, maybe those parallel lines don't intersect the circle. Wait, but the question is about the number of points on the circle that are at distance 1 from l. So, if the line l is such that the distance from the center of the circle to l is such that when you add and subtract 1/sqrt(2) (since the distance formula has denominator sqrt(2)), then the circle intersects the two parallel lines (distance 1 from l) in two points each, making four points in total. But the problem says there are exactly two distinct points on the circle at distance 1 from l. Therefore, maybe the line l is positioned such that only one of the two parallel lines (distance 1 above and below l) intersects the circle, each giving one point. Wait, but the circle is symmetric. If the line l is at a certain distance from the center, then moving it up and down by 1 unit (in terms of distance) would affect how many intersections there are.Wait, the circle x² + y² = 4 has center (0,0) and radius 2. The distance from the center to the line l: x + y + m = 0 is |0 + 0 + m| / sqrt(1² + 1²) = |m| / sqrt(2). So the distance from center to line l is |m| / sqrt(2). For there to be exactly two points on the circle at distance 1 from l, that implies that the two lines parallel to l at distance 1 from l each intersect the circle either once or twice, but overall only two points in total. Wait, but each of the two parallel lines would intersect the circle either 0, 1, or 2 points. So if the original line l is such that one of the parallel lines is tangent to the circle (distance 1) and the other intersects the circle at two points, then the total number of points at distance 1 would be three (two from one line, one tangent point). But the problem states exactly two points. So maybe both parallel lines are such that each intersects the circle once (i.e., both are tangent), but that would require the distance from the center to each of the parallel lines is equal to the radius. Wait, but the distance from the center to the original line l is |m| / sqrt(2), then the distance to the upper parallel line would be |m| / sqrt(2) + 1, and the lower one would be |m| / sqrt(2) - 1. Wait, actually, the distance from the center to the upper line is |m + sqrt(2)| / sqrt(2), and to the lower line is |m - sqrt(2)| / sqrt(2). Wait, maybe I need to compute that.Wait, the original line is x + y + m = 0. The lines at distance 1 from it would be x + y + m ± sqrt(2) = 0. Because the formula for parallel lines is changing the constant term by distance * sqrt(a² + b²). Since distance is 1, the change in constant term is 1 * sqrt(1 + 1) = sqrt(2). So the two lines are x + y + m + sqrt(2) = 0 and x + y + m - sqrt(2) = 0. Then, the distance from the center (0,0) to each of these lines is |m + sqrt(2)| / sqrt(2) and |m - sqrt(2)| / sqrt(2). For the original circle x² + y² = 4 with radius 2, the condition for a line to intersect the circle is that the distance from the center to the line is less than the radius. So, the upper line x + y + m + sqrt(2) = 0 would have distance |m + sqrt(2)| / sqrt(2) from the center. For this line to intersect the circle, we need |m + sqrt(2)| / sqrt(2) < 2. Similarly, the lower line x + y + m - sqrt(2) = 0 would have distance |m - sqrt(2)| / sqrt(2) < 2.But the problem states that there are exactly two points on the circle at distance 1 from line l. That would happen when one of the parallel lines intersects the circle in two points and the other doesn't intersect the circle. Because if both lines intersect the circle, then total points would be four (two on each line). If one line is tangent (distance equals radius) and the other line is outside, then there would be one point (the tangent point). If both lines are outside, then no points. So, for exactly two points, we need one line intersects the circle at two points and the other line does not intersect the circle. Therefore, one of the distances |m + sqrt(2)| / sqrt(2) < 2 and |m - sqrt(2)| / sqrt(2) >= 2, or vice versa.Wait, but the original line l is somewhere in the plane, so depending on m, the distance from the center to l is |m| / sqrt(2). Then, the two lines at distance 1 from l would be at distances |m| / sqrt(2) ± 1 from the center. Wait, actually, the distance from the center to each of the parallel lines is |m ± sqrt(2)| / sqrt(2). Let me verify that.The original line l: x + y + m = 0. A parallel line at distance 1 would be x + y + (m ± sqrt(2)) = 0. The distance from (0,0) to this line is |m ± sqrt(2)| / sqrt(2). So, for the upper line x + y + m + sqrt(2) = 0, distance from center is |m + sqrt(2)| / sqrt(2). For the lower line x + y + m - sqrt(2) = 0, distance is |m - sqrt(2)| / sqrt(2).To have exactly two points on the circle at distance 1 from l, we need exactly one of these two lines to intersect the circle (i.e., have distance less than 2), and the other line not to intersect the circle (distance >= 2). So either:1. |m + sqrt(2)| / sqrt(2) < 2 and |m - sqrt(2)| / sqrt(2) >= 2OR2. |m - sqrt(2)| / sqrt(2) < 2 and |m + sqrt(2)| / sqrt(2) >= 2.Let's solve these inequalities.First case:|m + sqrt(2)| / sqrt(2) < 2 => |m + sqrt(2)| < 2 sqrt(2) => -2 sqrt(2) < m + sqrt(2) < 2 sqrt(2) => subtract sqrt(2): -3 sqrt(2) < m < sqrt(2)And |m - sqrt(2)| / sqrt(2) >= 2 => |m - sqrt(2)| >= 2 sqrt(2) => m - sqrt(2) >= 2 sqrt(2) or m - sqrt(2) <= -2 sqrt(2) => m >= 3 sqrt(2) or m <= -sqrt(2)But combining with the first inequality, which is -3 sqrt(2) < m < sqrt(2), the intersection would be m <= -sqrt(2). But from the first inequality, m > -3 sqrt(2). So m is in (-3 sqrt(2), -sqrt(2)]Second case:|m - sqrt(2)| / sqrt(2) < 2 => |m - sqrt(2)| < 2 sqrt(2) => -2 sqrt(2) < m - sqrt(2) < 2 sqrt(2) => add sqrt(2): -sqrt(2) < m < 3 sqrt(2)And |m + sqrt(2)| / sqrt(2) >= 2 => |m + sqrt(2)| >= 2 sqrt(2) => m + sqrt(2) >= 2 sqrt(2) or m + sqrt(2) <= -2 sqrt(2) => m >= sqrt(2) or m <= -3 sqrt(2)Combining with the first inequality (-sqrt(2) < m < 3 sqrt(2)), the intersection is m >= sqrt(2) and m < 3 sqrt(2). So m is in [sqrt(2), 3 sqrt(2)).Therefore, combining both cases, the range of m is (-3 sqrt(2), -sqrt(2)] union [sqrt(2), 3 sqrt(2)). But the proposition A says the range is sqrt(2) < m < 3 sqrt(2). So that's only half of the possible range. So perhaps m can be positive or negative. But proposition A only mentions the positive range. However, maybe in the problem, they are considering m to be positive? Wait, the original line is x + y + m = 0. If m is positive, the line is in a certain position; if m is negative, it's in another. But the problem doesn't specify m being positive or negative. So actually, the correct range should be |m| in [sqrt(2), 3 sqrt(2)), but since m can be positive or negative, the range is -3 sqrt(2) < m < -sqrt(2) or sqrt(2) < m < 3 sqrt(2). However, proposition A only mentions sqrt(2) < m < 3 sqrt(2). Therefore, A is only partially correct. Because the range could also include negative m. Unless the problem assumes m is positive. But there's no indication of that. So proposition A is incorrect because it only considers the positive side. So A is incorrect.Wait, but maybe I made a mistake here. Let me double-check. The problem states "the range of m is sqrt(2) < m < 3√2". But m could be negative as well. So unless there's a restriction on m that's not mentioned, proposition A is only considering the positive m case. Therefore, A is incorrect.Moving on to proposition B: If point P(x1, y1) lies on the circle C: x² + y² = 4, then the maximum value of (x1 - 1)² + (y1 - 2√2)² is 25.Alright, so we need to find the maximum value of the expression (x - 1)^2 + (y - 2√2)^2 for points (x,y) on the circle x² + y² = 4.This is equivalent to finding the maximum distance squared from the point (1, 2√2) to a point on the circle x² + y² = 4. The maximum distance squared would be the square of the distance from (1, 2√2) to the center (0,0) plus the radius squared. Wait, the maximum distance from a point to a circle is the distance from the external point to the center plus the radius. Then, the maximum distance squared would be (distance + radius)^2.Let me compute the distance from (1, 2√2) to (0,0):sqrt(1^2 + (2√2)^2) = sqrt(1 + 8) = sqrt(9) = 3.So the distance is 3, the radius of the circle is 2. Therefore, the maximum distance from (1, 2√2) to a point on the circle is 3 + 2 = 5. Therefore, the maximum squared distance is 25. So the maximum value of (x1 - 1)^2 + (y1 - 2√2)^2 is indeed 25. Therefore, proposition B is correct.Proposition C: The equation of the line containing the common chord of the circles x² + y² + 4x - 4y = 0 and x² + y² + 2x - 12 = 0 is x - 2y + 6 = 0.To find the common chord, we can subtract the equations of the two circles. Let's write both equations:First circle: x² + y² + 4x - 4y = 0.Second circle: x² + y² + 2x - 12 = 0.Subtracting the second equation from the first:( x² + y² + 4x - 4y ) - ( x² + y² + 2x - 12 ) = 0 - 0.Simplify:4x - 4y - 2x + 12 = 0 => 2x - 4y + 12 = 0.Divide by 2: x - 2y + 6 = 0.So the equation of the common chord is x - 2y + 6 = 0. Therefore, proposition C is correct.Wait, but let me confirm. The common chord is the line where the two circles intersect. To find that, subtracting the equations gives the equation of the radical axis, which is the line containing the common chord. So yes, the radical axis here is x - 2y + 6 = 0, so proposition C is correct.Proposition D: The circles C: x² + y² + 2x = 0 and C2: x² + y² - 4x - 8y + 4 = 0 have exactly three common tangent lines.Three common tangents would imply that the circles are tangent to each other, so they have one common tangent at the point of contact, and two other common tangents. Wait, but usually, two circles can have four common tangents if they are separate: two direct and two transverse. If they intersect at two points, then they have two common tangents. If they are tangent, they have three common tangents (the tangent at the point of contact and two others). So proposition D says exactly three common tangents, which would mean the circles are tangent to each other.Let me check if the circles are tangent.First, rewrite both circles in standard form.Circle C: x² + y² + 2x = 0. Completing the square:x² + 2x + y² = 0 => (x + 1)^2 + y² = 1. So center at (-1, 0), radius 1.Circle C2: x² + y² - 4x - 8y + 4 = 0. Completing the square:x² - 4x + y² - 8y = -4.x² - 4x + 4 + y² - 8y + 16 = -4 + 4 + 16 => (x - 2)^2 + (y - 4)^2 = 16. So center at (2, 4), radius 4.Now, compute the distance between centers: sqrt[(2 - (-1))² + (4 - 0)^2] = sqrt[(3)^2 + (4)^2] = sqrt[9 + 16] = sqrt[25] = 5.The radii are 1 and 4. Sum of radii is 5, which is equal to the distance between centers. Therefore, the circles are externally tangent to each other. Therefore, they have exactly three common tangent lines: one at the point of contact, and two external tangents. Therefore, proposition D is correct.Wait, but let me confirm. If two circles are externally tangent, they have three common tangents: one common tangent at the point where they touch, and two separate common tangents that don't pass between the circles. So yes, three common tangents. Hence, D is correct.So summarizing:A: