answer：Okay, so I have this problem about a triangle ABC, and it says that the sum of two absolute values is zero: |2 sin A - 1| + |√2/2 - cos B| = 0. Hmm, I need to find angle C. Alright, let me think about how to approach this.First, I remember that absolute values are always non-negative, right? So if the sum of two absolute values is zero, each of them must individually be zero. That makes sense because if one was positive and the other was zero, the sum would still be positive, which isn't the case here. So, that means both |2 sin A - 1| and |√2/2 - cos B| must equal zero.Alright, so let me write that down:1. |2 sin A - 1| = 02. |√2/2 - cos B| = 0Starting with the first equation: |2 sin A - 1| = 0. If the absolute value is zero, then the expression inside must be zero. So, 2 sin A - 1 = 0. Let's solve for sin A:2 sin A - 1 = 0 2 sin A = 1 sin A = 1/2Okay, so sin A is 1/2. I know that sin 30° is 1/2, and also sin 150° is 1/2. But wait, in a triangle, all angles are between 0° and 180°, and the sum of angles is 180°. So, if angle A is 150°, then the other two angles would have to add up to 30°, which might be possible, but let's see what the second equation tells us.Moving on to the second equation: |√2/2 - cos B| = 0. Again, the expression inside the absolute value must be zero:√2/2 - cos B = 0 cos B = √2/2I remember that cos 45° is √2/2, and cos 315° is also √2/2, but in a triangle, angles are between 0° and 180°, so angle B must be 45°. Okay, so angle B is 45°.Now, let's go back to angle A. If angle B is 45°, and angle A is either 30° or 150°, let's see which one makes sense. If angle A were 150°, then angles A and B would already add up to 195°, which is more than 180°, which isn't possible in a triangle. So, angle A must be 30°.Alright, so angle A is 30°, angle B is 45°, and angle C is what we need to find. Since the sum of angles in a triangle is 180°, angle C is:C = 180° - A - B C = 180° - 30° - 45° C = 105°So, angle C is 105°. Let me just double-check my steps to make sure I didn't make a mistake.1. I set each absolute value to zero because their sum is zero.2. Solved for sin A and got 1/2, leading to A being 30° or 150°.3. Solved for cos B and got √2/2, leading to B being 45°.4. Checked that A can't be 150° because then the sum with B would exceed 180°.5. Subtracted A and B from 180° to get C as 105°.Everything seems to add up. I don't think I made any calculation errors. So, I'm confident that angle C is 105 degrees.**Final Answer**The measure of angle ( C ) is boxed{105^circ}.

answer：Okay, so I have this problem where a particle starts at the origin on the coordinate plane and hops along the x-axis. Each hop is one unit either to the positive or negative direction. After 5 hops, the particle ends up at the point (3,0). I need to find out how many different ways the particle can move to reach this position. Hmm, let me think about this step by step.First, the particle starts at (0,0). Each hop is either +1 or -1 on the x-axis. After 5 hops, it's at (3,0). So, the net movement is +3 units to the right. That means the particle must have moved more times to the right than to the left. Specifically, if we let R represent a move to the right and L represent a move to the left, the particle needs to have a net of 3 R's over L's.Let me denote the number of R's as 'r' and the number of L's as 'l'. Since each hop is either R or L, the total number of hops is r + l = 5. The net movement is r - l = 3. So, I have two