answer：Okay, so I have this problem here: A line ( l ) forms angles (alpha), (beta), and (gamma) with three pairwise perpendicular lines. I need to prove that (cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1). Hmm, interesting. I remember something about direction cosines from my geometry class, but I'm not entirely sure how to apply it here. Let me try to break it down step by step.First, let me visualize the scenario. There are three lines that are pairwise perpendicular, which means they form a sort of 3D coordinate system. So, imagine the x-axis, y-axis, and z-axis in a coordinate system. Line ( l ) is some arbitrary line that intersects these three axes, forming angles (alpha), (beta), and (gamma) with each of them, respectively.Now, I think direction cosines are related to the angles that a line makes with the coordinate axes. If I recall correctly, the direction cosines are the cosines of these angles, and they have a special property. Let me see... I think they must satisfy the equation ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ). But wait, that's exactly what I need to prove! So, maybe I just need to recall why this is true.Let me think about vectors. If I consider the line ( l ) as a vector in 3D space, then the direction cosines are the components of the unit vector in the direction of ( l ). So, if I have a vector ( vec{v} ) along line ( l ), I can write it as ( vec{v} = (v_x, v_y, v_z) ). The direction cosines are then ( cos alpha = frac{v_x}{|vec{v}|} ), ( cos beta = frac{v_y}{|vec{v}|} ), and ( cos gamma = frac{v_z}{|vec{v}|} ).Since ( vec{v} ) is a unit vector, its magnitude ( |vec{v}| ) is 1. That means ( v_x = cos alpha ), ( v_y = cos beta ), and ( v_z = cos gamma ). Now, for any unit vector, the sum of the squares of its components must equal 1. So, ( v_x^2 + v_y^2 + v_z^2 = 1 ). Substituting the direction cosines into this equation, we get ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ). Wait, that seems too straightforward. Did I just prove it? Maybe I should double-check. Let me go through it again.Starting with the vector ( vec{v} ) along line ( l ). If ( vec{v} ) is a unit vector, then its components satisfy ( v_x^2 + v_y^2 + v_z^2 = 1 ). The direction cosines are defined as the cosines of the angles between the vector and each axis, so ( cos alpha = v_x ), ( cos beta = v_y ), and ( cos gamma = v_z ). Therefore, substituting these into the equation, we have ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ). Yeah, that makes sense.But wait, what if the line ( l ) isn't a unit vector? Does that affect the proof? Hmm, no, because direction cosines are defined based on the unit vector in the direction of ( l ). So, even if ( l ) isn't a unit vector, we can always normalize it to get the direction cosines, and the property still holds.Let me think of a concrete example to test this. Suppose line ( l ) is along the x-axis. Then, the angle ( alpha ) with the x-axis is 0°, so ( cos alpha = 1 ). The angles with the y-axis and z-axis would be 90°, so ( cos beta = 0 ) and ( cos gamma = 0 ). Plugging into the equation: ( 1^2 + 0^2 + 0^2 = 1 ), which is correct.Another example: suppose line ( l ) makes equal angles with all three axes. Let's say each angle is 45°, so ( cos 45° = frac{sqrt{2}}{2} ). Then, ( cos^{2} 45° + cos^{2} 45° + cos^{2} 45° = frac{1}{2} + frac{1}{2} + frac{1}{2} = frac{3}{2} ). Wait, that's not equal to 1. Hmm, that's a problem. Did I make a mistake?Oh, right! If a line makes equal angles with all three axes, those angles can't all be 45°, because that would require the direction cosines to each be ( frac{sqrt{2}}{2} ), which squared and summed would give ( frac{3}{2} ), which is more than 1. That's impossible because the sum must be 1. So, what's the correct angle?Let me calculate. If ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ) and all angles are equal, then ( 3 cos^{2} alpha = 1 ), so ( cos^{2} alpha = frac{1}{3} ), which means ( cos alpha = frac{sqrt{3}}{3} ). Therefore, ( alpha = arccos left( frac{sqrt{3}}{3} right) approx 54.7° ). So, the angles can't be 45°, but approximately 54.7°. That makes sense because in 3D space, the angles with the axes have to be larger to satisfy the condition.Okay, so that example shows that the angles can't be arbitrary; they have to satisfy the equation ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ). Therefore, the proof seems solid.Let me try another approach to see if I can derive it differently. Maybe using the dot product. If I have two vectors, their dot product is equal to the product of their magnitudes and the cosine of the angle between them. So, if I take the unit vector along ( l ) and the unit vectors along the coordinate axes, their dot products should give the direction cosines.Let ( vec{u} ) be the unit vector along line ( l ), and let ( hat{i} ), ( hat{j} ), ( hat{k} ) be the unit vectors along the x, y, and z axes, respectively. Then, the dot product ( vec{u} cdot hat{i} = cos alpha ), ( vec{u} cdot hat{j} = cos beta ), and ( vec{u} cdot hat{k} = cos gamma ).Since ( vec{u} ) is a unit vector, ( vec{u} cdot vec{u} = 1 ). But ( vec{u} ) can also be expressed in terms of its components: ( vec{u} = (cos alpha, cos beta, cos gamma) ). Therefore, ( vec{u} cdot vec{u} = cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ). There, that's another way to see it.So, whether I approach it through the properties of unit vectors or through the dot product, I end up with the same conclusion. It seems like this identity is a fundamental property of direction cosines in three-dimensional space.I wonder if this extends to higher dimensions. For example, in four-dimensional space, would the sum of the squares of the direction cosines with four mutually perpendicular axes also equal 1? I think it would, because the same principle applies: the components of a unit vector in any number of dimensions must satisfy that the sum of their squares equals 1. So, in n-dimensional space, the sum of the squares of the direction cosines with n mutually perpendicular axes would be 1.But in this problem, we're specifically dealing with three dimensions, so the proof holds as is. It's a neat result because it shows how the geometry of angles in 3D space is constrained by the properties of vectors.Let me think about another aspect. If I have a line in space, its direction is uniquely determined by these three angles with the coordinate axes. But since the line is in three dimensions, knowing two angles would determine the third, right? Because if I know two of the direction cosines, the third is fixed by the equation ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ). So, there's a dependency among the angles.For example, if I know ( cos alpha ) and ( cos beta ), then ( cos gamma = sqrt{1 - cos^{2} alpha - cos^{2} beta} ). This shows that the angles aren't independent; they're linked through this equation. That makes sense because in three dimensions, you can't have complete freedom in choosing all three angles independently.This also ties into the concept of spherical coordinates, where a point on a sphere is determined by two angles (polar and azimuthal). But in this case, we're dealing with direction cosines, which are related but slightly different.I think I've covered the main points. The key takeaway is that the sum of the squares of the direction cosines equals 1 because they represent the components of a unit vector in three-dimensional space. This identity is a direct consequence of the properties of vectors and the geometry of angles in 3D.Just to wrap up, let me summarize the proof:1. Consider a unit vector ( vec{u} ) along line ( l ).2. The direction cosines are ( cos alpha ), ( cos beta ), and ( cos gamma ), which are the components of ( vec{u} ).3. Since ( vec{u} ) is a unit vector, the sum of the squares of its components equals 1.4. Therefore, ( cos^{2} alpha + cos^{2} beta + cos^{2} gamma = 1 ).This completes the proof.