Incorrect, because the range of m should also include negative values.B: Correct.C: Correct.D: Correct.Therefore, propositions B, C, D are correct. But let's double-check.Wait, proposition A: The original problem says "the range of m is √2 < m < 3√2". But according to my earlier analysis, m can be in (-3√2, -√2) or (√2, 3√2). However, the problem states "the range of m is √2 < m < 3√2", which is only the positive part. But since m can be negative as well, the correct range is two intervals. However, maybe the problem is considering m to be positive? But the original line equation is x + y + m = 0; m can be any real number. So the answer should include both positive and negative ranges. Therefore, proposition A is incorrect.Thus, the correct propositions are B, C, D.Final Answerboxed{BCD}</think>Analysis of Proposition A:The problem involves finding the range of ( m ) such that there are exactly two distinct points on the circle ( x^2 + y^2 = 4 ) at a distance of 1 from the line ( x + y + m = 0 ). The distance from the center of the circle (0,0) to the line is ( frac{|m|}{sqrt{2}} ). For there to be exactly two points on the circle at distance 1, the lines parallel to ( x + y + m = 0 ) at distance 1 must intersect the circle such that one line intersects the circle and the other does not. This results in the range of ( m ) being ( -3sqrt{2} < m < -sqrt{2} ) or ( sqrt{2} < m < 3sqrt{2} ). Since proposition A only considers the positive range, it is incorrect.Analysis of Proposition B:The maximum value of ( (x_1 - 1)^2 + (y_1 - 2sqrt{2})^2 ) for points on the circle ( x^2 + y^2 = 4 ) is calculated. The distance from the point (1, ( 2sqrt{2} )) to the center (0,0) is 3, and adding the radius 2 gives a maximum distance of 5. Thus, the maximum value of the expression is ( 25 ), making proposition B correct.Analysis of Proposition C:The common chord of the circles ( x^2 + y^2 + 4x - 4y = 0 ) and ( x^2 + y^2 + 2x - 12 = 0 ) is found by subtracting their equations, resulting in ( x - 2y + 6 = 0 ). This is correct, so proposition C is correct.Analysis of Proposition D:The circles ( x^2 + y^2 + 2x = 0 ) and ( x^2 + y^2 - 4x - 8y + 4 = 0 ) are rewritten in standard form. The distance between their centers is 5, which equals the sum of their radii (1 and 4), indicating they are externally tangent. Externally tangent circles have exactly three common tangents, making proposition D correct.Final Answerboxed{BCD},<think>Okay, let's try to figure out which of these propositions A, B, C, D are correct. I need to go through each one step by step.Starting with proposition A: If there are exactly two distinct points on the circle C: x² + y² = 4 that are at a distance of 1 from the line l: x + y + m = 0, then the range of m is √2 < m < 3√2.Hmm. So the circle C is centered at the origin with radius 2. The line is x + y + m = 0. We need to find m such that there are exactly two points on the circle at distance 1 from this line.First, the distance from the center of the circle (0,0) to the line l is |0 + 0 + m| / sqrt(1² + 1²) = |m| / √2. For there to be points on the circle at distance 1 from the line, the distance from the center to the line must be such that the line is within a certain range relative to the circle.If the distance from the center to the line is d, then the maximum distance from the line to any point on the circle would be d + 2 (radius), and the minimum would be |d - 2|. So, if we want exactly two points on the circle at distance 1 from the line, then 1 must lie between |d - 2| and d + 2. But actually, wait, maybe another approach: For a line and a circle, the set of points on the circle at a fixed distance from the line would form two points if the line is outside the circle but within a certain distance. Wait, maybe not. Let me recall that the distance from the center to the line is d. If we want points on the circle at distance 1 from the line, then those points lie on two lines parallel to the original line at distance 1. The intersection of those two lines with the circle will give the points. So, if the original line is at distance d from the center, then the two parallel lines are at distances d ± 1. For there to be exactly two intersection points with the circle, each of these parallel lines must intersect the circle exactly once? Wait, no. Wait, the original line is at distance d, and the two lines parallel to it at distance 1 would each have their own distance from the center: d + 1 and |d - 1|. For there to be exactly two points on the circle at distance 1 from the original line, each of these two parallel lines must intersect the circle, but each line can intersect the circle in 0, 1, or 2 points. Wait, actually, no. If the original line is l, then the points on the circle at distance 1 from l lie on two lines parallel to l, each at distance 1 from l. So the number of such points depends on how these two parallel lines intersect the circle. For there to be exactly two points, each of the two lines must intersect the circle exactly once (so each line is tangent), but that would give two points total. Wait, but if a line intersects a circle twice, that would be two points, so two lines each intersecting once would total two points, or each line could intersect twice, giving four points. So, the number of points at distance 1 from l on the circle would be 0, 1, 2, 3, or 4. So, if we want exactly two points, that could happen if one of the parallel lines cuts the circle twice and the other doesn't intersect, but that would give two points. Or, each line is tangent, giving one point each, totaling two. Wait, but tangent lines would give exactly one point each, so two points total. If the lines are such that one cuts the circle twice and the other once, that would be three points. Hmm, maybe.Wait, perhaps the correct approach is as follows. The set of points at distance 1 from the line l: x + y + m = 0 are two lines: x + y + m ± 1*sqrt(1² + 1²) = 0. Wait, no. The formula for the distance from a point to a line is |Ax + By + C| / sqrt(A² + B²). So, the lines at distance 1 from l: x + y + m = 0 would be x + y + m ± 1*sqrt(2) = 0. Wait, because the distance is 1, so the numerator must be 1*sqrt(2). So the equations of the two lines are x + y + (m ± sqrt(2)) = 0. These two lines are parallel to l.Now, the number of intersection points between these two lines and the circle x² + y² = 4. For each line, we can compute the distance from the center (0,0) to the line. If that distance is less than 2, the line intersects the circle at two points; if equal to 2, tangent (one point); if greater than 2, no intersection.So, for the lines x + y + (m ± sqrt(2)) = 0, the distance from (0,0) to each line is |0 + 0 + (m ± sqrt(2))| / sqrt(2) = |m ± sqrt(2)| / sqrt(2).To have exactly two points on the circle at distance 1 from line l, we need exactly one of these two lines to intersect the circle, and the other not. Because if both lines intersect the circle, each could intersect twice, leading to four points. If one line intersects twice and the other doesn't, that would give two points. If one line is tangent (one point) and the other doesn't intersect, that's one point. So, to have exactly two points, we need one line to intersect the circle twice and the other line not to intersect. Therefore, the distance from the center to one line must be less than 2, and the distance to the other line must be greater than 2.So, for the lines x + y + (m + sqrt(2)) = 0 and x + y + (m - sqrt(2)) = 0:Case 1: |m + sqrt(2)| / sqrt(2) < 2 and |m - sqrt(2)| / sqrt(2) > 2ORCase 2: |m + sqrt(2)| / sqrt(2) > 2 and |m - sqrt(2)| / sqrt(2) < 2Let's analyze Case 1:|m + sqrt(2)| / sqrt(2) < 2 => |m + sqrt(2)| < 2 sqrt(2) => -2 sqrt(2) < m + sqrt(2) < 2 sqrt(2) => subtract sqrt(2): -3 sqrt(2) < m < sqrt(2)Similarly, |m - sqrt(2)| / sqrt(2) > 2 => |m - sqrt(2)| > 2 sqrt(2) => m - sqrt(2) > 2 sqrt(2) or m - sqrt(2) < -2 sqrt(2) => m > 3 sqrt(2) or m < -sqrt(2)So in Case 1, we have m between -3 sqrt(2) and sqrt(2) (from first inequality) and m > 3 sqrt(2) or m < -sqrt(2). The overlap here is m < -sqrt(2) but also m > -3 sqrt(2). So in Case 1, m ∈ (-3 sqrt(2), -sqrt(2))Similarly, Case 2:|m + sqrt(2)| / sqrt(2) > 2 => |m + sqrt(2)| > 2 sqrt(2) => m + sqrt(2) > 2 sqrt(2) or m + sqrt(2) < -2 sqrt(2) => m > sqrt(2) or m < -3 sqrt(2)And |m - sqrt(2)| / sqrt(2) < 2 => |m - sqrt(2)| < 2 sqrt(2) => -2 sqrt(2) < m - sqrt(2) < 2 sqrt(2) => adding sqrt(2): -sqrt(2) < m < 3 sqrt(2)So in Case 2, m must be either m > sqrt(2) or m < -3 sqrt(2) (from first inequality), and also m ∈ (-sqrt(2), 3 sqrt(2)). The overlap here is sqrt(2) < m < 3 sqrt(2)Therefore, combining both cases, the range for m is (-3 sqrt(2), -sqrt(2)) union (sqrt(2), 3 sqrt(2))But proposition A states that the range is sqrt(2) < m < 3 sqrt(2). So part of the actual range. But since m can also be negative between -3 sqrt(2) and -sqrt(2), proposition A only considers the positive side. So depending on how the problem is phrased, maybe m is assumed positive? Wait, the original problem didn't specify m being positive. So proposition A is only giving half of the possible range. Therefore, proposition A is incorrect.Wait, but maybe the problem didn't specify direction, so maybe the user just wrote sqrt(2) < m < 3√2, but m could also be negative. So the actual range is |m| ∈ (sqrt(2), 3 sqrt(2)). So the proposition A is only considering positive m, so it's incomplete. Therefore, A is incorrect.Wait, but maybe I made a mistake here. Let me check again.The problem states: If there are exactly two distinct points on the circle C: x² + y² = 4 that are at a distance of 1 from the line l: x + y + m = 0, then the range of m is √2 < m < 3√2.But from our analysis, the possible m can be in (-3 sqrt(2), -sqrt(2)) or (sqrt(2), 3 sqrt(2)). Therefore, the correct range is |m| between sqrt(2) and 3 sqrt(2). So if the problem says the range is sqrt(2) < m < 3 sqrt(2), then it's missing the negative part. Therefore, A is incorrect.But maybe the person who made the problem considered m as a positive number? The question didn't specify, so I think A is incorrect.Moving on to proposition B: If point P(x₁, y₁) lies on the circle C: x² + y² = 4, then the maximum value of (x₁ - 1)² + (y₁ - 2√2)² is 25.So, this is asking for the maximum distance squared from the point P on the circle x² + y² = 4 to the point (1, 2√2). So, the maximum value of the squared distance is 25. Let's verify.The distance squared from P to (1, 2√2) is (x - 1)² + (y - 2√2)². To find the maximum value of this expression given that x² + y² = 4.We can use geometry here. The maximum distance from a point on a circle to another fixed point is equal to the distance from the fixed point to the center plus the radius. Similarly, the minimum is the distance minus the radius.First, find the distance between the center of the circle (0,0) and the point (1, 2√2). That distance is sqrt((1 - 0)^2 + (2√2 - 0)^2) = sqrt(1 + 8) = sqrt(9) = 3.Since the radius of the circle is 2, the maximum distance from a point on the circle to (1, 2√2) is 3 + 2 = 5. Therefore, the squared distance is 5² = 25. So the maximum value is indeed 25. Therefore, proposition B is correct.Now, proposition C: The equation of the line containing the common chord of the circles x² + y² + 4x - 4y = 0 and x² + y² + 2x - 12 = 0 is x - 2y + 6 = 0.To find the equation of the common chord, we can subtract the equations of the two circles. Let's write both equations:First circle: x² + y² + 4x - 4y = 0Second circle: x² + y² + 2x - 12 = 0Subtracting the second equation from the first:( x² + y² + 4x - 4y ) - ( x² + y² + 2x - 12 ) = 0 - 0Simplify:(4x - 4y) - (2x - 12) = 0 => 4x -4y -2x +12 = 0 => 2x -4y +12 = 0Divide by 2: x - 2y + 6 = 0. So the equation of the common chord is x - 2y +6 =0. Therefore, proposition C is correct.Wait, but let's verify if this is indeed the common chord. The line obtained by subtracting the two circle equations is indeed the radical axis, which is the line containing the common chord. So yes, the answer should be correct. Therefore, C is correct.Proposition D: The circles C: x² + y² + 2x = 0 and C₂: x² + y² -4x -8y +4=0 have exactly three common tangent lines.So, three common tangents. For two circles, the number of common tangents depends on their positions:- If the circles are separate and not overlapping, there are 4 common tangents: 2 direct and 2 transverse.