equations:1. r + l = 52. r - l = 3I can solve these equations to find the values of r and l. Adding both equations:(r + l) + (r - l) = 5 + 32r = 8r = 4Substituting back into the first equation:4 + l = 5l = 1So, the particle must have moved 4 times to the right and 1 time to the left in some order. Now, the question is, how many different sequences of R's and L's can we have with 4 R's and 1 L?This is a combinatorial problem. The number of different sequences is the number of ways to arrange 4 R's and 1 L in 5 hops. This is equivalent to choosing 1 position out of 5 for the L (or equivalently, choosing 4 positions out of 5 for the R's).The formula for combinations is:C(n, k) = frac{n!}{k!(n - k)!}Where 'n' is the total number of items, and 'k' is the number of items to choose.In this case, n = 5 (total hops) and k = 1 (number of L's). So,C(5, 1) = frac{5!}{1!(5 - 1)!} = frac{5!}{1!4!} = frac{120}{1 times 24} = 5Alternatively, if I think about it as choosing positions for R's, it would be:C(5, 4) = frac{5!}{4!(5 - 4)!} = frac{120}{24 times 1} = 5Either way, the number of different sequences is 5. So, there are 5 different ways the particle can move to reach the point (3,0) after 5 hops.Wait a second, let me make sure I didn't miss anything. The particle is allowed to pass through (3,0) more than once, but it only needs to end up there after 5 hops. So, does that affect the number of ways? Hmm, actually, no. Because regardless of how many times it passes through (3,0) during the hops, as long as it ends there after 5 hops, the count remains the same. The key is the net movement and the number of R's and L's, which we've already accounted for.Let me think of an example. Suppose the particle goes R, R, R, R, L. That would end at 3. Another way is R, R, R, L, R. Similarly, R, R, L, R, R; R, L, R, R, R; and L, R, R, R, R. That's 5 different sequences. Yep, that matches the calculation.I think I've covered all the possibilities. So, the answer should be 5 different ways.

answer：To find the coefficient of the ( x^5 ) term in the expansion of ( (1 - x + x^2)(1 + x)^6 ), we can use the distributive property and focus on the terms that will contribute to ( x^5 ).First, we expand ( (1 + x)^6 ) using the binomial theorem:[(1 + x)^6 = sum_{k=0}^{6} binom{6}{k} x^k]We need the coefficients of ( x^5 ), ( x^4 ), and ( x^3 ) from this expansion because multiplying these by the corresponding terms in ( (1 - x + x^2) ) will give us the ( x^5 ) term.The coefficients are:[binom{6}{5} = 6, quad binom{6}{4} = 15, quad binom{6}{3} = 20]Now, we multiply these coefficients by the respective terms in ( (1 - x + x^2) ):[1 cdot binom{6}{5} x^5 + (-1) cdot binom{6}{4} x^5 + 1 cdot binom{6}{3} x^5][= 1 cdot 6x^5 - 1 cdot 15x^5 + 1 cdot 20x^5][= (6 - 15 + 20)x^5][= 11x^5]Therefore, the coefficient of the ( x^5 ) term is 11.

answer：First, I need to determine the total number of children in the classroom based on the given information.I know that one-third of the children are girls, which means two-thirds of the children are boys.Given that there are 30 boys, I can set up an equation where two-thirds of the total number of children equals 30.Let ( C ) represent the total number of children.So, (frac{2}{3} times C = 30).To solve for ( C ), I'll multiply both sides of the equation by the reciprocal of (frac{2}{3}), which is (frac{3}{2}).This gives me ( C = 30 times frac{3}{2} ).Calculating this, ( C = 45 ).Therefore, there are 45 children in the classroom.