answer：Alright, so I have this problem here: Given the function ( f(x) = sqrt{3}sin(2017x) + cos(2017x) ), its maximum value is ( A ). If there exist real numbers ( x_1 ) and ( x_2 ) such that for any real number ( x ), ( f(x_1) leq f(x) leq f(x_2) ) holds true, then the minimum value of ( 2A|x_1 - x_2| ) is ( ).The options are:A: ( frac{pi}{2017} )B: ( frac{2pi}{2017} )C: ( frac{4pi}{2017} )D: ( frac{pi}{4034} )Hmm, okay. Let me try to break this down step by step.First, I need to find the maximum value ( A ) of the function ( f(x) ). The function is a combination of sine and cosine functions with the same argument ( 2017x ). I remember that such functions can be rewritten using the amplitude-phase form, which is ( Rsin(2017x + phi) ) or ( Rcos(2017x + phi) ), where ( R ) is the amplitude and ( phi ) is the phase shift.To find ( R ), I can use the formula:[ R = sqrt{a^2 + b^2} ]where ( a ) and ( b ) are the coefficients of sine and cosine respectively. In this case, ( a = sqrt{3} ) and ( b = 1 ).Calculating ( R ):[ R = sqrt{(sqrt{3})^2 + 1^2} = sqrt{3 + 1} = sqrt{4} = 2 ]So, the maximum value ( A ) is 2 because the sine function oscillates between -1 and 1, and multiplying by 2 scales it to between -2 and 2.Now, the function can be rewritten as:[ f(x) = 2sin(2017x + phi) ]where ( phi ) is some phase shift. I might not need to find ( phi ) explicitly, but it's good to know that the function is a sine wave with amplitude 2.Next, the problem mentions ( x_1 ) and ( x_2 ) such that for any real number ( x ), ( f(x_1) leq f(x) leq f(x_2) ). This means ( f(x_1) ) is the minimum value of the function, and ( f(x_2) ) is the maximum value. Since we already found the maximum value ( A = 2 ), the minimum value would be ( -2 ).But wait, the problem says "for any real number ( x )", ( f(x_1) leq f(x) leq f(x_2) ). So, ( x_1 ) is a point where the function attains its minimum, and ( x_2 ) is where it attains its maximum.Now, I need to find the minimum value of ( 2A|x_1 - x_2| ). Let's break this down:- ( A = 2 )- ( |x_1 - x_2| ) is the distance between the points where the function attains its minimum and maximum.So, ( 2A|x_1 - x_2| = 2 times 2 times |x_1 - x_2| = 4|x_1 - x_2| ).Therefore, I need to find the minimum possible value of ( 4|x_1 - x_2| ).To find ( |x_1 - x_2| ), I need to understand the period of the function ( f(x) ). The function is ( 2sin(2017x + phi) ), so its period ( T ) is given by:[ T = frac{2pi}{2017} ]This is because the period of ( sin(kx) ) is ( frac{2pi}{k} ).Now, in a sine wave, the distance between a maximum and the next minimum is half the period. So, the distance between ( x_2 ) (where the function is maximum) and ( x_1 ) (where it's minimum) is ( frac{T}{2} ).Calculating ( |x_1 - x_2| ):[ |x_1 - x_2| = frac{T}{2} = frac{pi}{2017} ]Therefore, substituting back into ( 4|x_1 - x_2| ):[ 4 times frac{pi}{2017} = frac{4pi}{2017} ]So, the minimum value of ( 2A|x_1 - x_2| ) is ( frac{4pi}{2017} ), which corresponds to option C.Wait, let me just double-check if I interpreted the problem correctly. It says "the minimum value of ( 2A|x_1 - x_2| )". I found ( |x_1 - x_2| ) as half the period, which seems right because between a maximum and the next minimum, the function covers half its cycle. So, multiplying that by ( 2A ) (which is 4) gives ( frac{4pi}{2017} ). That seems correct.I don't think I made any mistakes in the calculations. The key steps were recognizing the function can be rewritten as a single sine function with amplitude 2, finding the period, and then realizing that the distance between a maximum and minimum is half the period. Multiplying that by 4 gives the answer.**Final Answer**boxed{C}

answer：Okay, so I have this problem here: I need to evaluate the integer nearest to 500 times the sum from n equals 4 to 20000 of 1 over (n minus 3)(n plus 3). Hmm, let me try to figure this out step by step.First, the expression inside the sum is 1 divided by (n-3)(n+3). That looks like a product of two terms, n-3 and n+3. Maybe I can simplify this fraction somehow. I remember that sometimes fractions like this can be broken down into partial fractions. Let me try that.So, I want to write 1/((n-3)(n+3)) as A/(n-3) + B/(n+3). To find A and B, I can set up the equation:1 = A(n+3) + B(n-3)Let me solve for A and B. If I plug in n = 3, then the equation becomes:1 = A(3+3) + B(3-3) => 1 = 6A => A = 1/6Similarly, if I plug in n = -3, the equation becomes:1 = A(-3+3) + B(-3-3) => 1 = -6B => B = -1/6So, I have:1/((n-3)(n+3)) = (1/6)/(n-3) - (1/6)/(n+3)That simplifies to:1/((n-3)(n+3)) = (1/6)[1/(n-3) - 1/(n+3)]Okay, so now I can rewrite the original sum using this partial fraction decomposition. Let me write that out:Sum from n=4 to 20000 of 1/((n-3)(n+3)) = (1/6) * Sum from n=4 to 20000 of [1/(n-3) - 1/(n+3)]Hmm, this looks like a telescoping series. I remember that in telescoping series, a lot of terms cancel out when you write out the partial sums. Let me try to write out the first few terms and see what happens.When n=4: 1/(4-3) - 1/(4+3) = 1/1 - 1/7When n=5: 1/(5-3) - 1/(5+3) = 1/2 - 1/8When n=6: 1/(6-3) - 1/(6+3) = 1/3 - 1/9When n=7: 1/(7-3) - 1/(7+3) = 1/4 - 1/10...When n=20000: 1/(20000-3) - 1/(20000+3) = 1/19997 - 1/20003So, if I write out the entire sum, it would look like:(1/1 - 1/7) + (1/2 - 1/8) + (1/3 - 1/9) + (1/4 - 1/10) + ... + (1/19997 - 1/20003)Now, let's see how this telescopes. The negative terms will cancel with positive terms a few steps ahead. Specifically, the -1/7 from the first term will cancel with the +1/7 somewhere later, but wait, in this case, the denominators are increasing by 1 each time, so the negative terms are actually offset by 6 in the denominator.Wait, maybe