- If they are externally tangent, 3 common tangents.- If they intersect at two points, 2 common tangents.- If one is inside the other, no common tangents.- If internally tangent, 1 common tangent.So, first we need to find the positions of the two circles.First, write both equations in standard form.Circle C: x² + y² + 2x =0. Completing the square:x² + 2x + y² =0 => (x +1)^2 -1 + y² =0 => (x +1)^2 + y² =1. So center at (-1,0), radius 1.Circle C₂: x² + y² -4x -8y +4=0. Completing the squares:x² -4x + y² -8y = -4 => (x -2)^2 -4 + (y -4)^2 -16 = -4 => (x -2)^2 + (y -4)^2 -20 = -4 => (x -2)^2 + (y -4)^2 =16. So center at (2,4), radius 4.Now, compute the distance between centers: sqrt( (2 - (-1))² + (4 -0)^2 ) = sqrt(3² +4²) = sqrt(9 +16)=sqrt(25)=5.Sum of radii:1 +4=5. Difference of radii:4 -1=3.Since the distance between centers (5) is equal to the sum of the radii (5), the circles are externally tangent. Therefore, they have exactly three common tangent lines: two external tangents and the common tangent at the point of contact. Therefore, proposition D is correct.Wait, but let me confirm. If two circles are externally tangent, they have exactly three common tangents: the common tangent at the point where they touch, and two separate external tangents. So yes, three common tangents. Therefore, D is correct.Wait, but let me check the calculation again. The distance between centers is 5, sum of radii is 1 +4=5. So yes, they are externally tangent, so exactly three common tangents. Therefore, D is correct.So, summarizing:A: Incorrect (range is both positive and negative, but proposition only gives positive)B: CorrectC: CorrectD: CorrectTherefore, propositions B, C, D are correct. But let me double-check.Wait, in proposition C, the equation x -2y +6=0. Let me check if this is indeed the common chord.Take the two circles:First circle: center (-2, 2), radius 2√2? Wait, no. Wait, original equations:Wait, first circle: x² + y² +4x -4y=0. Let's redo completing the square.First circle: x² +4x + y² -4y=0. Completing the square:(x² +4x +4) + (y² -4y +4) =0 +4 +4=8So (x +2)^2 + (y -2)^2=8. So center (-2,2), radius 2√2.Second circle: x² + y² +2x -12=0. Completing the square:x² +2x + y²=12(x +1)^2 -1 + y²=12(x +1)^2 + y²=13. So center (-1,0), radius √13.Wait, wait! Wait, I made a mistake earlier. Let me redo the centers and radii for proposition C.Original circles:First circle: x² + y² +4x -4y=0.Completing the square:x² +4x + y² -4y =0(x² +4x +4) + (y² -4y +4) =0 +4 +4=8So, (x +2)^2 + (y -2)^2=8. Center at (-2, 2), radius 2√2.Second circle: x² + y² +2x -12=0.x² +2x + y²=12(x +1)^2 -1 + y²=12(x +1)^2 + y²=13. Center at (-1, 0), radius √13.Distance between centers: sqrt[ (-1 - (-2))² + (0 - 2)² ]=sqrt[(1)^2 + (-2)^2]=sqrt(1 +4)=sqrt(5)≈2.236.Now, sum of radii: 2√2 + √13≈2.828 +3.605≈6.433. Difference of radii: |2√2 - √13|≈|2.828 -3.605|≈0.777.Since the distance between centers is sqrt(5)≈2.236, which is less than the sum of radii (≈6.433), so the circles intersect at two points. Therefore, the common chord is the radical axis, which is the line found by subtracting the two equations.So, original equations:First circle: x² + y² +4x -4y=0Second circle: x² + y² +2x -12=0Subtracting second from first:(4x -4y) - (2x -12) =0 - (-12)Wait, wait. Let's do it step by step.First equation minus second equation:(x² + y² +4x -4y) - (x² + y² +2x -12)=0 -0Which gives: 4x -4y -2x +12=0 => 2x -4y +12=0 => Divide by 2: x -2y +6=0.Therefore, the radical axis is x -2y +6=0, which is the line of the common chord. Since the circles intersect at two points, this line is the common chord. So proposition C is correct.Wait, but the user wrote proposition C as "the equation of the line containing the common chord ... is x -2y +6=0." So yes, correct.But let's check if x -2y +6=0 is indeed the radical axis. Let me plug in the centers into the line equation.Center of first circle: (-2, 2). Plug into x -2y +6=0: (-2) -2*(2) +6= -2 -4 +6=0. So the center (-2,2) lies on the line. Wait, but the radical axis is supposed to be perpendicular to the line connecting the centers. Wait, the line connecting the centers is from (-2,2) to (-1,0). The slope is (0 -2)/(-1 +2)= (-2)/1= -2. The radical axis's slope is 1/2 (from equation x -2y +6=0, rearranged to y = (1/2)x +3). Since the line connecting the centers has slope -2, and the radical axis has slope 1/2, which are negative reciprocals, so they are perpendicular. That checks out. Also, the radical axis should pass through the midpoint of the common chord. Wait, but we already found the radical axis by subtracting the equations, so it's correct. Therefore, proposition C is correct.So, in conclusion:A: Incorrect (range is both positive and negative, but proposition A only gives the positive part)B: CorrectC: CorrectD: Correct (since the circles are externally tangent, so three common tangents)Wait, wait, no. Wait in proposition D, the circles are:C: x² + y² +2x =0, which we had earlier converted to (x +1)^2 + y²=1, center (-1,0), radius 1.C₂: x² + y² -4x -8y +4=0 converted to (x -2)^2 + (y -4)^2=16, center (2,4), radius 4.Distance between centers: sqrt( (2 - (-1))² + (4 -0)^2 )=sqrt(9 +16)=sqrt(25)=5.Sum of radii:1 +4=5. Therefore, the circles are externally tangent. Therefore, three common tangent lines. Hence, D is correct.Yes, so propositions B, C, D are correct. A is incorrect.But let me check proposition A again because sometimes there might be a miscalculation.Proposition A: If exactly two points on the circle x² + y²=4 are at distance 1 from line x + y + m=0, then m is between sqrt(2) and 3 sqrt(2).From earlier analysis, the possible m is |m| between sqrt(2) and 3 sqrt(2). So m can be in (-3 sqrt(2), -sqrt(2)) or (sqrt(2), 3 sqrt(2)). Therefore, proposition A only gives the positive interval. But the problem didn't specify that m is positive, so proposition A is only partially correct, hence incorrect.Yes, so final answer: B, C, D are correct.Final Answerboxed{BCD}</think>To determine which propositions are correct, we analyze each one step by step:Proposition A: The range of ( m ) is ( sqrt{2} < m < 3sqrt{2} ).- The circle ( C: x^2 + y^2 = 4 ) has radius 2 and center at the origin.- The distance from the center to the line ( l: x + y + m = 0 ) is ( frac{|m|}{sqrt{2}} ).- For there to be exactly two points on the circle at distance 1 from the line, the parallel lines ( x + y + (m pm sqrt{2}) = 0 ) must intersect the circle such that one line intersects twice and the other does not intersect.- The valid ranges for ( m ) are ( -3sqrt{2} < m < -sqrt{2} ) and ( sqrt{2} < m < 3sqrt{2} ).- Proposition A only considers the positive range, so it is incorrect.Proposition B: The maximum value of ( (x_1 - 1)^2 + (y_1 - 2sqrt{2})^2 ) is 25.- The circle ( C: x^2 + y^2 = 4 ) has radius 2.- The distance from the center (0,0) to the point (1, 2√2) is 3.- The maximum distance from a point on the circle to (1, 2√2) is ( 3 + 2 = 5 ), so the squared distance is 25.- Proposition B is correct.Proposition C: The common chord equation is ( x - 2y + 6 = 0 ).- Subtracting the equations of the circles ( x^2 + y^2 + 4x - 4y = 0 ) and ( x^2 + y^2 + 2x - 12 = 0 ) gives ( x - 2y + 6 = 0 ).- This line is the radical axis and common chord, so proposition C is correct.Proposition D: The circles have exactly three common tangent lines.- Circle ( C_1: (x + 1)^2 + y^2 = 1 ) and circle ( C_2: (x - 2)^2 + (y - 4)^2 = 16 ).- The distance between centers is 5, equal to the sum of the radii (1 + 4), making them externally tangent.- Externally tangent circles have three common tangents, so proposition D is correct.Final answer: boxed{BCD}]

🔑:To analyze each proposition step by step, we proceed as follows:Proposition A:Given the circle C: x^{2}+y^{2}=4 and the line l: x+y+m=0, we are interested in the condition where there are exactly two distinct points on the circle at a distance of 1 from the line. The distance d from the center of the circle (0,0) to the line is given by the formula d=frac{|0+0+m|}{sqrt{1^2+1^2}}=frac{|m|}{sqrt{2}}. For there to be exactly two distinct points at a distance of 1 from the line, we require 1<d<3 (the radius plus the distance from the line to the circle), leading to:[1<frac{|m|}{sqrt{2}}<3]Solving this inequality, we get:[sqrt{2}<|m|<3sqrt{2}]This implies:[sqrt{2}<m<3sqrt{2} quad text{or} quad -3sqrt{2}<m<-sqrt{2}]Therefore, proposition A is incorrect because it does not account for the negative range of m.Proposition B:For any point P(x_{1}, y_{1}) on the circle C: x^{2}+y^{2}=4, we consider the expression (x_1-1)^2+(y_1-2sqrt{2})^2. This represents the square of the distance from P to the point (1, 2sqrt{2}). The distance from the center of the circle (0,0) to (1, 2sqrt{2}) is:[d=sqrt{(1-0)^2+(2sqrt{2}-0)^2}=sqrt{1+8}=sqrt{9}=3]The maximum distance from P to (1, 2sqrt{2}) occurs when P is on the opposite side of the circle from (1, 2sqrt{2}), giving a maximum distance of d+r=3+2=5. Therefore, the maximum value of (x_1-1)^2+(y_1-2sqrt{2})^2 is 5^2=25.Proposition C:Subtracting the equations of the two given circles:[x^{2}+y^{2}+4x-4y=0][x^{2}+y^{2}+2x-12=0]yields:[2x-4y+12=0]which simplifies to:[x-2y+6=0]Thus, the equation of the line containing the common chord is indeed x-2y+6=0, making proposition C correct.Proposition D:For the circle C: x^{2}+y^{2}+2x=0, converting to standard form gives:[(x+1)^{2}+y^{2}=1]indicating a center at (-1,0) and a radius r_{1}=1. For the circle C_{2}: x^{2}+y^{2}-4x-8y+4=0, converting to standard form yields:[(x-2)^{2}+(y-4)^{2}=16]with a center at (2,4) and a radius r_{2}=4. The distance between the centers is:[|C{C}_{2}|=sqrt{(2+1)^{2}+(4-0)^{2}}=sqrt{9+16}=sqrt{25}=5]Since |C{C}_{2}|=r_{1}+r_{2}=5, the circles are externally tangent to each other, implying they have exactly 3 common tangent lines. Thus, proposition D is correct.Hence, the correct propositions are: boxed{BCD}.