answer：Alright, so I've got this equation to solve: |x - 2| = |x - 1| + |x - 3| + |x - 4|. Hmm, absolute value equations can be tricky because of the different cases depending on the value of x. I remember that when dealing with absolute values, it's helpful to break the problem into different intervals based on the critical points where the expressions inside the absolute values change their signs. Looking at the equation, the expressions inside the absolute values are |x - 2|, |x - 1|, |x - 3|, and |x - 4|. Each of these will change their sign at x = 1, x = 2, x = 3, and x = 4 respectively. So, these points divide the real number line into different intervals where the expressions inside the absolute values will either be positive or negative. Therefore, I should consider the following intervals:1. x < 12. 1 ≤ x < 23. 2 ≤ x < 34. 3 ≤ x < 45. x ≥ 4In each of these intervals, I can rewrite the absolute value expressions without the absolute value signs by considering their sign in that interval. Then, I can solve the equation in each interval and check if the solution lies within that interval.Let me start with the first interval: x < 1.**Case 1: x < 1**In this interval, x is less than 1, so:- |x - 2| = 2 - x (since x - 2 is negative)- |x - 1| = 1 - x (since x - 1 is negative)- |x - 3| = 3 - x (since x - 3 is negative)- |x - 4| = 4 - x (since x - 4 is negative)So, substituting these into the equation:2 - x = (1 - x) + (3 - x) + (4 - x)Simplify the right side:1 - x + 3 - x + 4 - x = (1 + 3 + 4) + (-x - x - x) = 8 - 3xSo, the equation becomes:2 - x = 8 - 3xLet me solve for x:2 - x = 8 - 3xAdd 3x to both sides:2 + 2x = 8Subtract 2 from both sides:2x = 6Divide both sides by 2:x = 3Wait, x = 3. But in this case, we're considering x < 1. So, x = 3 doesn't lie in this interval. Therefore, there is no solution in this case.**Case 2: 1 ≤ x < 2**In this interval, x is between 1 and 2, so:- |x - 2| = 2 - x (since x - 2 is negative)- |x - 1| = x - 1 (since x - 1 is non-negative)- |x - 3| = 3 - x (since x - 3 is negative)- |x - 4| = 4 - x (since x - 4 is negative)Substituting into the equation:2 - x = (x - 1) + (3 - x) + (4 - x)Simplify the right side:x - 1 + 3 - x + 4 - x = (x - x - x) + (-1 + 3 + 4) = (-x) + 6So, the equation becomes:2 - x = -x + 6Wait, let's write that out:2 - x = (-x) + 6Hmm, let me solve for x:2 - x = -x + 6Add x to both sides:2 = 6Wait, that can't be right. 2 equals 6? That doesn't make sense. So, this implies that there's no solution in this interval as well.Wait, did I make a mistake in simplifying?Let me double-check the right side:(x - 1) + (3 - x) + (4 - x)= x - 1 + 3 - x + 4 - x= (x - x - x) + (-1 + 3 + 4)= (-x) + 6Yes, that seems correct. So, 2 - x = -x + 6Adding x to both sides:2 = 6Which is a contradiction. Therefore, no solution in this interval either.**Case 3: 2 ≤ x < 3**In this interval, x is between 2 and 3, so:- |x - 2| = x - 2 (since x - 2 is non-negative)- |x - 1| = x - 1 (since x - 1 is non-negative)- |x - 3| = 3 - x (since x - 3 is negative)- |x - 4| = 4 - x (since x - 4 is negative)Substituting into the equation:x - 2 = (x - 1) + (3 - x) + (4 - x)Simplify the right side:x - 1 + 3 - x + 4 - x= (x - x - x) + (-1 + 3 + 4)= (-x) + 6So, the equation becomes:x - 2 = -x + 6Let me solve for x:x - 2 = -x + 6Add x to both sides:2x - 2 = 6Add 2 to both sides:2x = 8Divide both sides