I should think about how many terms are left after cancellation. Let me consider the positive terms and the negative terms separately.Positive terms: 1/1, 1/2, 1/3, 1/4, ..., 1/19997Negative terms: -1/7, -1/8, -1/9, ..., -1/20003So, the positive terms start at 1/1 and go up to 1/19997, while the negative terms start at 1/7 and go up to 1/20003.Therefore, the positive terms that don't get canceled are the first six terms: 1/1, 1/2, 1/3, 1/4, 1/5, 1/6.Similarly, the negative terms that don't get canceled are the last six terms: -1/19998, -1/19999, -1/20000, -1/20001, -1/20002, -1/20003.Wait, let me check that. If the negative terms start at 1/7 and go up to 1/20003, and the positive terms go up to 1/19997, then the negative terms beyond 1/19997 won't be canceled. So, the negative terms from 1/19998 to 1/20003 are not canceled.Similarly, the positive terms before 1/7 are not canceled, which are 1/1, 1/2, 1/3, 1/4, 1/5, 1/6.So, the sum simplifies to:Sum = (1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)So, putting it all together, the original sum is (1/6) times this expression.Therefore, the sum S is:S = (1/6)[(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)]Now, I need to compute this expression and then multiply by 500 to get the final value.First, let's compute the positive part: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6.Let me calculate that:1 = 11/2 = 0.51/3 ≈ 0.33333331/4 = 0.251/5 = 0.21/6 ≈ 0.1666667Adding these up:1 + 0.5 = 1.51.5 + 0.3333333 ≈ 1.83333331.8333333 + 0.25 = 2.08333332.0833333 + 0.2 = 2.28333332.2833333 + 0.1666667 ≈ 2.45So, the positive part is approximately 2.45.Now, the negative part: 1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003.These are all very small numbers. Let me approximate each term.Since 1/20000 is 0.00005, and the other terms are slightly larger or smaller, but all around that magnitude.Let me compute each term:1/19998 ≈ 0.0000500051/19999 ≈ 0.00005000251/20000 = 0.000051/20001 ≈ 0.00004999751/20002 ≈ 0.0000499951/20003 ≈ 0.0000499925Adding these up:0.000050005 + 0.0000500025 ≈ 0.00010000750.0001000075 + 0.00005 = 0.00015000750.0001500075 + 0.0000499975 ≈ 0.0001999950.000199995 + 0.000049995 ≈ 0.000249990.00024999 + 0.0000499925 ≈ 0.0002999825So, the negative part is approximately 0.0002999825.Therefore, the entire expression inside the brackets is approximately:2.45 - 0.0002999825 ≈ 2.4497000175Now, multiply this by 1/6:(1/6) * 2.4497000175 ≈ 0.40828333625So, the sum S is approximately 0.40828333625.Now, multiply this by 500:500 * 0.40828333625 ≈ 204.141668125Wait, that can't be right because earlier I thought the answer was 153. Did I make a mistake somewhere?Wait, let me double-check my calculations.Wait, when I broke down the sum, I had:Sum = (1/6)[(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)]But when I calculated the positive part, I got approximately 2.45, and the negative part was approximately 0.0003, so the difference was approximately 2.4497.Then, multiplying by 1/6 gives approximately 0.408283, and then multiplying by 500 gives approximately 204.141668.Wait, but the initial problem was to evaluate 500 times the sum, so 500 * S, where S is approximately 0.408283, which is about 204.141668.But in the initial thought process, the user got 153. So, I must have made a mistake somewhere.Wait, let me go back to the partial fraction decomposition.I had:1/((n-3)(n+3)) = (1/6)[1/(n-3) - 1/(n+3)]Yes, that seems correct.Then, when I wrote out the terms, I had:(1/1 - 1/7) + (1/2 - 1/8) + (1/3 - 1/9) + ... + (1/19997 - 1/20003)So, the positive terms are 1, 1/2, 1/3, ..., 1/19997The negative terms are -1/7, -1/8, ..., -1/20003Therefore, the sum is:Sum = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)Wait, but in my earlier calculation, I only subtracted six terms, but actually, the negative terms start at 1/7 and go up to 1/20003, which is 20003 - 7 + 1 = 19997 terms? Wait, no, that's not correct.Wait, the negative terms are from n=4 to n=20000, so when n=4, it's -1/7, and when n=20000, it's -1/20003. So, the negative terms are from 1/7 to 1/20003, which is 20003 - 7 + 1 = 19997 terms.Similarly, the positive terms are from n=4 to n=20000, which is 20000 - 4 + 1 = 19997 terms, starting at 1/1 (when n=4, 1/(4-3)=1/1) up to 1/19997 (when n=20000, 1/(20000-3)=1/19997).Therefore, the positive terms are 1, 1/2, 1/3, ..., 1/19997The negative terms are -1/7, -1/8, ..., -1/20003So, when we subtract, the positive terms that don't get canceled are the first six terms: 1, 1/2, 1/3, 1/4, 1/5, 1/6And the negative terms that don't get canceled are the last six terms: -1/19998, -1/19999, -1/20000, -1/20001, -1/20002, -1/20003Wait, that makes sense because the negative terms start at 1/7, so the first six positive terms (1 to 1/6) don't have corresponding negative terms to cancel them. Similarly, the negative terms go up to 1/20003, while the positive terms only go up to 1/19997, so the last six negative terms don't get canceled.Therefore, the sum is:(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)So, that part is correct.Now, calculating the positive part: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6Let me compute this more accurately.1 = 11 + 1/2 = 1.51.5 + 1/3 ≈ 1.5 + 0.3333333 = 1.83333331.8333333 + 1/4 = 1.8333333 + 0.25 = 2.08333332.0833333 + 1/5 = 2.0833333 + 0.2 = 2.28333332.2833333 + 1/6 ≈ 2.2833333 + 0.1666667 = 2.45So, the positive part is exactly 2.45.Now, the negative part: 1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003These are very small numbers. Let me compute each term more accurately.1/19998 ≈ 0.0000500051/19999 ≈ 0.00005000251/20000 = 0.000051/20001 ≈ 0.00004999751/20002 ≈ 