🧠:[<think>Okay, so I need to evaluate the integral from -1 to 1 of (sqrt(1 - x²) + x) dx. Hmm, let me think about how to approach this. First, I remember that integrals can be split up into separate terms. So maybe I can split this into two integrals: one involving sqrt(1 - x²) and the other involving x. That is, ∫ from -1 to 1 of sqrt(1 - x²) dx plus ∫ from -1 to 1 of x dx. Let me check if that's allowed. Yeah, the integral of a sum is the sum of the integrals, so that should work.Starting with the first integral: ∫ sqrt(1 - x²) dx from -1 to 1. This looks familiar. Isn't this the equation for a semicircle? The function y = sqrt(1 - x²) is the upper half of the circle with radius 1 centered at the origin. So, integrating this from -1 to 1 would give the area of the semicircle. The area of a full circle is πr², so a semicircle would be (1/2)π(1)² = π/2. Therefore, the integral of sqrt(1 - x²) from -1 to 1 is π/2. That seems right. Let me confirm. If we do a substitution, like x = sinθ, then dx = cosθ dθ. Then when x = -1, θ = -π/2, and x = 1, θ = π/2. The integral becomes ∫ from -π/2 to π/2 of sqrt(1 - sin²θ) * cosθ dθ. That simplifies to ∫ cosθ * cosθ dθ = ∫ cos²θ dθ. Then using the identity cos²θ = (1 + cos2θ)/2, the integral becomes (1/2)∫ (1 + cos2θ) dθ. Integrating from -π/2 to π/2 gives (1/2)[θ + (1/2)sin2θ] evaluated at those limits. Plugging in π/2: (1/2)[π/2 + (1/2)sin(π)] = (1/2)(π/2 + 0) = π/4. Plugging in -π/2: (1/2)[-π/2 + (1/2)sin(-π)] = (1/2)(-π/2 + 0) = -π/4. Subtracting: π/4 - (-π/4) = π/2. Yep, that checks out. So the first integral is π/2.Now the second integral: ∫ from -1 to 1 of x dx. Hmm, that's straightforward. The integral of x is (1/2)x². Evaluated from -1 to 1. So plugging in 1: (1/2)(1)² = 1/2. Plugging in -1: (1/2)(-1)² = 1/2. Subtracting: 1/2 - 1/2 = 0. Wait, so the second integral is zero. Is that right? Let's think. The function x is an odd function because f(-x) = -f(x). The interval from -1 to 1 is symmetric around zero. For odd functions, the integral over symmetric limits is zero. Yeah, that makes sense. So the area on the left cancels the area on the right. So indeed, the second integral is zero.So putting it all together: the original integral is π/2 + 0 = π/2. Therefore, the answer should be π/2. Let me just verify once more. If I graph sqrt(1 - x²), it's a semicircle above the x-axis from -1 to 1, area π/2. The function x from -1 to 1 is a straight line passing through the origin, and since it's odd, the integral is zero. So adding them gives π/2. Seems solid.I don't see any mistakes in my reasoning. The key points are recognizing the semicircle integral and the odd function property. Both steps check out. So I think the value of the integral is π/2.Final AnswerThe value of the integral is boxed{dfrac{pi}{2}}.</think>To evaluate the integral (int_{-1}^{1} left( sqrt{1 - x^2} + x right) , dx), we can split it into two separate integrals:[int_{-1}^{1} sqrt{1 - x^2} , dx + int_{-1}^{1} x , dx]First, consider the integral (int_{-1}^{1} sqrt{1 - x^2} , dx). This represents the area of a semicircle with radius 1. The area of a full circle is (pi r^2), so the area of a semicircle is (frac{1}{2} pi (1)^2 = frac{pi}{2}). Therefore,[int_{-1}^{1} sqrt{1 - x^2} , dx = frac{pi}{2}]Next, consider the integral (int_{-1}^{1} x , dx). The function (x) is an odd function, and the integral of an odd function over a symmetric interval ([-a, a]) is zero. Thus,[int_{-1}^{1} x , dx = 0]Adding the results of the two integrals, we get:[frac{pi}{2} + 0 = frac{pi}{2}]Therefore, the value of the integral is (boxed{dfrac{pi}{2}}).,<think>Okay, so I need to evaluate this integral from -1 to 1 of the function sqrt(1 - x²) + x dx. Hmm, let me start by breaking this down. The integrand is the sum of two functions: the square root term and just x. Maybe I can split the integral into two separate integrals? That usually works because integration is linear, right? So like, the integral of (A + B) is the integral of A plus the integral of B. So let's try that.First, let me write it out: the integral from -1 to 1 of sqrt(1 - x²) dx plus the integral from -1 to 1 of x dx. That should simplify things. Now, let's tackle each part one by one.Starting with the first integral: sqrt(1 - x²) dx from -1 to 1. Hmm, that looks familiar. Isn't that the equation for the upper half of a circle with radius 1? Because if you square both sides, you get y = sqrt(1 - x²), which becomes y² = 1 - x², so x² + y² = 1. Yeah, that's a circle centered at the origin with radius 1. So the integral of sqrt(1 - x²) from -1 to 1 is actually the area of a semicircle. Since the full circle area is πr², and r is 1 here, the area would be π(1)² = π. But since it's a semicircle, the area is (1/2)π. So that first integral should be (1/2)π. Let me confirm that. If I integrate sqrt(1 - x²) from -a to a, it's (π a²)/2. Here a is 1, so yeah, (π/2). Cool, so that part is straightforward.Now the second integral: integral from -1 to 1 of x dx. Hmm, integrating x dx is simple. The antiderivative of x is (1/2)x². So evaluating from -1 to 1 would be (1/2)(1)² - (1/2)(-1)². Let's compute that. (1/2)(1) - (1/2)(1) = 1/2 - 1/2 = 0. Wait, so the integral of x from -1 to 1 is zero? That makes sense because x is an odd function, and the limits are symmetric around zero. So yeah, integrating an odd function over symmetric limits gives zero. So that checks out.So putting it all together, the original integral is (π/2) + 0, which is just π/2. Is that right? Let me double-check.First integral: sqrt(1 - x²) from -1 to 1. If I use a substitution, like x = sinθ, then dx = cosθ dθ. Then sqrt(1 - x²) becomes sqrt(1 - sin²θ) = cosθ. The limits when x = -1, θ = -π/2, and when x = 1, θ = π/2. So the integral becomes integral from -π/2 to π/2 of cosθ * cosθ dθ = integral of cos²θ dθ from -π/2 to π/2. Using the identity cos²θ = (1 + cos2θ)/2. So integrating that gives (1/2)θ + (1/4)sin2θ evaluated from -π/2 to π/2. Plugging in π/2: (1/2)(π/2) + (1/4)sin(π) = π/4 + 0. Plugging in -π/2: (1/2)(-π/2) + (1/4)sin(-π) = -π/4 + 0. Subtracting the lower limit from the upper: (π/4) - (-π/4) = π/2. So that confirms the first integral is indeed π/2. So that's solid.The second integral, as I thought, is zero. So the total integral is π/2. So I think that's the answer. Let me just visualize this. The sqrt(1 - x²) part is the top half of the circle, so integrating from -1 to 1 gives the area of the semicircle, which is π/2, and the x part cancels out. So yeah, that seems right.Wait, but just to be thorough, maybe I should check if there's any mistake in splitting the integral. The integral of a sum is the sum of the integrals, right? Yes, that's a basic property of integrals. So integrating term by term is allowed here. So no problem there.Another way to check: if I compute the integral numerically. Let's see, the first part is π/2, which is approximately 1.5708. The second part is zero, so the total integral is approximately 1.5708. If I tried to compute this numerically, say using some approximation method, like the trapezoidal rule or something, but since I know the exact answer, maybe I can just verify with a simple midpoint. Wait, maybe at x=0, the function sqrt(1 - 0) + 0 = 1. At x=1, it's sqrt(0) +1 =1. At x=-1, sqrt(0) + (-1) = -1. Wait, but the integral is from -1 to 1, so the function is sqrt(1 -x²) + x. Hmm, when x is 0, it's 1. At x=1, it's 0 +1 =1. At x=-1, it's 0 + (-1) = -1. But the integral is the area under the curve from -1 to 1. The sqrt(1 - x²) is symmetric about the y-axis, and the x term is an odd function. So when you add them, the graph is the semicircle shifted up by x. But since the x term is odd, its integral over symmetric limits cancels out. So the area is just the semicircle. So yeah, that makes sense. So the answer should indeed be π/2. So I think that's solid. The answer is π/2.Final AnswerThe value of the integral is boxed{dfrac{pi}{2}}.</think>To evaluate the integral (int_{-1}^{1} left( sqrt{1 - x^2} + x right) , dx), we can split it into two separate integrals:[int_{-1}^{1} sqrt{1 - x^2} , dx + int_{-1}^{1} x , dx]First, consider the integral (int_{-1}^{1} sqrt{1 - x^2} , dx). This integral represents the area of a semicircle with radius 1. The area of a full circle is (pi r^2), so the area of a semicircle is (frac{1}{2} pi r^2). Here, (r = 1), so the area is:[frac{1}{2} pi (1)^2 = frac{pi}{2}]Next, consider the integral (int_{-1}^{1} x , dx). The function (x) is an odd function, and the integral of an odd function over symmetric limits ([-a, a]) is zero. Therefore:[int_{-1}^{1} x , dx = 0]Adding the results of the two integrals, we get:[frac{pi}{2} + 0 = frac{pi}{2}]Thus, the value of the integral is (boxed{dfrac{pi}{2}}).]

🔑:To solve int _{ -1 }^{ 1 } ( sqrt {1-x^{2}}+x)dx, we separate it into two parts: int _{ -1 }^{ 1 } sqrt {1-x^{2}}dx+ int _{ -1 }^{ 1 } xdx.The geometric meaning of int _{ -1 }^{ 1 } sqrt {1-x^{2}}dx is the area enclosed by the arc of a circle with center at (0,0) and radius 1 in the first and second quadrants and the coordinate axes.Thus, int _{ -1 }^{ 1 } sqrt {1-x^{2}}dx = frac {1}{2}pi times 1 = frac {pi }{2}.Additionally, int _{ -1 }^{ 1 }xdx = frac {1}{2}x^{2} |_{ -1 }^{ 1 } = 0.Therefore, the original expression equals frac {pi }{2} + 0 = frac {pi }{2}.Hence, the answer is: boxed{frac {pi }{2}}.

🧠:[<think>Okay, so I have this problem here where I need to find the maximum number of natural numbers that can be chosen from the interval [1353, 2134] such that the sum of any two chosen numbers isn't divisible by 11. Hmm, interesting. Let me try to break this down step by step.First, the problem is about selecting numbers from a range, with a condition on their sums modulo 11. Since the sum of any two numbers shouldn't be divisible by 11, that means if I take any two numbers from my selected set, their sum modulo 11 shouldn't equal 0. So, in other words, for any two numbers x and y in the set, (x + y) mod 11 ≠ 0.To approach this, maybe I can look at each number modulo 11 and categorize them based on their remainders when divided by 11. Let me recall that if two numbers are congruent to r and s modulo 11, then their sum is congruent to (r + s) modulo 11. So, for their sum not to be 0 modulo 11, we need to ensure that r + s ≠ 11 mod 11, which simplifies to r + s ≠ 11. But since we're working modulo 11, the possible remainders are 0 through 10. So, pairs of remainders that add up to 11 would be (1,10), (2,9), (3,8), (4,7), (5,6). Also, if a number is congruent to 0 modulo 11, adding another number congruent to 0 modulo 11 would give a sum congruent to 0 modulo 11, which is not allowed. So, pairs of 0s are also problematic.Therefore, the strategy here might be to group the numbers in the interval [1353, 2134] by their remainder modulo 11. Then, for each pair of remainders that add up to 11, we can only choose numbers from one of the remainders in the pair. Additionally, we can have at most one number with remainder 0 modulo 11 because adding two 0s would give 0 modulo 11. Wait, actually, even one number with remainder 0 can't be paired with any other number because if there's another number with remainder r, then 0 + r = r, which is only 0 modulo 11 if r is 0. So, if we have a number with remainder 0, we can only have at most one such number, and no other numbers with remainder 0. But if we include a 0, then other numbers can have any remainder except 0? Wait, no. If we have a number with remainder 0, and another number with remainder r, then their sum is r modulo 11, which isn't 0 unless r is 0. So actually, if we include a 0, we can include numbers from other remainders as long as those remainders don't conflict with each other. Wait, maybe I need to think more carefully.Let me rephrase. The problem states that the sum of any two numbers shouldn't be divisible by 11. So, if I include a number x ≡ 0 mod 11, then any other number y ≡ 0 mod 11 would make x + y ≡ 0 + 0 ≡ 0 mod 11, which is not allowed. Therefore, we can include at most one number congruent to 0 mod 11.For the other remainders, if I have numbers with remainder r, I cannot have any numbers with remainder 11 - r mod 11. Because r + (11 - r) = 11 ≡ 0 mod 11. Therefore, for each pair of remainders (r, 11 - r), we can choose numbers from one remainder or the other, but not both. The goal is to maximize the total count.So, the plan is:1. For each number in [1353, 2134], compute its remainder modulo 11.2. Count how many numbers there are for each remainder.3. For each complementary pair (r, 11 - r), choose the one with the higher count.4. For the remainder 0, we can take at most 1 number.5. Sum all these chosen counts.But we also need to check if the interval [1353, 2134] is a complete set of residues modulo 11. If it's not, maybe some remainders have more numbers than others. Let me verify the total numbers first.The interval is from 1353 to 2134 inclusive. The number of numbers is 2134 - 1353 + 1. Let's compute that: 2134 - 1353 = 781, so total numbers is 782. So, 782 numbers. Divided into 11 remainders. Let's see, 782 divided by 11 is 71 with a remainder of 1. So, 11*71=781, so there will be 71 numbers for each remainder from 0 to 10, except one remainder which will have 71 +1=72 numbers. So, one remainder will have an extra number.But first, let's check if the interval [1353, 2134] is aligned such that each remainder occurs exactly 71 or 72 times. Let's compute 1353 mod 11 and 2134 mod 11 to find out how the numbers are distributed.First, compute 1353 mod 11. Let's divide 1353 by 11:11*122 = 1342, so 1353 - 1342 = 11. So, 1353 is 11*123. Therefore, 1353 mod 11 is 0.Wait, 11*123 = 1353. So, 1353 is divisible by 11. Then, 2134 mod 11: Let's compute 2134 divided by 11.11*193 = 2123, so 2134 - 2123 = 11. Therefore, 2134 is 11*194. So, 2134 mod 11 is 0 as well. Wait, so both endpoints are divisible by 11. Therefore, the interval [1353, 2134] includes numbers from 1353 to 2134, both divisible by 11. So, how many numbers are in this interval? As before, 2134 - 1353 +1 = 782 numbers. Since both endpoints are divisible by 11, the interval is 1353, 1354, ..., 2134. Since 1353 is divisible by 11, the next number 1354 is 1 mod 11, ..., up to 2134 which is 0 mod 11. Therefore, each remainder from 0 to 10 occurs exactly floor(782 / 11) or ceil(782 / 11) times.But 782 divided by 11 is 71.09, so 71*11=781, so 782=71*11 +1. Therefore, there are 71 full cycles of 11 numbers, and one extra number. Since the interval starts at 0 mod 11 and ends at 0 mod 11, the extra number would be at the end, which is 0 mod 11. Wait, but starting at 0 mod 11, each block of 11 numbers will cover remainders 0 through 10. So, 71 full blocks (71*11=781 numbers) will cover remainders 0-10 exactly 71 times each. Then, the last number, which is 2134, is another 0 mod 11, so remainder 0 will have 71 +1=72 numbers, and the other remainders have 71 each.Therefore, in the interval [1353, 2134], remainder 0 occurs 72 times, and remainders 1 through 10 occur 71 times each.Wait, but let's verify that. Let me count the number of numbers congruent to 0 mod 11 in the interval. Since 1353 is the first number, which is 0 mod 11, and the last number is 2134, which is also 0 mod 11. The numbers divisible by 11 in this interval are 1353, 1364, ..., 2134. The number of terms in this arithmetic sequence is ((2134 - 1353)/11) +1. Let's compute (2134 - 1353)/11: 781/11 = 71. So, 71 +1 =72 numbers. So, remainder 0 has 72 numbers. Then, the other remainders from 1 to 10 each have (782 -72)/10 = (710)/10=71. So, yes, each remainder 1-10 has 71 numbers, remainder 0 has 72.Therefore, the counts per remainder are:- Remainder 0: 72- Remainders 1-10: 71 each.Now, going back to the problem. We need to select numbers such that no two numbers sum to a multiple of 11. As discussed earlier, this means we can't have two numbers where their remainders add up to 11. So, for each pair (r, 11 - r), we can choose numbers from either r or 11 - r, but not both. Also, we can have at most one number from remainder 0.So, let's list the complementary pairs:- (1,10)- (2,9)- (3,8)- (4,7)- (5,6)And remainder 0 is alone.For each of these pairs, we choose the remainder with the higher count. Since all the non-zero remainders have the same count (71), except remainder 0. Wait, no, for each pair (r, 11 - r), both have 71 numbers. So, for each pair, we can take all 71 numbers from one remainder. Since both are equal, it doesn't matter which one we choose. So, for each of the five pairs, we can take 71 numbers. Then, for remainder 0, we can take at most 1 number.Therefore, total maximum numbers would be 5*71 +1 = 355 +1 = 356.Wait, but hold on. Let me check.Each of the five pairs (1,10), (2,9), (3,8), (4,7), (5,6) contributes 71 numbers. Since for each pair, the counts are equal, we just take one side. So 5 pairs, each contributing 71, so 5*71=355. Then remainder 0: we can take at most 1. So total 355 +1=356.But wait, but in the counts, remainder 0 has 72 numbers. But we can only take at most 1. So we are leaving 71 numbers on the table. Wait, but if we can take at most 1 from remainder 0, then even though there are 72 numbers with remainder 0, we can only take 1. But then, the rest of the numbers (355) are from the other pairs. So total is 355 +1=356.But let me confirm if this is correct. Let me think again.Each complementary pair (r, s) where r + s =11, we can take all numbers from r or all from s. Since both have 71 numbers, we take 71 from each pair, and there are 5 pairs. So 5*71=355. Then, remainder 0: 72 numbers, but only 1 can be taken. So total is 355 +1=356.But wait, is there a way to take more numbers? Let me think.Suppose instead of taking 1 from remainder 0, can we take some other combination?But if we take any number from remainder 0, we can only take 1. Because if we take two numbers from remainder 0, their sum is 0+0=0 mod11, which is not allowed. So even though there are 72 numbers, maximum we can take is 1.Alternatively, if we don't take any numbers from remainder 0, can we take more numbers from other pairs? But no, because the other pairs are already maxed out. For each pair, we take all 71 numbers from one side. So even if we don't take the 1 from remainder 0, the total would be 355. Which is less than 356. So 356 is better.Therefore, the maximum number of numbers is 355 +1=356.Wait, but let me check if there's any mistake here.First, total numbers in the interval: 782. If we take 356 numbers, that's less than half. But given the restriction, maybe that's correct.Alternatively, perhaps there's a way to optimize further. Let me see.Wait, if I take 1 number from remainder 0, then all other numbers must not conflict with it. But a number from remainder 0 added to any other number x will be x mod11. Since x is from remainder r (1-10), the sum will be r mod11, which is not 0. So, if we take 1 number from remainder 0, we can still take all the numbers from the other remainders as long as they don't conflict with each other. Wait, but the other numbers are already chosen such that no two are from complementary pairs. So, actually, taking a 0 doesn't interfere with the others. Because 0 + r = r, which is allowed as long as you don't have two rs that sum to 11. But the selection from the pairs is already such that no two numbers in a pair are taken. So, adding a 0 is safe.Therefore, the maximum number is indeed 5*71 +1=356.But let's verify the counts again.Total numbers in each remainder:- 0:72- 1-10:71 each.For each pair (r, 11 - r), we take 71 from one side. Since both are 71, we take 71 from each pair, so 5 pairs give 5*71=355.Plus 1 from remainder 0: total 356.So, 356 is the maximum.But let me double-check if there's a way to include more numbers. Suppose instead of taking 1 from remainder 0, maybe we can take some from remainder 0 and adjust the other counts. For example, if we take 1 from remainder 0, can we take more than 71 from any pair? No, because each pair has 71. If you take more than 71, you have to take from both sides of a pair, which is not allowed. So, 71 is the maximum per pair. So, 5*71 +1=356.Alternatively, if we don't take any from remainder 0, can we take more from the pairs? But since each pair is already being maxed out, you can't take more. So, no. So 355 vs 356. So 356 is better.Therefore, the answer should be 356.But let me check once again with a small example to see if my logic holds.Suppose we have a smaller case: numbers from 1 to 22, N=11. So A=1, B=22, N=11. The remainders 1-10 and 11-22 would be 1-10,11,12-22. Wait, 11 mod11=0, 12=1 mod11, etc. So in 1-22, remainders 1-10 have two numbers each, and remainder 0 has two numbers (11 and 22). So counts: remainders 0:2, 1-10:2 each.To maximize the set where no two sum to 11. Then, for each pair (1,10), (2,9), etc., take the higher count. But since all are equal (2 each), take 2 from each pair. Wait, no. Wait, no, because if you take 2 from remainder 1, then you can't take any from remainder 10. But in reality, if we take 2 from remainder 1 and 2 from remainder 10, then their sums would be 1+10=11, which is not allowed. So actually, per pair, you can take all from one side. So for each pair, take 2. So 5 pairs, 2 each, total 10, plus 1 from remainder 0. Total 11. But in 1-22, the maximum such set would be 11 numbers. Let's see: take remainders 1,2,3,4,5,6,7,8,9,10, and 0. Wait, no. If we take one from each non-complementary remainder? Wait, no. Let me think.Wait, in this smaller example, if you take all numbers from remainders 1-5 and 0. Then, the sum of any two numbers from 1-5 will be 2-10, which is not 0 mod11. Adding 0, the sum with any number is the number itself, so not 0. So, maximum size would be 5*2 +1=11. But actually, let's check. Numbers 1-10 (each remainder 1-10 appears twice), and 11,22 (remainder 0). So, if we take one remainder from each pair:Take remainders 1,2,3,4,5. Each has 2 numbers. That's 10 numbers. Plus 1 from remainder 0. Total 11. Indeed, this works because no two numbers sum to 11. Alternatively, take remainders 6-10 and 0. Similarly, 11 numbers. So 11 is the maximum.Alternatively, if you tried to take more, you would have conflicts. So, the logic holds here. Therefore, the same logic applies to the original problem.Therefore, in the original problem, with 5 pairs each contributing 71 numbers, and remainder 0 contributing 1, the total is 5*71 +1=356.But just to confirm, let's calculate 5*71=355, 355+1=356. Yes.But wait, another thought: in the original problem, remainder 0 has 72 numbers. But we can only take 1. So, even though there are 72 numbers, we can take at most 1. If there was a way to take more, but the problem says the sum of any two numbers should not be divisible by 11. So, even if you take one number from remainder 0, you can't take another because 0+0=0. So, maximum 1. So, that's correct.Therefore, the maximum number of numbers is 356.I think that's the answer.Final Answerboxed{356}</think>To solve the problem of finding the maximum number of natural numbers that can be chosen from the interval ([1353, 2134]) such that the sum of any two chosen numbers is not divisible by 11, we need to consider the remainders of these numbers modulo 11. 