by 2:x = 4Wait, x = 4. But in this case, we're considering 2 ≤ x < 3. So, x = 4 doesn't lie in this interval. Therefore, no solution in this case.Wait, that seems odd. Let me double-check the substitution and simplification.Left side: |x - 2| = x - 2Right side: |x - 1| + |x - 3| + |x - 4| = (x - 1) + (3 - x) + (4 - x)Simplify:x - 1 + 3 - x + 4 - x = (x - x - x) + (-1 + 3 + 4) = (-x) + 6So, equation is x - 2 = -x + 6Adding x to both sides: 2x - 2 = 62x = 8x = 4Yes, that's correct. But x = 4 is outside of this interval, so no solution here.Hmm, maybe I made a mistake in setting up the equation. Let me check.Wait, in this interval, 2 ≤ x < 3, so |x - 2| = x - 2, |x - 1| = x - 1, |x - 3| = 3 - x, |x - 4| = 4 - x. So, substitution seems correct.So, equation is x - 2 = (x - 1) + (3 - x) + (4 - x)Simplify: x - 2 = x - 1 + 3 - x + 4 - x= (x - x - x) + (-1 + 3 + 4)= (-x) + 6So, x - 2 = -x + 6Adding x: 2x - 2 = 62x = 8x = 4Yes, same result. So, no solution in this interval.**Case 4: 3 ≤ x < 4**In this interval, x is between 3 and 4, so:- |x - 2| = x - 2 (since x - 2 is positive)- |x - 1| = x - 1 (since x - 1 is positive)- |x - 3| = x - 3 (since x - 3 is non-negative)- |x - 4| = 4 - x (since x - 4 is negative)Substituting into the equation:x - 2 = (x - 1) + (x - 3) + (4 - x)Simplify the right side:x - 1 + x - 3 + 4 - x= (x + x - x) + (-1 - 3 + 4)= x + 0= xSo, the equation becomes:x - 2 = xSubtract x from both sides:-2 = 0Wait, that's not possible. So, again, a contradiction. Therefore, no solution in this interval.Wait, let me verify the simplification.Right side: (x - 1) + (x - 3) + (4 - x)= x - 1 + x - 3 + 4 - x= (x + x - x) + (-1 - 3 + 4)= x + 0= xYes, that's correct. So, x - 2 = x leads to -2 = 0, which is impossible. So, no solution here.**Case 5: x ≥ 4**In this interval, x is greater than or equal to 4, so:- |x - 2| = x - 2 (since x - 2 is positive)- |x - 1| = x - 1 (since x - 1 is positive)- |x - 3| = x - 3 (since x - 3 is positive)- |x - 4| = x - 4 (since x - 4 is non-negative)Substituting into the equation:x - 2 = (x - 1) + (x - 3) + (x - 4)Simplify the right side:x - 1 + x - 3 + x - 4= (x + x + x) + (-1 - 3 - 4)= 3x - 8So, the equation becomes:x - 2 = 3x - 8Let me solve for x:x - 2 = 3x - 8Subtract x from both sides:-2 = 2x - 8Add 8 to both sides:6 = 2xDivide both sides by 2:x = 3But in this case, we're considering x ≥ 4. So, x = 3 doesn't lie in this interval. Therefore, no solution in this interval either.Wait, so after checking all intervals, I didn't find any solutions? That seems strange because the original equation is |x - 2| = |x - 1| + |x - 3| + |x - 4|. Maybe I made a mistake in one of the cases.Let me go back and check each case again.**Rechecking Case 1: x < 1**Equation: 2 - x = (1 - x) + (3 - x) + (4 - x) = 8 - 3xSo, 2 - x = 8 - 3xAdding 3x: 2 + 2x = 82x = 6x = 3But x < 1, so no solution. Correct.**Rechecking Case 2: 1 ≤ x < 2**Equation: 2 - x = (x - 1) + (3 - x) + (4 - x) = (-x) + 6So, 2 - x = -x + 6Adding x: 2 = 6Contradiction. Correct.**Rechecking Case 3: 2 ≤ x < 3**Equation: x - 2 = (x - 1) + (3 - x) + (4 - x) = (-x) + 6So, x - 2 = -x + 6Adding x: 2x - 2 = 62x = 8x = 4But x < 3, so no solution. Correct.**Rechecking Case 4: 3 ≤ x < 4**Equation: x - 2 = (x - 1) + (x - 3) + (4 - x) = xSo, x - 2 = xSubtract x: -2 = 0Contradiction. Correct.