0.0000499951/20003 ≈ 0.0000499925Adding these up:0.000050005 + 0.0000500025 = 0.00010000750.0001000075 + 0.00005 = 0.00015000750.0001500075 + 0.0000499975 = 0.0001999950.000199995 + 0.000049995 = 0.000249990.00024999 + 0.0000499925 = 0.0002999825So, the negative part is approximately 0.0002999825.Therefore, the expression inside the brackets is:2.45 - 0.0002999825 ≈ 2.4497000175Now, multiply this by 1/6:(1/6) * 2.4497000175 ≈ 0.40828333625So, the sum S is approximately 0.40828333625.Now, multiply this by 500:500 * 0.40828333625 ≈ 204.141668125Wait, that's different from the initial thought process where the answer was 153. Did I make a mistake in the partial fraction decomposition?Wait, let me double-check the partial fraction decomposition.I had:1/((n-3)(n+3)) = (1/6)[1/(n-3) - 1/(n+3)]Yes, that seems correct because:(1/6)[1/(n-3) - 1/(n+3)] = (1/6)[(n+3 - (n-3))/((n-3)(n+3))] = (1/6)[6/((n-3)(n+3))] = 1/((n-3)(n+3))So, that part is correct.Wait, but in the initial thought process, the user had:Sum from n=4 to 20000 of 1/((n-3)(n+3)) = (1/6)[(1 + 1/2 + 1/3) - (1/19998 + 1/19999 + 1/20000)]But that's not correct because the negative terms go up to 1/20003, not 1/20000. So, the user only subtracted three terms, but actually, there are six terms to subtract.Therefore, the user's calculation was incorrect because they only subtracted three terms, leading to an overestimation of the sum.So, in my calculation, I correctly accounted for six terms in both the positive and negative parts, leading to a more accurate result.Therefore, the correct value is approximately 204.141668125, which is approximately 204.1417.So, the integer nearest to this value is 204.Wait, but the initial thought process had 153 as the answer, which is significantly different. So, I must have made a mistake in my calculation.Wait, let me go back to the sum.Wait, the sum is from n=4 to n=20000, which is 19997 terms.But when I broke it down, I had:Sum = (1/6)[(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)]But wait, actually, the negative terms are from 1/7 to 1/20003, which is 20003 - 7 + 1 = 19997 terms.Similarly, the positive terms are from 1/1 to 1/19997, which is also 19997 terms.Therefore, when we subtract, the positive terms that don't get canceled are the first six terms: 1, 1/2, 1/3, 1/4, 1/5, 1/6And the negative terms that don't get canceled are the last six terms: -1/19998, -1/19999, -1/20000, -1/20001, -1/20002, -1/20003So, the sum is indeed:(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)Therefore, my calculation seems correct.But then why did the initial thought process get 153?Wait, perhaps the initial thought process made a mistake in the partial fraction decomposition.Wait, let me check the initial thought process again.The user wrote:Sum from n=4 to 20000 of 1/((n-3)(n+3)) = (1/6)[(1 + 1/2 + 1/3) - (1/19998 + 1/19999 + 1/20000)]But that's incorrect because they only subtracted three terms instead of six.Therefore, their calculation was wrong, leading to an incorrect result.So, in my calculation, I correctly accounted for six terms in both the positive and negative parts, leading to a more accurate result.Therefore, the correct value is approximately 204.1417, so the nearest integer is 204.Wait, but let me double-check my calculation of the negative part.I had:1/19998 ≈ 0.0000500051/19999 ≈ 0.00005000251/20000 = 0.000051/20001 ≈ 0.00004999751/20002 ≈ 0.0000499951/20003 ≈ 0.0000499925Adding these up:0.000050005 + 0.0000500025 = 0.00010000750.0001000075 + 0.00005 = 0.00015000750.0001500075 + 0.0000499975 = 0.0001999950.000199995 + 0.000049995 = 0.000249990.00024999 + 0.0000499925 = 0.0002999825So, the negative part is approximately 0.0002999825.Therefore, the expression inside the brackets is:2.45 - 0.0002999825 ≈ 2.4497000175Multiplying by 1/6:2.4497000175 / 6 ≈ 0.40828333625Then, multiplying by 500:0.40828333625 * 500 ≈ 204.141668125So, approximately 204.1417.Therefore, the integer nearest to this value is 204.Wait, but the initial thought process had 153 as the answer, which is significantly different. So, I must have made a mistake in my calculation.Wait, let me check the partial fraction decomposition again.I had:1/((n-3)(n+3)) = (1/6)[1/(n-3) - 1/(n+3)]Yes, that's correct.Then, the sum from n=4 to 20000 is:Sum = (1/6)[(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) - (1/19998 + 1/19999 + 1/20000 + 1/20001 + 1/20002 + 1/20003)]Yes, that's correct.So, the positive part is 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 2.45The negative part is approximately 0.0002999825So, the difference is approximately 2.4497000175Multiply by 1/6: 2.4497000175 / 6 ≈ 0.40828333625Multiply by 500: 0.40828333625 * 500 ≈ 204.141668125So, the integer nearest is 204.Therefore, the correct answer should be 204, not 153.Wait, but in the initial thought process, the user got 153. So, perhaps they made a mistake in their calculation.Alternatively, maybe I made a mistake in the partial fraction decomposition.Wait, let me check the partial fraction decomposition again.Given:1/((n-3)(n+3)) = A/(n-3) + B/(n+3)Multiplying both sides by (n-3)(n+3):1 = A(n+3) + B(n-3)Expanding:1 = An + 3A + Bn - 3BCombine like terms:1 = (A + B)n + (3A - 3B)Therefore, we have the system of equations:A + B = 03A - 3B = 1From the first equation, B = -ASubstitute into the second equation:3A - 3(-A) = 1 => 3A + 3A = 1 => 6A = 1 => A = 1/6Therefore, B = -1/6So, the partial fraction decomposition is correct.Therefore, the initial thought process must have made a mistake in the number of terms subtracted.In the initial thought process, the user only subtracted three terms, leading to an overestimation of the sum.Therefore, the correct answer is 204.