1. Identify the interval and remainders: - The interval ([1353, 2134]) contains 782 numbers. - Both endpoints (1353 and 2134) are divisible by 11, so the interval starts and ends with remainder 0 modulo 11. - The numbers in the interval are distributed such that each remainder from 1 to 10 occurs 71 times, and remainder 0 occurs 72 times.2. Group numbers by remainders: - Remainder 0: 72 numbers. - Remainders 1-10: 71 numbers each.3. Complementary pairs: - The pairs of remainders that add up to 11 (mod 11) are ((1, 10)), ((2, 9)), ((3, 8)), ((4, 7)), and ((5, 6)). - For each pair, we can choose numbers from one remainder in the pair but not both. Since each remainder in a pair has 71 numbers, we choose 71 numbers from each pair.4. Maximize the count: - For each of the 5 pairs, we take 71 numbers, contributing (5 times 71 = 355) numbers. - For remainder 0, we can take at most 1 number to avoid sums divisible by 11.5. Total maximum numbers: - Summing these contributions gives (355 + 1 = 356).Thus, the maximum number of numbers that can be chosen such that the sum of any two is not divisible by 11 is (boxed{356}).,<think>Okay, so I need to figure out the maximum number of numbers that can be chosen from the interval [1353, 2134] such that the sum of any two chosen numbers isn't divisible by 11. Hmm, let's start by understanding the problem better.First, the key constraint here is that the sum of any two numbers shouldn't be divisible by 11. So, if I pick two numbers, say x and y, then (x + y) mod 11 shouldn't be 0. Which also means that x mod 11 and y mod 11 shouldn't add up to 11 or 22, but since we're working modulo 11, 22 is equivalent to 0. So, essentially, we need to avoid pairs where x mod 11 + y mod 11 ≡ 0 mod 11. Therefore, the residues modulo 11 of the numbers we choose should be such that no two residues add up to 11 or 0 modulo 11.So, maybe the first step is to compute the residues modulo 11 for all numbers in the given interval. Then, group the numbers by their residues. Once we have that, the problem reduces to selecting residues in such a way that no two residues are complementary (i.e., their sum is 11). Also, we have to be careful with the residue 0, because adding two numbers with residue 0 would give 0 modulo 11. So, for residue 0, we can have at most one number in our selection.Similarly, for residues 1 to 10, we need to make sure that if we include numbers with residue r, we don't include any numbers with residue (11 - r). For example, if we include numbers with residue 1, we can't include numbers with residue 10, because 1 + 10 = 11 ≡ 0 mod 11. Similarly for 2 and 9, 3 and 8, 4 and 7, 5 and 6.Therefore, for each pair of complementary residues (1 and 10, 2 and 9, etc.), we have to choose the one with more numbers in the interval [1353, 2134]. Then, for residue 0, we can take at most one number. Then, sum up the counts of the chosen residues. That should give the maximum possible numbers we can select without violating the sum condition.So, the steps I need to do:1. Find how many numbers in [1353, 2134] have each residue modulo 11. That is, for each r in 0 to 10, compute the count of numbers x in [1353, 2134] where x mod 11 = r.2. For each pair of complementary residues (r and 11 - r), choose the one with the higher count. For example, if residue 1 has 100 numbers and residue 10 has 90, pick residue 1.3. For residue 0, since adding two such numbers would give a sum divisible by 11, we can take at most 1 number.4. Sum all the selected counts from the pairs and the residue 0 (either 0 or 1).But before that, let me verify if that's the correct approach.Wait, another way to think is that in order to maximize the number of numbers, we need to avoid having two numbers whose residues are complementary. So, for each complementary pair, we can choose the larger group, and for residue 0, take at most one. So, that's correct. So, the strategy is indeed to take the maximum of each complementary pair and add 1 if we take a residue 0.Therefore, the first step is to compute the counts for each residue mod 11 in the interval [1353, 2134].How do we compute the count of numbers congruent to r mod 11 in [A, B]?The formula is as follows:The number of integers x in [A, B] such that x ≡ r mod 11 is equal to floor((B - r)/11) - floor((A - 1 - r)/11) + 1 if A ≡ r mod 11, otherwise floor((B - r)/11) - floor((A - 1 - r)/11).Wait, maybe a better way is:First, compute the first number >= A that is ≡ r mod 11.Then compute the last number <= B that is ≡ r mod 11.If the first number exists (i.e., <= B), then the count is ((last - first)/11) + 1.Otherwise, the count is 0.Alternatively, the formula can be written as:count = floor((B - r)/11) - floor((A - r - 1)/11)But maybe let's test this with an example.Suppose A = 1, B = 22, r = 0.Numbers divisible by 11 in [1,22]: 11,22. So count = 2.Using the formula: floor((22 - 0)/11) - floor((1 - 0 -1)/11) = floor(22/11) - floor(-1/11) = 2 - (-1) = 3. Wait, that gives 3, which is wrong. So, maybe the formula isn't correct.Alternatively, another formula: count = floor((B - r)/11) - floor((A - 1 - r)/11) + 1.Wait, let's see for the above example:floor((22 - 0)/11) = 2, floor((1 -1 -0)/11) = floor(0/11) = 0. Then 2 - 0 +1 = 3. Still wrong.Wait, perhaps another approach.The number of terms in arithmetic progression from a to l with common difference d is ((l - a)/d) + 1.So, to compute the number of numbers congruent to r mod 11 in [A, B], we need to find the first term >= A that is congruent to r mod 11, and the last term <= B that is congruent to r mod 11.If such terms exist, then the count is ((last - first)/11) + 1.If there's no such term, the count is 0.So, let's formalize that.First term: If A mod 11 <= r, then first = A + (r - A mod 11). Otherwise, first = A + (11 - A mod 11) + r.Wait, maybe not. Let's think. For example, A = 1353. Let's compute 1353 mod 11.Wait, perhaps first I need to compute A mod 11 and B mod 11.But maybe let's take a concrete example.Suppose A=1353, B=2134.First, compute 1353 mod 11.To compute 1353 mod 11: 11*123 = 1353, so 1353 is divisible by 11. Therefore, 1353 mod 11 = 0.Similarly, 2134 mod 11: Let's compute 2134 ÷ 11. 11*193 = 2123, 2134 - 2123 = 11, so 2134 = 11*194, so 2134 mod 11 = 0. Wait, is that correct?Wait, 11*190 = 2090, 2134 - 2090 = 44, which is 11*4, so 2134 = 11*190 + 44 = 11*194. So yes, 2134 mod 11 is 0. So both A and B are divisible by 11. So the interval [1353, 2134] includes numbers from 1353 to 2134 inclusive.Since both ends are divisible by 11, the numbers congruent to 0 mod 11 in this interval are 1353, 1364, ..., 2134. The count would be ((2134 - 1353)/11) + 1 = (781)/11 +1 = 71 +1 = 72 numbers. So, residue 0 has 72 numbers.Similarly, for residue 1: The first number >=1353 that is ≡1 mod 11. Since 1353 ≡0 mod 11, the next number is 1353 +1 =1354 ≡1 mod 11. Then the last number <=2134 ≡0 mod 11, so the last number ≡1 mod 11 would be 2134 -10 = 2124. Let's check 2124 mod11: 2124 ÷11=193*11=2123, 2124=2123+1, so 2124 mod11=1. Therefore, the numbers are 1354,1365,...,2124. The count is ((2124 -1354)/11)+1 = (770)/11 +1=70 +1=71 numbers.Similarly, for residue 2: First number is 1353 +2 =1355. Last number is 2134 -9=2125. 2125 mod11: 2125=11*193 +2=2123 +2=2125, so 2125 mod11=2. Then count is ((2125 -1355)/11)+1=(770)/11 +1=70 +1=71.Wait, same as residue 1?Wait, but 2125 -1355 =770. 770 ÷11=70. So, 70 intervals, so 71 numbers. So yes, same for residues 1 through 10?Wait, but let's check residue 10.First number ≡10 mod11: Since 1353 ≡0, so previous number would be 1353 -1=1352≡10 mod11, but that's less than A=1353. So the first number in [1353,2134] ≡10 mod11 is 1353 +10=1363.Wait, 1353 +10=1363. 1363 mod11: 1363 ÷11=123*11=1353, so 1363=1353+10, so 1363 mod11=10. Then the last number ≡10 mod11 would be 2134 -1=2133, which is 2134-1=2133≡10 mod11. Let's check 2133 mod11: 2133=11*193 +10=2123 +10=2133. So, 2133 mod11=10. So numbers are 1363,1374,...,2133. The count is ((2133 -1363)/11)+1= (770)/11 +1=70 +1=71. So same as residues 1 through 10. Therefore, each residue from 1 to 10 has 71 numbers. And residue 0 has 72 numbers. So total numbers in [1353,2134] is 72 +10*71=72+710=782. Let's check: 2134 -1353 +1=781 +1=782. So correct. So each residue from 1 to 10 has 71 numbers, residue 0 has 72.So, now, back to the problem. We need to choose numbers such that no two residues add up to 11 mod11. So, complementary residues are (1,10), (2,9), (3,8), (4,7), (5,6). For each pair, we can take the maximum of the two counts. Since each residue from 1 to 10 has 71 numbers, each pair (r, 11 -r) has counts 71 and 71. So maximum is 71. Therefore, for each pair, we can take 71. There are 5 such pairs: (1,10), (2,9), (3,8), (4,7), (5,6). So 5 pairs, each contributing 71. Then residue 0 has 72 numbers, but we can take at most 1. So total maximum numbers would be 5*71 +1=355 +1=356.Wait, but wait. Each pair is (1,10), (2,9), etc. Each of those has two residues, each with 71 numbers. So, if we take 71 from one residue in the pair, but if both residues have the same count, which one to pick? Since they are symmetric, it doesn't matter. For maximum, we pick one of them, which gives 71.So, 5 pairs, each gives 71. So 5*71=355. Then residue 0: can take 1. So total 356.But hold on. Wait, in this case, since all residues 1-10 have the same count (71), we can choose either residue from each complementary pair, but since they are equal, we can take either one, but since they are the same, the total would be 5*71 +1=356.But let me verify this reasoning.Suppose we have pairs (1,10), (2,9), ..., (5,6). Each has 71 numbers. So, if we take all residues 1,2,3,4,5, each of which has 71 numbers, and avoid their complements. Then residue 0: take 1. So total would be 5*71 +1=356. Alternatively, if we take residues 10,9,8,7,6, each has 71, plus 1 from residue 0: same total.Alternatively, if for some pairs we take one residue and others take the complement, but since all counts are equal, it doesn't matter. The total is 5*71 +1=356.Therefore, the maximum number is 356.But let's check if that's correct.Wait, but let's make sure that there are indeed 71 numbers in each residue 1-10.Earlier, we saw that residues 1-10 each have 71 numbers, and residue 0 has 72. So, the count seems accurate.So, if we take 5 residues (each from a complementary pair) each with 71 numbers, and 1 number from residue 0, total 5*71 +1=356. But let me check if this is indeed the maximum.Alternatively, could we take more numbers if we exclude residue 0 entirely and take more from other residues? For example, if we don't take residue 0, then for each pair, we can take 71 from each. But if we take 71 from both residues in a pair, that's not allowed because their sum would be 11. Wait, no. For each pair, we can take only one residue. So, even if we exclude residue 0, the maximum would still be 5*71. Which is 355, which is less than 355 +1=356. So including residue 0 is better.Therefore, the maximum is 356.Wait, but let me make another point. If we include residue 0, we can only include 1 number. If we don't include residue 0, we can include all 5*71=355. But adding 1 gives 356, which is better.Alternatively, if there is some way to include more numbers from other residues by not including residue 0? But since residue 0 can only add 1, and other residues are fixed, I don't think so.Therefore, the answer should be 356.Wait, but let me check again the counts.Total numbers from residue 0: 72. So if we take 1, then 71 are excluded. Whereas if we don't take any from residue 0, we have 355. So 355 vs 355 +1=356. So definitely 356 is better.Therefore, maximum is 356.But wait, but in the problem statement, A=1353 and B=2134. Let's verify the counts once again.Given that both A and B are divisible by 11. So A=1353=11*123, B=2134=11*194.Therefore, the numbers from 1353 to 2134 inclusive are 2134 -1353 +1=782 numbers.Numbers congruent to 0 mod11: every 11th number. From 1353 to 2134, step 11. The first number is 1353, the last is 2134. The count is ((2134 -1353)/11) +1= (781)/11 +1=71 +1=72. So 72 numbers.For residues 1-10, let's take residue 1. The first number is 1353 +1=1354. Then next is 1354 +11=1365, etc. The last number would be 2134 -10=2124. Let's compute how many terms: (2124 -1354)/11 +1= (770)/11 +1=70 +1=71. Similarly for residue 2: first 1355, last 2125. Same calculation: 770/11 +1=71. Similarly for residues up to 10. So each residue 1-10 has 71 numbers. Therefore, total numbers 72 +10*71=72+710=782, which matches. So counts are accurate.Therefore, the maximum numbers are 5*71 +1=356. Therefore, the answer is 356. So boxed{356}.But wait, let me think again. Wait, is there any other constraint? For example, when you choose numbers from residue r and residue s, where r + s ≠0 mod11. But since we have already partitioned the residues into complementary pairs and selected only one from each pair, then any two numbers chosen will have residues that are not complementary, so their sum won't be 0 mod11. So, the selection is valid.Therefore, the answer is 356.Final Answerboxed{356}</think>To solve the problem of finding the maximum number of numbers that can be chosen from the interval ([1353, 2134]) such that the sum of any two chosen numbers is not divisible by 11, we need to consider the residues of these numbers modulo 11. 1. Calculate residues modulo 11: - The interval ([1353, 2134]) includes numbers from 1353 to 2134. - Both 1353 and 2134 are divisible by 11, so they are congruent to 0 modulo 11. - For residues 1 to 10, each residue class contains 71 numbers. This is calculated by finding the first and last numbers in the interval for each residue and using the arithmetic progression formula. - The residue 0 modulo 11 contains 72 numbers.2. Select residues to avoid complementary pairs: - For each pair of complementary residues ((r, 11-r)), we can choose the one with the higher count. Since each residue from 1 to 10 has 71 numbers, we can choose one residue from each pair (1,10), (2,9), (3,8), (4,7), (5,6), resulting in 5 pairs. - For residue 0, we can take at most 1 number to avoid the sum being divisible by 11.3. Sum the counts: - Each of the 5 pairs contributes 71 numbers. - Adding 1 number from residue 0 gives the total count.Thus, the maximum number of numbers that can be chosen is:[5 times 71 + 1 = 355 + 1 = 356]The final answer is (boxed{356}).]