**Rechecking Case 5: x ≥ 4**Equation: x - 2 = (x - 1) + (x - 3) + (x - 4) = 3x - 8So, x - 2 = 3x - 8Subtract x: -2 = 2x - 8Add 8: 6 = 2xx = 3But x ≥ 4, so no solution. Correct.Wait, so according to this, there are no solutions? But that can't be right because when I plug in x = 2, let's see:Left side: |2 - 2| = 0Right side: |2 - 1| + |2 - 3| + |2 - 4| = 1 + 1 + 2 = 4So, 0 ≠ 4. So, x = 2 is not a solution.Wait, but earlier in Case 2, I thought x = 2 might be a solution, but it turned out not to be. Hmm.Wait, maybe I missed something. Let me try plugging in x = 2.25, which was mentioned earlier.Left side: |2.25 - 2| = 0.25Right side: |2.25 - 1| + |2.25 - 3| + |2.25 - 4| = 1.25 + 0.75 + 1.75 = 3.75So, 0.25 ≠ 3.75. So, x = 2.25 is not a solution either.Wait, that's confusing. So, according to my calculations, there are no solutions, but the initial thought was that x = 2 and x = 2.25 are solutions. Maybe I made a mistake in the initial approach.Let me try a different method. Maybe graphing both sides of the equation.The left side is |x - 2|, which is a V-shaped graph with vertex at (2, 0).The right side is |x - 1| + |x - 3| + |x - 4|. This is a sum of absolute values, which will also be a piecewise linear function with possible vertices at x = 1, x = 3, and x = 4.Let me analyze the right side function.Define f(x) = |x - 1| + |x - 3| + |x - 4|We can consider the same intervals as before.**For x < 1:**f(x) = (1 - x) + (3 - x) + (4 - x) = 8 - 3x**For 1 ≤ x < 3:**f(x) = (x - 1) + (3 - x) + (4 - x) = (-x) + 6**For 3 ≤ x < 4:**f(x) = (x - 1) + (x - 3) + (4 - x) = x**For x ≥ 4:**f(x) = (x - 1) + (x - 3) + (x - 4) = 3x - 8So, f(x) is a piecewise linear function with different slopes in each interval.Now, let's compare f(x) with |x - 2|.We can plot both functions or analyze their intersections.But since I can't plot here, let me analyze the slopes and see where they might intersect.Left side: |x - 2| has a slope of -1 for x < 2 and +1 for x ≥ 2.Right side f(x):- For x < 1: slope = -3- For 1 ≤ x < 3: slope = -1- For 3 ≤ x < 4: slope = +1- For x ≥ 4: slope = +3So, let's see where the slopes of f(x) and |x - 2| could be equal or where they might intersect.In the interval x < 1:f(x) has slope -3, |x - 2| has slope -1. Since f(x) is steeper, they might intersect somewhere.But earlier, solving for x < 1 gave x = 3, which is outside this interval.In the interval 1 ≤ x < 2:f(x) has slope -1, |x - 2| has slope -1. So, both have the same slope. Therefore, they are parallel in this interval. If they are parallel, they either coincide or don't intersect. Let's see.At x = 1:f(1) = |1 - 1| + |1 - 3| + |1 - 4| = 0 + 2 + 3 = 5Left side: |1 - 2| = 1So, f(1) = 5 > 1At x = 2:f(2) = |2 - 1| + |2 - 3| + |2 - 4| = 1 + 1 + 2 = 4Left side: |2 - 2| = 0So, f(2) = 4 > 0Since both functions have the same slope (-1) in this interval and f(x) is always above |x - 2|, they don't intersect here.In the interval 2 ≤ x < 3:f(x) has slope -1, |x - 2| has slope +1. So, f(x) is decreasing while |x - 2| is increasing. They might intersect somewhere in this interval.Let me set up the equation in this interval:|x - 2| = f(x)Which is:x - 2 = (-x) + 6Wait, that's the same as before. Solving gives x = 4, which is outside this interval.Wait, but in this interval, x is between 2 and 3, so x = 4 is outside. So, no solution here.In the interval 3 ≤ x < 4:f(x) has slope +1, |x - 2| has slope +1. So, both have the