answer：Okay, so I have this function f(x) = 2x³ - 3(a+1)x² + 6ax, and there are three parts to the problem. Let me try to tackle each part step by step.Starting with part (1): The slope of the tangent line to the curve y = f(x) at x = 0 is 3. I need to find the value of a.Hmm, the slope of the tangent line at a point is given by the derivative of the function at that point. So, I need to find f'(x) and then evaluate it at x = 0.Let's compute the derivative f'(x). The derivative of 2x³ is 6x², the derivative of -3(a+1)x² is -6(a+1)x, and the derivative of 6ax is 6a. So, putting it all together:f'(x) = 6x² - 6(a+1)x + 6aNow, evaluate this at x = 0:f'(0) = 6*(0)² - 6(a+1)*(0) + 6a = 0 - 0 + 6a = 6aAccording to the problem, this slope is 3. So,6a = 3Solving for a:a = 3 / 6 = 1/2Okay, so a is 1/2. That seems straightforward.Moving on to part (2): For any x ∈ (0, +∞), f(x) + f(-x) ≥ 12 ln x always holds. I need to find the range of values for a.First, let's compute f(x) + f(-x). Let's substitute -x into f(x):f(-x) = 2(-x)³ - 3(a+1)(-x)² + 6a(-x) = -2x³ - 3(a+1)x² - 6axNow, add f(x) and f(-x):f(x) + f(-x) = [2x³ - 3(a+1)x² + 6ax] + [-2x³ - 3(a+1)x² - 6ax]Let's combine like terms:2x³ - 2x³ = 0-3(a+1)x² - 3(a+1)x² = -6(a+1)x²6ax - 6ax = 0So, f(x) + f(-x) simplifies to -6(a+1)x²So, the inequality becomes:-6(a+1)x² ≥ 12 ln xLet me write that as:-6(a+1)x² ≥ 12 ln xDivide both sides by 6 to simplify:-(a+1)x² ≥ 2 ln xSo,-(a+1) ≥ (2 ln x) / x²Because x² is positive for x > 0, so dividing both sides by x² doesn't change the inequality direction.So, we have:-(a+1) ≥ (2 ln x) / x²Which can be rewritten as:a + 1 ≤ - (2 ln x) / x²So,a ≤ -1 - (2 ln x) / x²This inequality must hold for all x > 0. Therefore, the maximum value of the right-hand side (RHS) must be less than or equal to a.Wait, no. Since the inequality is a ≤ -1 - (2 ln x)/x² for all x > 0, then a must be less than or equal to the minimum value of -1 - (2 ln x)/x² over x > 0.Wait, actually, let me think again.We have:-(a + 1) ≥ (2 ln x)/x² for all x > 0.Which means:(a + 1) ≤ - (2 ln x)/x² for all x > 0.So, a + 1 must be less than or equal to the minimum value of - (2 ln x)/x² over x > 0.Therefore, a + 1 ≤ min_{x > 0} [ - (2 ln x)/x² ]Which implies:a ≤ -1 + min_{x > 0} [ - (2 ln x)/x² ]Wait, that doesn't seem right. Let me rephrase.Wait, if we have:-(a + 1) ≥ (2 ln x)/x² for all x > 0.Then, -(a + 1) must be greater than or equal to the maximum value of (2 ln x)/x² over x > 0.Because for the inequality to hold for all x, -(a + 1) must be at least as large as the maximum of (2 ln x)/x².So, let me denote:g(x) = (2 ln x)/x²We need to find the maximum value of g(x) over x > 0.Once we find that maximum value, say M, then:-(a + 1) ≥ MWhich implies:a + 1 ≤ -MHence,a ≤ -M - 1So, first, let's find the maximum of g(x) = (2 ln x)/x².To find the maximum, we can take the derivative of g(x) and set it equal to zero.Compute g'(x):g'(x) = [2*(1/x)*x² - 2 ln x * 2x] / x⁴Wait, that's using the quotient rule: (numerator derivative * denominator - numerator * denominator derivative) / denominator squared.Wait, let me compute it step by step.g(x) = (2 ln x) / x²Let me write it as 2 ln x * x^{-2}Then, using the product rule:g'(x) = 2*(1/x)*x^{-2} + 2 ln x*(-2)x^{-3}Simplify:= 2 x^{-3} - 4 ln x x^{-3}= (2 - 4 ln x) / x³Set g'(x) = 0:(2 - 4 ln x) / x³ = 0The numerator must be zero:2 - 4 ln x = 0So,4 ln x = 2ln x = 2 / 4 = 1/2Therefore,x = e^{1/2} = sqrt(e)So, the critical point is at x = sqrt(e).Now, let's check the second derivative or use test points to confirm it's a maximum.Alternatively, we can analyze the behavior of g(x).As x approaches 0+, ln x approaches -infty, so g(x) approaches -infty.As x approaches +infty, ln x grows slower than x², so g(x) approaches 0.At x = sqrt(e), we have a critical point. Let's compute g(sqrt(e)):g(sqrt(e)) = (2 ln sqrt(e)) / (sqrt(e))²ln sqrt(e) = (1/2) ln e = 1/2(sqrt(e))² = eSo,g(sqrt(e)) = (2*(1/2)) / e = 1 / eTherefore, the maximum value of g(x) is 1/e, achieved at x = sqrt(e).Therefore, going back to our inequality:-(a + 1) ≥ 1/eSo,a + 1 ≤ -1/eThus,a ≤ -1 - 1/eSo, the range of a is (-∞, -1 - 1/e]Hmm, that seems correct. Let me double-check.We had f(x) + f(-x) = -6(a + 1)x², and the inequality is -6(a + 1)x² ≥ 12 ln x.We divided both sides by 6, getting -(a + 1)x² ≥ 2 ln x.Then, since x² > 0, we can divide both sides by x² without changing the inequality direction:-(a + 1) ≥ (2 ln x)/x²Which is equivalent to:(a + 1) ≤ - (2 ln x)/x²Since this must hold for all x > 0, the maximum of the RHS is 1/e, so a + 1 ≤ -1/e, leading to a ≤ -1 - 1/e.Yes, that seems correct.Now, part (3): If a > 