🔑:Answer: 356.Solution: For k=0,1, ldots, 10, let I_{k} be the set of all numbers in [A, B) that give a remainder of k when divided by 11. Since A and B are multiples of 11, all sets I_{k} contain an equal number of elements. Therefore, all numbers in [A, B) that are not multiples of 11 can be paired as (x, y), where x in I_{k}, y in I_{11-k} for some k from {1, ldots, 5}. The number of such pairs is frac{2134-1353}{11} cdot 5=355. Clearly, from each pair, at most one number can be included in the final set. Additionally, this set can contain at most one number that is a multiple of 11. Thus, the solution includes no more than 356 numbers.Now, we will show that a set of 356 numbers is possible. Include A and all numbers that give an odd remainder when divided by 11 in this set. Take two distinct numbers x and y from this set, different from each other and from A. Then (x bmod 11) + (y bmod 11) is an even number between 2 and 18. Therefore, it is not divisible by 11, and thus x+y is also not divisible by 11. Clearly, A+x is also notdivisible by 11. This means that such a set satisfies the problem's conditions. It remains to note that this set contains frac{2134-1353}{11} cdot 5 + 1 = 356 numbers.

🧠:[<think>Okay, let's see. I have this problem here where I need to find the vector OC such that the dot product of vectors CA and CB is minimized. Then, using that point C, I have to calculate the cosine of the angle ACB. Hmm, okay. Let me try to break this down step by step.First, let me visualize the problem. We have points O, which is the origin, A at (1,7), B at (5,1), and P at (2,1). Point C lies somewhere on the line OP. So, OP is a line from the origin to point P (2,1). Therefore, any point C on OP can be represented parametrically. Maybe I can parametrize point C as a scalar multiple of vector OP. Since OP is (2,1), then OC would be t*(2,1) where t is a real number. That makes sense because as t varies, C moves along the line OP.So, let me write that down. Let’s let C = (2t, t) for some parameter t. Then, vector CA would be OA - OC, right? Because CA is from point C to point A. So OA is (1,7) and OC is (2t, t), so CA = (1 - 2t, 7 - t). Similarly, vector CB would be OB - OC, which is (5 - 2t, 1 - t).Now, the dot product of CA and CB is (1 - 2t)(5 - 2t) + (7 - t)(1 - t). Let me compute that. Let's expand the terms.First, multiply (1 - 2t)(5 - 2t):1*5 + 1*(-2t) + (-2t)*5 + (-2t)*(-2t) = 5 - 2t -10t + 4t² = 5 -12t +4t².Then, multiply (7 - t)(1 - t):7*1 + 7*(-t) + (-t)*1 + (-t)*(-t) = 7 -7t -t + t² = 7 -8t + t².Now, add those two results together for the dot product:5 -12t +4t² +7 -8t + t² = (5+7) + (-12t -8t) + (4t² + t²) = 12 -20t +5t².So the dot product CA · CB is 5t² -20t +12. Now, we need to find the value of t that minimizes this quadratic expression.Since this is a quadratic in t, and the coefficient of t² is positive (5), the parabola opens upwards, so the minimum occurs at the vertex. The vertex of a quadratic at² + bt + c is at t = -b/(2a). Here, a = 5, b = -20. So t = -(-20)/(2*5) = 20/10 = 2. So t = 2.Wait, so t = 2. Therefore, point C is (2t, t) = (4, 2). But wait, point P is (2,1). So OP is from the origin to (2,1), but if t = 2, then C is (4,2), which is beyond point P on the line OP extended past P. Is that allowed?Hmm, the problem says "C is a point on the line OP (where O is the origin of coordinates)". So the line OP is infinite in both directions? Or is it the line segment OP? Wait, in math, when they say "a point on the line OP", it usually refers to the entire line, not just the segment. So even though OP is from O to P, the line OP extends infinitely in both directions. Therefore, t can be any real number, positive or negative. So t=2 would put C at (4,2), which is on the line beyond point P. That's acceptable.But wait, let me check if my calculation is correct. If t=2, then CA is (1 - 4, 7 - 2) = (-3,5), and CB is (5 - 4, 1 - 2) = (1, -1). Then, the dot product is (-3)(1) + (5)(-1) = -3 -5 = -8. Let's plug t=2 into the quadratic equation: 5*(4) -20*(2) +12 = 20 -40 +12 = -8. That matches. But if I take t=0, which would be point O, then the dot product is 5*0 -20*0 +12 =12. If I take t=1, which is halfway to P, then the dot product is 5*1 -20*1 +12 =5 -20 +12= -3. At t=2, it's -8. Then, if we go further, say t=3, the dot product is 5*9 -20*3 +12=45 -60 +12= -3. So, the minimum occurs at t=2. So even though it's beyond point P, it's still valid.Wait, but maybe there's a mistake here. Let me confirm. The quadratic function 5t² -20t +12 has its minimum at t=2, so that's correct. Therefore, OC is (4,2). So the answer for part (1) is OC=(4,2). But let me verify again with coordinates.Alternatively, maybe I should parametrize the line OP differently. Since OP is the line from O(0,0) to P(2,1), parametric equations can be written as x = 2s, y = s, where s is a real number. So point C is (2s, s). So if I use s instead of t, same thing. Then, vectors CA and CB would be (1 - 2s, 7 - s) and (5 - 2s, 1 - s). The dot product is (1 - 2s)(5 - 2s) + (7 - s)(1 - s). Which is the same as before. So regardless of variable name, s=2 gives point (4,2). Therefore, that's correct.Okay, so part (1) answer is vector OC = (4,2). But let me check once more. If point C is (4,2), then vectors CA and CB are (-3,5) and (1,-1), as above. Their dot product is -3*1 +5*(-1)= -8. If I take a point before P, say t=1, which is (2,1), which is point P. Then, vectors CA = (1-2,7-1)=(-1,6), CB=(5-2,1-1)=(3,0). Their dot product is (-1)(3) +6*0 = -3. Which is higher than -8. So indeed, t=2 gives a lower dot product. If we take t=3, which is (6,3), then CA=(1-6,7-3)=(-5,4), CB=(5-6,1-3)=(-1,-2). Dot product is (-5)(-1) +4*(-2)=5 -8= -3, same as t=1. So t=2 gives the minimum. So that's correct.So part (1) is done. The answer is OC=(4,2). Therefore, boxed{(4,2)} for part (1).Now, part (2): Calculate cos∠ACB using the point C found in part (1). So angle ACB is the angle at point C between points A and B. To find the cosine of this angle, we can use the formula:cosθ = (CA · CB) / (|CA| |CB|)We already have CA · CB from part (1) when t=2, which is -8. So we need to compute |CA| and |CB|.First, vector CA is from C(4,2) to A(1,7): (1-4,7-2)=(-3,5). The magnitude of CA is sqrt((-3)^2 +5^2)=sqrt(9+25)=sqrt(34).Vector CB is from C(4,2) to B(5,1): (5-4,1-2)=(1,-1). The magnitude of CB is sqrt(1^2 + (-1)^2)=sqrt(1+1)=sqrt(2).Therefore, cosθ = (-8)/(sqrt(34)*sqrt(2)) = (-8)/sqrt(68) = (-8)/(2*sqrt(17)) = (-4)/sqrt(17).But cosine can be negative. However, we might need to rationalize the denominator. So, multiplying numerator and denominator by sqrt(17):(-4*sqrt(17))/(17). So cosθ = -4√17 /17.But let me verify the calculation again. Let's recheck vectors:From C(4,2) to A(1,7): Δx = 1-4 = -3, Δy=7-2=5. So CA=(-3,5). Correct.From C(4,2) to B(5,1): Δx=5-4=1, Δy=1-2=-1. So CB=(1,-1). Correct.Dot product: (-3)(1) + (5)(-1) = -3 -5 = -8. Correct.|CA| = sqrt(9 +25)=sqrt(34). Correct.|CB| = sqrt(1 +1)=sqrt(2). Correct.Therefore, cosθ = -8/(sqrt(34)*sqrt(2)) = -8/sqrt(68) = -8/(2*sqrt(17)) = -4/sqrt(17). Rationalizing, that's -4√17/17.But angles in a triangle can be between 0 and 180 degrees, so cosine can be negative, which would mean the angle is obtuse. So that's acceptable.Therefore, the answer is -4√17/17. But let me check if the question wants the answer in a particular form. The problem says to put the final answer in boxed notation. So, I can write it as boxed{-dfrac{4sqrt{17}}{17}}.But let me check once more:CA · CB = -8.|CA| = sqrt(34), |CB|=sqrt(2). So product of magnitudes is sqrt(34*2)=sqrt(68)=2*sqrt(17). Therefore, cosθ = -8/(2√17)= -4/√17. Rationalizing gives -4√17/17. Yes, that's correct.Therefore, part (2) answer is -4√17/17. So box that.But wait, the problem says "calculate cos∠ACB". So just to confirm, angle at C, between points A, C, B. Yes, so vectors CA and CB emanate from C, so the angle between them is indeed angle ACB. So the formula is correct.Yes, so I think that's it.Final Answer(1) boxed{(4, 2)}(2) boxed{-dfrac{4sqrt{17}}{17}}</think>Given the vectors (overrightarrow{OP} = (2,1)), (overrightarrow{OA} = (1,7)), (overrightarrow{OB} = (5,1)), we need to find the vector (overrightarrow{OC}) such that (overrightarrow{CA} cdot overrightarrow{CB}) is minimized and then calculate (cosangle ACB).1. Finding (overrightarrow{OC}) such that (overrightarrow{CA} cdot overrightarrow{CB}) is minimized: - Parametrize point (C) on the line (OP) as (C = (2t, t)) where (t) is a real number. - Calculate vectors (overrightarrow{CA}) and (overrightarrow{CB}): [ overrightarrow{CA} = overrightarrow{OA} - overrightarrow{OC} = (1 - 2t, 7 - t) ] [ overrightarrow{CB} = overrightarrow{OB} - overrightarrow{OC} = (5 - 2t, 1 - t) ] - Compute the dot product (overrightarrow{CA} cdot overrightarrow{CB}): [ (1 - 2t)(5 - 2t) + (7 - t)(1 - t) = 5 - 12t + 4t^2 + 7 - 8t + t^2 = 5t^2 - 20t + 12 ] - The quadratic expression (5t^2 - 20t + 12) is minimized at (t = frac{20}{2 cdot 5} = 2). - Therefore, (overrightarrow{OC} = (2 cdot 2, 2) = (4, 2)).2. Calculating (cosangle ACB): - Vectors (overrightarrow{CA}) and (overrightarrow{CB}) at (C(4, 2)): [ overrightarrow{CA} = (1 - 4, 7 - 2) = (-3, 5) ] [ overrightarrow{CB} = (5 - 4, 1 - 2) = (1, -1) ] - Dot product (overrightarrow{CA} cdot overrightarrow{CB}): [ (-3)(1) + (5)(-1) = -3 - 5 = -8 ] - Magnitudes of (overrightarrow{CA}) and (overrightarrow{CB}): [ |overrightarrow{CA}| = sqrt{(-3)^2 + 5^2} = sqrt{34} ] [ |overrightarrow{CB}| = sqrt{1^2 + (-1)^2} = sqrt{2} ] - Calculate (cosangle ACB): [ cosangle ACB = frac{overrightarrow{CA} cdot overrightarrow{CB}}{|overrightarrow{CA}| |overrightarrow{CB}|} = frac{-8}{sqrt{34} cdot sqrt{2}} = frac{-8}{2sqrt{17}} = frac{-4}{sqrt{17}} = frac{-4sqrt{17}}{17} ] Final Answer(1) (boxed{(4, 2)})(2) (boxed{-dfrac{4sqrt{17}}{17}}),<think>Okay, let's tackle this problem step by step. First, I need to understand what's being asked here. So, we have points O, A, B, and P with their respective position vectors given. Point C is somewhere on the line OP, and we need to find the position vector of C such that the dot product of vectors CA and CB is minimized. Then, using that point C, we have to find the cosine of the angle ACB. Starting with part (1). Let me recall that the dot product of two vectors CA and CB is given by the sum of the products of their corresponding components. The formula for the dot product is (overrightarrow{CA} cdot overrightarrow{CB} = |overrightarrow{CA}||overrightarrow{CB}| cos theta), where θ is the angle between them. But since we need to minimize this dot product, maybe I can express it in terms of coordinates and then use calculus to find the minimum.First, I need to parameterize point C on the line OP. The line OP goes from the origin O to point P(2,1). So, any point C on OP can be written as a scalar multiple of the vector OP. Let's let t be a scalar parameter such that (overrightarrow{OC} = t cdot overrightarrow{OP} = (2t, t)). So, the coordinates of point C are (2t, t) where t is a real number. This makes sense because when t=0, we're at the origin, and when t=1, we're at point P. But since the problem says "the line OP", t can be any real number, so C can be anywhere along that line, even extended beyond O or P.Now, vectors CA and CB can be expressed in terms of t. Let's find their components. First, vector CA is from point C to point A, which is (overrightarrow{OA} - overrightarrow{OC}). Similarly, vector CB is (overrightarrow{OB} - overrightarrow{OC}).Given that OA is (1,7) and OB is (5,1), and OC is (2t, t). Therefore:(overrightarrow{CA} = (1 - 2t, 7 - t))(overrightarrow{CB} = (5 - 2t, 1 - t))The dot product of CA and CB is:[(1 - 2t)(5 - 2t) + (7 - t)(1 - t)]Let me compute that step by step.First, expand the first product: (1 - 2t)(5 - 2t)Multiply 1 by 5: 51 by -2t: -2t-2t by 5: -10t-2t by -2t: 4t²So, adding up: 5 - 2t -10t +4t² = 5 -12t +4t²Now the second product: (7 - t)(1 - t)Multiply 7 by 1: 77 by -t: -7t-t by 1: -t-t by -t: t²Adding up: 7 -7t -t +t² = 7 -8t +t²Now sum both parts:First part: 5 -12t +4t²Second part: 7 -8t +t²Total: (5+7) + (-12t -8t) + (4t² + t²) = 12 -20t +5t²So, the dot product (overrightarrow{CA} cdot overrightarrow{CB} = 5t² -20t +12)Wait, that's a quadratic in terms of t. To find its minimum value, since the coefficient of t² is positive (5), the parabola opens upwards, so the minimum occurs at the vertex. The formula for the vertex of a quadratic ax² + bx + c is at t = -b/(2a). Here, a = 5, b = -20, so t = -(-20)/(2*5) = 20/10 = 2. So, t=2.Therefore, the value of t that minimizes the dot product is 2. Therefore, the coordinates of point C are (2*2, 2) = (4, 2). So, (overrightarrow{OC} = (4,2)).Wait, but hold on. The line OP is from O to P(2,1), but here we're getting t=2, which would be at (4,2), which is beyond point P. So, point C is not between O and P but beyond P on the line OP. But the problem says "C is a point on the line OP", which includes the entire line, not just the segment OP. So, that's acceptable. So, part (1) answer is (4,2).But let me verify my calculations again to be sure.Let me recompute the dot product:First term: (1 - 2t)(5 - 2t)Let me compute this again:1*5 = 51*(-2t) = -2t(-2t)*5 = -10t(-2t)*(-2t) = 4t²Total: 5 -12t +4t². That's correct.Second term: (7 - t)(1 - t)7*1 = 77*(-t) = -7t(-t)*1 = -t(-t)*(-t) = t²Total: 7 -8t +t². Correct.Adding both parts:5 +7 =12-12t -8t = -20t4t² + t² =5t²Total expression:5t² -20t +12. Correct.Therefore, the quadratic is 5t² -20t +12. To minimize this, take derivative with respect to t: 10t -20 =0 => t=2. So, indeed, t=2 is the critical point. Since the quadratic opens upwards, that's the minimum. So, OC is (4,2). Therefore, part (1) answer is (4,2). Wait a moment. The problem says "where O is the origin of coordinates". So, OP is a line passing through O and P(2,1). So, parameterizing as t*(2,1) gives points along OP. But when t=2, we get (4,2), which is twice the vector OP from O. So, yes, that's correct. So, point C is (4,2). But just to be thorough, let's check if t=2 gives the minimum. Let's compute the dot product at t=2 and at some other points.At t=2: 5*(4) -20*(2) +12 =20 -40 +12= -8At t=1:5*(1) -20*(1)+12=5-20+12=-3Wait, but at t=1, the value is -3, which is higher than at t=2. Hmm, but according to the quadratic, the minimum is at t=2. But when t increases beyond 2, the value increases again? Wait, but when t=3: 5*9 -20*3 +12=45-60+12=-3. So, at t=3, the value is -3 again.Wait, so the quadratic has a minimum at t=2, but when we check t=2, the value is -8, and at t=1 and t=3, it's -3. So, actually, the minimum is indeed at t=2 with value -8, which is lower than -3. So, that's correct.But wait, how is the value going down from t=1 to t=2 and then back up at t=3? Let's check another value. Let's take t=0. At t=0: 0 -0 +12=12. So, higher. t=2 is indeed the minimum. So, all right.Therefore, part (1) is solved with OC=(4,2).Now, moving on to part (2): calculate cos∠ACB.To find the cosine of the angle at point C between points A and B, we can use the formula:[cos theta = frac{overrightarrow{CA} cdot overrightarrow{CB}}{|overrightarrow{CA}| |overrightarrow{CB}|}]We already have the dot product from part (1) when t=2, which is -8. So, we need to compute the magnitudes of vectors CA and CB at t=2.First, let's find vectors CA and CB when t=2.From earlier, vector CA is (1 - 2t, 7 - t). At t=2:CA = (1 -4, 7 -2) = (-3,5)Similarly, vector CB is (5 - 2t, 1 - t). At t=2:CB = (5 -4, 1 -2) = (1,-1)Now, compute |CA| and |CB|.|CA| = sqrt((-3)^2 +5^2) = sqrt(9 +25) = sqrt(34)|CB| = sqrt(1^2 + (-1)^2) = sqrt(1 +1) = sqrt(2)Therefore,cosθ = (-3)(1) + (5)(-1) divided by (sqrt(34)*sqrt(2))Wait, no, wait. Wait, we already computed the dot product CA ⋅ CB when t=2 as -8. Wait, but let me confirm. Wait, when t=2, vector CA is (-3,5), CB is (1,-1). So, their dot product is (-3)(1) + (5)(-1) = -3 -5 = -8. Correct. So, that matches part (1).Therefore, cosθ = (-8)/(sqrt(34)*sqrt(2)) = -8/(sqrt(68)) = -8/(2*sqrt(17)) = -4/sqrt(17) = -4√17 /17But since cosine can be negative, we can rationalize the denominator.Therefore, cos∠ACB = -4√17 /17But let me check this again.First, vectors CA = (-3,5), CB = (1,-1)Compute their magnitudes:|CA| = sqrt((-3)^2 +5^2) = sqrt(9 +25) = sqrt(34)|CB| = sqrt(1 +1) = sqrt(2)Then, their dot product is (-3)(1) +5*(-1) = -3 -5 = -8Thus,cosθ = (-8)/(sqrt(34)*sqrt(2)) = (-8)/sqrt(68) = (-8)/(2*sqrt(17)) = -4/sqrt(17)Rationalizing the denominator:-4/sqrt(17) = (-4√17)/17Therefore, the answer is -4√17/17.But let's verify once more.Alternatively, I can write sqrt(34)*sqrt(2) = sqrt(68) = sqrt(4*17) = 2*sqrt(17). Hence, yes, denominator is 2*sqrt(17). Then, -8 divided by 2*sqrt(17) is -4/sqrt(17), which is -4√17/17. Correct.Therefore, cos∠ACB is -4√17/17.But I need to confirm if the angle is acute or obtuse. Since the cosine is negative, the angle is obtuse, which makes sense given the vectors. But the problem just asks for the cosine, so the answer is negative.So, putting it all together, part (1) is OC=(4,2) and part (2) is -4√17/17.Just to make sure, let me check if there's any possible mistake in part (1). Suppose I made an error in the parameterization. Wait, we parametrized OC as t*(2,1), so OC=(2t, t). Then, OA is (1,7), so CA is OA - OC = (1-2t,7 - t). Similarly, OB is (5,1), so CB is (5-2t,1 - t). Then, their dot product is computed correctly as 5t² -20t +12. The derivative is 10t -20, set to zero gives t=2. So, OC=(4,2). That seems correct.Alternatively, maybe another approach can be used. For example, using projection formulas or calculus. But I think the current method is solid.Another way to think about part (1): since we need to minimize CA ⋅ CB, which is equivalent to minimizing the expression we derived. Since it's a quadratic in t, the calculus approach gives the minimum. Alternatively, we can consider the expression as a function of t and complete the square. Let's try that:5t² -20t +12 =5(t² -4t) +12 =5(t² -4t +4 -4) +12=5(t -2)^2 -20 +12=5(t-2)^2 -8Therefore, the minimum value is -8 when t=2. So, this confirms that the minimum is achieved at t=2. Therefore, OC=(4,2). So, all checks out.Thus, the answers are:1) (boxed{(4, 2)})2) (boxed{-dfrac{4sqrt{17}}{17}})Final Answer(1) (boxed{(4, 2)})(2) (boxed{-dfrac{4sqrt{17}}{17}})</think>Given (overrightarrow{OP} = (2,1)), (overrightarrow{OA} = (1,7)), (overrightarrow{OB} = (5,1)), we need to find the point (C) on the line (OP) such that (overrightarrow{CA} cdot overrightarrow{CB}) is minimized and then calculate (cosangle ACB).1. Finding (overrightarrow{OC}) such that (overrightarrow{CA} cdot overrightarrow{CB}) is minimized: - Parameterize point (C) on the line (OP) as (overrightarrow{OC} = t cdot overrightarrow{OP} = (2t, t)). - Express vectors (overrightarrow{CA}) and (overrightarrow{CB}) in terms of (t): [ overrightarrow{CA} = (1 - 2t, 7 - t) ] [ overrightarrow{CB} = (5 - 2t, 1 - t) ] - Compute the dot product (overrightarrow{CA} cdot overrightarrow{CB}): [ (1 - 2t)(5 - 2t) + (7 - t)(1 - t) ] Expanding and simplifying: [ 5t^2 - 20t + 12 ] - This quadratic function in (t) has its minimum at (t = frac{-b}{2a} = frac{20}{10} = 2). - Therefore, (overrightarrow{OC} = (2 cdot 2, 2) = (4, 2)).2. Calculating (cosangle ACB): - For (t = 2), the coordinates of (C) are ((4, 2)). - Vectors (overrightarrow{CA}) and (overrightarrow{CB}) are: [ overrightarrow{CA} = (-3, 5) ] [ overrightarrow{CB} = (1, -1) ] - Compute the magnitudes: [ |overrightarrow{CA}| = sqrt{(-3)^2 + 5^2} = sqrt{34} ] [ |overrightarrow{CB}| = sqrt{1^2 + (-1)^2} = sqrt{2} ] - The dot product (overrightarrow{CA} cdot overrightarrow{CB}) is already found to be (-8). - Calculate (cosangle ACB): [ cosangle ACB = frac{-8}{sqrt{34} cdot sqrt{2}} = frac{-8}{2sqrt{17}} = frac{-4}{sqrt{17}} = -frac{4sqrt{17}}{17} ]Final answers:1. (boxed{(4, 2)})2. (boxed{-dfrac{4sqrt{17}}{17}})]

🔑:(1) Since overrightarrow{OP}=(2,1), the equation of line OP is y = frac{1}{2}x.Since C is a point on line OP, let's denote point C as C(x, frac{1}{2}x).Thus, overrightarrow{CA} = overrightarrow{OA} - overrightarrow{OC} = (1-x, 7 - frac{1}{2}x),and overrightarrow{CB} = overrightarrow{OB} - overrightarrow{OC} = (5-x, 1 - frac{1}{2}x).The product of the two vectors is:overrightarrow{CA} cdot overrightarrow{CB} = (1-x)(5-x) + left(7 - frac{1}{2}xright)left(1 - frac{1}{2}xright) = frac{5}{4}x^2 - 10x + 12 = frac{5}{4}(x-4)^2 - 8.Therefore, when x = 4, overrightarrow{CA} cdot overrightarrow{CB} has the minimum value. At this point, C is at (4,2),so overrightarrow{OC} = (4,2).(2) From (1), we know that C is at (4,2),thus overrightarrow{CA} = (-3, 5) and overrightarrow{CB} = (1, -1).To find cosangle ACB, we use the following formula:cosangle ACB = frac{overrightarrow{CA} cdot overrightarrow{CB}}{|overrightarrow{CA}| cdot |overrightarrow{CB}|} = frac{-3 cdot 1 + 5 cdot (-1)}{sqrt{(-3)^2 + 5^2} sqrt{1^2 + (-1)^2}} = -frac{4sqrt{17}}{17}.Thus, cosangle ACB = boxed{-frac{4sqrt{17}}{17}}.