1, let the maximum and minimum values of f(x) on the interval [1, 2] be M(a) and m(a) respectively, and let h(a) = M(a) - m(a). Find the minimum value of h(a).Alright, so we need to analyze the function f(x) on [1, 2] for a > 1, find its maximum and minimum, compute h(a), and then find the minimum of h(a).First, let's recall f(x) = 2x³ - 3(a + 1)x² + 6a x.We can compute its derivative to find critical points:f'(x) = 6x² - 6(a + 1)x + 6aFactor out 6:f'(x) = 6(x² - (a + 1)x + a)Factor the quadratic:x² - (a + 1)x + a = (x - 1)(x - a)So,f'(x) = 6(x - 1)(x - a)Therefore, critical points at x = 1 and x = a.Since a > 1, and our interval is [1, 2], we need to consider where a is in relation to 2.Case 1: a ≤ 2Case 2: a > 2Wait, but since a > 1, it can be either between 1 and 2 or greater than 2.So, let's analyze both cases.First, let's compute f(1) and f(2):f(1) = 2(1)³ - 3(a + 1)(1)² + 6a(1) = 2 - 3(a + 1) + 6a = 2 - 3a - 3 + 6a = 3a - 1f(2) = 2(8) - 3(a + 1)(4) + 6a(2) = 16 - 12(a + 1) + 12a = 16 - 12a - 12 + 12a = 4So, f(1) = 3a - 1, f(2) = 4.Now, let's analyze the critical points.Since f'(x) = 6(x - 1)(x - a), the critical points are at x = 1 and x = a.But on the interval [1, 2], x = a is inside the interval only if a ∈ (1, 2]. If a > 2, then x = a is outside the interval.Therefore, we have two cases:Case 1: 1 < a ≤ 2Case 2: a > 2Let's handle each case.Case 1: 1 < a ≤ 2In this case, the critical points inside [1, 2] are x = 1 and x = a.But x = 1 is the endpoint, so we need to evaluate f at x = 1, x = a, and x = 2.Compute f(a):f(a) = 2a³ - 3(a + 1)a² + 6a*a = 2a³ - 3(a³ + a²) + 6a² = 2a³ - 3a³ - 3a² + 6a² = (-a³) + 3a²So, f(a) = -a³ + 3a²Now, let's determine the behavior of f(x) on [1, 2].Since f'(x) = 6(x - 1)(x - a), and a ∈ (1, 2], the sign of f'(x) changes as follows:- For x ∈ (1, a): (x - 1) > 0, (x - a) < 0, so f'(x) < 0. Thus, f is decreasing on (1, a).- For x ∈ (a, 2): (x - 1) > 0, (x - a) > 0, so f'(x) > 0. Thus, f is increasing on (a, 2).Therefore, on [1, 2], f(x) decreases from x = 1 to x = a, then increases from x = a to x = 2.Therefore, the minimum occurs at x = a, and the maximum occurs at one of the endpoints, either x = 1 or x = 2.We need to compare f(1) and f(2):f(1) = 3a - 1f(2) = 4So, which one is larger?Set 3a - 1 compared to 4.3a - 1 > 4 => 3a > 5 => a > 5/3 ≈ 1.6667So, if a > 5/3, then f(1) > f(2); else, f(1) ≤ f(2).Therefore, within Case 1 (1 < a ≤ 2), we can further divide into:Subcase 1a: 1 < a ≤ 5/3Subcase 1b: 5/3 < a ≤ 2Subcase 1a: 1 < a ≤ 5/3Here, f(1) ≤ f(2), so maximum is at x = 2, which is 4, and minimum is at x = a, which is -a³ + 3a².Thus, h(a) = M(a) - m(a) = 4 - (-a³ + 3a²) = 4 + a³ - 3a²Subcase 1b: 5/3 < a ≤ 2Here, f(1) > f(2), so maximum is at x = 1, which is 3a - 1, and minimum is at x = a, which is -a³ + 3a².Thus, h(a) = M(a) - m(a) = (3a - 1) - (-a³ + 3a²) = 3a - 1 + a³ - 3a² = a³ - 3a² + 3a - 1Case 2: a > 2In this case, the critical point x = a is outside the interval [1, 2]. So, on [1, 2], the derivative f'(x) = 6(x - 1)(x - a). Since a > 2, for x ∈ [1, 2], (x - a) < 0. Therefore, f'(x) = 6(x - 1)(negative). So, for x ∈ (1, 2):- When x ∈ (1, 2), (x - 1) > 0, (x - a) < 0, so f'(x) < 0.Thus, f(x) is decreasing on [1, 2].Therefore, maximum at x = 1, which is f(1) = 3a - 1, and minimum at x = 2, which is f(2) = 4.Thus, h(a) = M(a) - m(a) = (3a - 1) - 4 = 3a - 5So, summarizing:- For 1 < a ≤ 5/3: h(a) = a³ - 3a² + 4- For 5/3 < a ≤ 2: h(a) = a³ - 3a² + 3a - 1- For a > 2: h(a) = 3a - 5Now, we need to find the minimum value of h(a) over a > 1.So, let's analyze each interval.First, for 1 < a ≤ 5/3: h(a) = a³ - 3a² + 4Compute its derivative to find minima:h'(a) = 3a² - 6aSet h'(a) = 0:3a² - 6a = 0 => 3a(a - 2) = 0 => a = 0 or a = 2But in this interval, a ∈ (1, 5/3], so the critical point at a = 2 is outside this interval.Thus, in (1, 5/3], h'(a) = 3a² - 6a. Let's evaluate h'(a) at a = 5/3:h'(5/3) = 3*(25/9) - 6*(5/3) = (75/9) - (30/3) = (25/3) - 10 = (25/3 - 30/3) = (-5/3) < 0So, h(a) is decreasing on (1, 5/3]. Therefore, its minimum in this interval is at a = 5/3.Compute h(5/3):h(5/3) = (5/3)³ - 3*(5/3)² + 4Compute each term:(5/3)³ = 125 / 273*(5/3)² = 3*(25/9) = 75/9 = 25/3So,h(5/3) = 125/27 - 25/3 + 4Convert all to 27 denominators:125/27 - 225/27 + 108/27 = (125 - 225 + 108)/27 = (8)/27So, h(5/3) = 8/27 ≈ 0.296Now, moving to the next interval: 5/3 < a ≤ 2, h(a) = a³ - 3a² + 3a - 1Compute its derivative:h'(a) = 3a² - 6a + 3 = 3(a² - 2a + 1) = 3(a - 1)²Since (a - 1)² is always non-negative, h'(a) ≥ 0 for all a. So, h(a) is increasing on (5/3, 2].Therefore, the minimum in this interval is at a = 5/3.Compute h(5/3):h(5/3) = (5/3)³ - 3*(5/3)² + 3*(5/3) - 1Compute each term:(5/3)³ = 125/273*(5/3)² = 3*(25/9) = 75/9 = 25/33*(5/3) = 5So,h(5/3) = 125/27 - 25/3 + 5 - 1Convert all to 27 denominators:125/27 - 225/27 + 135/27 - 27/27 = (125 - 225 + 135 - 27)/27 = (125 + 135 - 225 - 27)/27 = (260 - 252)/27 = 8/27Same as before, h(5/3) = 8/27Now, for a > 2, h(a) = 3a - 5This is a linear function with slope 3, so it's increasing for a > 2. Therefore, its minimum occurs at a = 2.Compute h(2):h(2) = 3*2 - 5 = 6 - 5 = 1So, comparing the minimums from each interval:- For 1 < a ≤ 5/3: minimum h(a) = 8/27 at a = 5/3- For 5/3 < a ≤ 2: minimum h(a) = 8/27 at a = 5/3- For a > 2: minimum h(a) = 1 at a = 2Therefore, the overall minimum of h(a) is 8/27.So, the minimum value of h(a) is 8/27.**Final Answer**(1) boxed{dfrac{1}{2}}(2) boxed{(-infty, -1 - dfrac{1}{e}]}(3) boxed{dfrac{8}{27}}

answer：Okay, so I have this problem here: I need to calculate the sum of the highest powers of 12 and 18 that divide 20!. The options are A) 10, B) 12, C) 14, D) 16. Hmm, let me try to figure this out step by step.First, I remember that to find the highest power of a number that divides a factorial like 20!, I need to use Legendre's formula. This formula helps determine the exponent of a prime number in the prime factorization of a factorial. So, I think I should start by breaking down 12 and 18 into their prime factors.Alright, 12 can be factored into 2^2 * 3^1, and 18 can be factored into 2^1 * 3^2. So, both 12 and 18 are composed of the primes 2 and 3. That means I need to find out how many times 2 and 3 appear in the prime factorization of 20!.Let me recall how Legendre's formula works. For a prime number p and a factorial n!, the exponent of p in n! is given by the sum of the floor division of n by p, n by p^2, n by p^3, and so on, until the division results in zero. So, I need to apply this formula for both primes 2 and 3 in 20!.Starting with prime 2:- 20 divided by 2 is 10.- 20 divided by 4 (which is 2^2) is 5.- 20 divided by 8 (which is 2^3) is 2.- 20 divided by 16 (which is 2^4) is 1.- 20 divided by 32 (which is 2^5) is 0, so we can stop here.Adding these up: 10 + 5 + 2 + 1 = 18. So, there are 18 factors of 2 in 20!.Now, for prime 3:- 20 divided by 3 is approximately 6.666, but we take the floor, so 6.- 20 divided by 9 (which is 3^2) is approximately 2.222, so floor is 2.- 20 divided by 27 (which is 3^3) is less than 1, so 0. We can stop here.Adding these up: 6 + 2 = 8. So, there are 8 factors of 3 in 20!.Alright, now that I have the exponents for 2 and 3 in 20!, I can figure out the highest powers of 12 and 18 that divide 20!.Starting with 12, which is 2^2 * 3^1. To find the highest power of 12 that divides 20!, I need to see how many times 12 can be formed from the available factors of 2 and 3.Each 12 requires 2^2 and 3^1. So, the number of times we can form 12 is limited by the number of pairs of 2s and the number of 3s.We have 18 factors of 2, so the number of 2^2 pairs is 18 divided by 2, which is 9. We have 8 factors of 3, so the number of 3^1 is 8.The limiting factor here is the number of 3s, which is 8. So, the highest power of 12 that divides 20! is 12^8.Now, moving on to 18, which is 2^1 * 3^2. Similarly, I need to find how many times 18 can be formed from the available factors of 2 and 3.Each 18 requires 2^1 and 3^2. So, the number of times we can form 18 is limited by the number of single 2s and the number of pairs of 3s.We have 18 factors of 2, so the number of 2^1 is 18. We have 8 factors of 3, so the number of 3^2 pairs is 8 divided by 2, which is 4.The limiting factor here is the number of 3^2 pairs, which is 4. So, the highest power of 18 that divides 20! is 18^4.Now, the problem asks for the sum of these highest powers. So, I need to add the exponents from 12^8 and 18^4.8 (from 12^8) + 4 (from 18^4) = 12.Looking back at the options, B) 12 is the correct answer.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For prime 2 in 20!:- 20 / 2 = 10- 20 / 4 = 5- 20 / 8 = 2- 20 / 16 = 1Total: 10 + 5 + 2 + 1 = 18. That seems correct.For prime 3 in 20!:- 20 / 3 = 6- 20 / 9 = 2Total: 6 + 2 = 8. That also seems correct.For 12^8:- Each 12 needs 2^2 and 3^1.- 18 factors of 2 can make 9 pairs of 2^2.- 8 factors of 3 can make 8 single 3s.- Minimum of 9 and 8 is 8. So, 12^8 is correct.For 18^4:- Each 18 needs 2^1 and 3^2.- 18 factors of 2 can make 18 single 2s.- 8 factors of 3 can make 4 pairs of 3^2.- Minimum of 18 and 4 is 4. So, 18^4 is correct.Sum of exponents: 8 + 4 = 12. Yep, that's correct.I think I've got it right. The sum of the highest powers of 12 and 18 that divide 20! is 12.

answer：Alright, so I have this equation to solve: [x] × {x} + x = 2{x} + 9. Hmm, okay. I remember that [x] is the integer part of x, and {x} is the fractional part. So, for example, if x is 3.14, then [x] is 3 and {x} is 0.14. Got it.Let me try to break this down. Maybe I can express x in terms of its integer and fractional parts. Let's say x = a + b, where a is the integer part [x] and b is the fractional part {x}. So, a is an integer, and b is between 0 (inclusive) and 1 (exclusive).Substituting x = a + b into the equation, we have:a × b + (a + b) = 2b + 9Let me simplify this step by step. First, distribute the a in the first term:a*b + a + b = 2b + 9Now, let's combine like terms. I see a*b, a, b on the left side, and 2b on the right side. Let me subtract 2b from both sides to get all terms on one side:a*b + a + b - 2b = 9Simplify the b terms:a*b + a - b = 9Hmm, okay. So, a*b + a - b = 9. Maybe I can factor something out here. Let's see, I can factor out a from the first two terms:a*(b + 1) - b = 9That looks a bit better. Now, maybe I can isolate terms involving b. Let me rearrange:a*(b + 1) = b + 9Hmm, interesting. So, a*(b + 1) = b + 9. Let's try to solve for b. Maybe I can express b in terms of a.First, expand the left side:a*b + a = b + 9Now, let's get all terms involving b on one side and constants on the other:a*b - b = 9 - aFactor out b on the left side:b*(a - 1) = 9 - aSo, b = (9 - a)/(a - 1)Wait, that seems a bit messy. Let me double-check my steps to make sure I didn't make a mistake.Starting from:a*(b + 1) = b + 9Expanding:a*b + a = b + 9Subtract b from both sides:a*b + a - b = 9Factor out b:b*(a - 1) + a = 9Wait, maybe I should rearrange differently. Let's try:From a*b + a - b = 9Bring the a to the other side:a*b - b = 9 - aFactor out b:b*(a - 1) = 9 - aSo, b = (9 - a)/(a - 1)Yes, that's correct. Now, since b is the fractional part of x, it must satisfy 0 ≤ b < 1. Also, a is the integer part, so a is an integer.So, b = (9 - a)/(a - 1) must be between 0 and 1.Let me write that as:0 ≤ (9 - a)/(a - 1) < 1Now, I need to find integer values of a such that this inequality holds.Let me consider the expression (9 - a)/(a - 1). Let's denote this as f(a) = (9 - a)/(a - 1)I need to find integer a where f(a) is between 0 and 1.First, let's analyze f(a):f(a) = (9 - a)/(a - 1)Let me simplify this expression:f(a) = -(a - 9)/(a - 1) = -(a - 1 - 8)/(a - 1) = -1 + 8/(a - 1)Wait, that might not be helpful. Alternatively, let's consider the sign of the numerator and denominator.Since a is an integer, and a - 1 is in the denominator, a cannot be 1, because that would make the denominator zero, which is undefined.So, a ≠ 1.Now, let's consider the possible values of a.Since b must be less than 1, and b = (9 - a)/(a - 1), let's see for which a this fraction is positive and less than 1.First, let's find when (9 - a)/(a - 1) is positive.The fraction is positive when numerator and denominator have the same sign.Case 1: Both numerator and denominator positive.So, 9 - a > 0 and a - 1 > 0Which implies:9 - a > 0 ⇒ a < 9a - 1 > 0 ⇒ a > 1So, in this case, a is an integer greater than 1 and less than 9. So, a can be 2, 3, 4, 5, 6, 7, 8.Case 2: Both numerator and denominator negative.So, 9 - a < 0 and a - 1 < 0Which implies:9 - a < 0 ⇒ a > 9a - 1 < 0 ⇒ a < 1But a cannot be both greater than 9 and less than 1 at the same time. So, this case is impossible.Therefore, the only possible values for a are integers from 2 to 8.Now, let's check for each a from 2 to 8 whether b = (9 - a)/(a - 1) is less than 1.Compute b for each a:a = 2:b = (9 - 2)/(2 - 1) = 7/1 = 7But b must be less than 1, so this is invalid.a = 3:b = (9 - 3)/(3 - 1) = 6/2 = 3Again, b = 3 is not less than 1. Invalid.a = 4:b = (9 - 4)/(4 - 1) = 5/3 ≈ 1.666...Still greater than 1. Invalid.a = 5:b = (9 - 5)/(5 - 1) = 4/4 = 1But b must be less than 1, so this is invalid.a = 6:b = (9 - 6)/(6 - 1) = 3/5 = 0.6Okay, this is less than 1. Valid.a = 7:b = (9 - 7)/(7 - 1) = 2/6 ≈ 0.333...Also less than 1. Valid.a = 8:b = (9 - 8)/(8 - 1) = 1/7 ≈ 0.142...Less than 1. Valid.a = 9:Wait, earlier we considered a < 9, but let's check a = 9 just in case.b = (9 - 9)/(9 - 1) = 0/8 = 0Which is valid since b = 0 is allowed (as fractional part can be zero).So, a = 9 is also valid.Wait, but earlier when I considered a from 2 to 8, I didn't include a = 9. But since a = 9 gives b = 0, which is valid, we should include it.So, the possible values of a are 6, 7, 8, 9.Now, let's compute x for each valid a:For a = 6:b = 3/5 = 0.6So, x = a + b = 6 + 0.6 = 6.6For a = 7:b = 2/6 = 1/3 ≈ 0.333...So, x = 7 + 1/3 ≈ 7.333...For a = 8:b = 1/7 ≈ 0.142...So, x = 8 + 1/7 ≈ 8.142...For a = 9:b = 0So, x = 9 + 0 = 9Now, let's verify each solution in the original equation to make sure they work.First, x = 6.6:[x] = 6, {x} = 0.6Left side: [x] × {x} + x = 6 × 0.6 + 6.6 = 3.6 + 6.6 = 10.2Right side: 2{x} + 9 = 2 × 0.6 + 9 = 1.2 + 9 = 10.2They are equal, so x = 6.6 is a solution.Next, x ≈ 7.333... (which is 7 + 1/3):[x] = 7, {x} = 1/3Left side: 7 × (1/3) + 7 + 1/3 = 7/3 + 22/3 = 29/3 ≈ 9.666...Right side: 2 × (1/3) + 9 = 2/3 + 9 = 29/3 ≈ 9.666...They are equal, so x = 7 + 1/3 is a solution.Next, x ≈ 8.142... (which is 8 + 1/7):[x] = 8, {x} = 1/7Left side: 8 × (1/7) + 8 + 1/7 = 8/7 + 57/7 = 65/7 ≈ 9.285...Right side: 2 × (1/7) + 9 = 2/7 + 63/7 = 65/7 ≈ 9.285...They are equal, so x = 8 + 1/7 is a solution.Finally, x = 9:[x] = 9, {x} = 0Left side: 9 × 0 + 9 = 0 + 9 = 9Right side: 2 × 0 + 9 = 0 + 9 = 9They are equal, so x = 9 is a solution.Therefore, the solutions are x = 6.6, x ≈ 7.333..., x ≈ 8.142..., and x = 9.But let me express them as exact fractions for clarity:x = 6 + 3/5 = 6.6x = 7 + 1/3 ≈ 7.333...x = 8 + 1/7 ≈ 8.142...x = 9So, in boxed form, the solutions are:x = 6 3/5, x = 7 1/3, x = 8 1/7, and x = 9.Wait, but earlier when I considered a = 9, b = 0, which is valid, but when I checked a = 9, b = 0, and x = 9, it worked. So, that's another solution.But in my initial analysis, I thought a could be from 2 to 8, but a = 9 also works because b = 0 is allowed.So, the complete set of solutions is x = 6 3/5, x = 7 1/3, x = 8 1/7, and x = 9.I think that's all. Let me just double-check if there are any other possible a values.Wait, when a = 10:b = (9 - 10)/(10 - 1) = (-1)/9 ≈ -0.111...But b cannot be negative, so a = 10 is invalid.Similarly, a = 0:b = (9 - 0)/(0 - 1) = 9/(-1) = -9Negative, invalid.a = -1:b = (9 - (-1))/(-1 - 1) = 10/(-2) = -5Negative, invalid.So, no other integer a values will give b in [0,1).Therefore, the solutions are x = 6 3/5, x = 7 1/3, x = 8 1/7, and x = 9.