same slope. Let's check their values.At x = 3:f(3) = |3 - 1| + |3 - 3| + |3 - 4| = 2 + 0 + 1 = 3Left side: |3 - 2| = 1So, f(3) = 3 > 1At x = 4:f(4) = |4 - 1| + |4 - 3| + |4 - 4| = 3 + 1 + 0 = 4Left side: |4 - 2| = 2So, f(4) = 4 > 2Since both functions have the same slope and f(x) is always above |x - 2| in this interval, they don't intersect here.In the interval x ≥ 4:f(x) has slope +3, |x - 2| has slope +1. So, f(x) is steeper. Let's see if they intersect.Set up the equation:x - 2 = 3x - 8Solving:x - 2 = 3x - 8-2x = -6x = 3But x ≥ 4, so no solution here.Wait, so according to this analysis, there are no solutions? But that contradicts the initial thought that x = 2 and x = 2.25 are solutions. Maybe I made a mistake in the initial approach.Wait, let me try plugging in x = 2 again.Left side: |2 - 2| = 0Right side: |2 - 1| + |2 - 3| + |2 - 4| = 1 + 1 + 2 = 4So, 0 ≠ 4. So, x = 2 is not a solution.Similarly, x = 2.25:Left side: |2.25 - 2| = 0.25Right side: |2.25 - 1| + |2.25 - 3| + |2.25 - 4| = 1.25 + 0.75 + 1.75 = 3.750.25 ≠ 3.75. So, x = 2.25 is not a solution.Wait, so maybe there are no solutions? But that seems odd because the equation is |x - 2| = |x - 1| + |x - 3| + |x - 4|. Let me try plugging in x = 0.Left side: |0 - 2| = 2Right side: |0 - 1| + |0 - 3| + |0 - 4| = 1 + 3 + 4 = 82 ≠ 8x = 5:Left side: |5 - 2| = 3Right side: |5 - 1| + |5 - 3| + |5 - 4| = 4 + 2 + 1 = 73 ≠ 7x = 1.5:Left side: |1.5 - 2| = 0.5Right side: |1.5 - 1| + |1.5 - 3| + |1.5 - 4| = 0.5 + 1.5 + 2.5 = 4.50.5 ≠ 4.5x = 3.5:Left side: |3.5 - 2| = 1.5Right side: |3.5 - 1| + |3.5 - 3| + |3.5 - 4| = 2.5 + 0.5 + 0.5 = 3.51.5 ≠ 3.5Hmm, so in all these test points, the left side is less than the right side. Maybe the equation has no solution because |x - 2| is always less than |x - 1| + |x - 3| + |x - 4| for all real x.But let me think about the behavior as x approaches infinity.As x approaches positive infinity:|x - 2| ≈ x|x - 1| + |x - 3| + |x - 4| ≈ x + x + x = 3xSo, x ≈ 3x, which implies x ≈ 0, but as x approaches infinity, this doesn't hold. So, the left side grows linearly with slope 1, while the right side grows with slope 3. So, for large x, right side is much larger.Similarly, as x approaches negative infinity:|x - 2| ≈ -x + 2|x - 1| + |x - 3| + |x - 4| ≈ (-x + 1) + (-x + 3) + (-x + 4) = -3x + 8So, -x + 2 ≈ -3x + 8Adding 3x: 2x + 2 ≈ 82x ≈ 6x ≈ 3But as x approaches negative infinity, this doesn't hold. So, again, the right side grows faster.Therefore, it's possible that the equation has no solution because the right side is always greater than the left side.Wait, but let me check at x = 2.5:Left side: |2.5 - 2| = 0.5Right side: |2.5 - 1| + |2.5 - 3| + |2.5 - 4| = 1.5 + 0.5 + 1.5 = 3.50.5 < 3.5x = 2.1:Left: 0.1Right: 1.1 + 0.9 + 1.9 = 3.9Still, left < right.x = 2.9:Left: 0.9Right: 1.9 + 0.1 + 1.1 = 3.1Left < right.Wait, so in all these points, left side is less than right side. So, maybe the equation has no solution.But then, why did the initial thought suggest x = 2 and x = 2.25 are solutions? Maybe that was incorrect.Alternatively, perhaps I made a mistake in the initial case analysis.Wait, let me try solving the equation again, but this time, instead of breaking into intervals, maybe I can square both sides to eliminate the absolute values. Although, squaring can introduce extraneous solutions, so I'd have to check.But before that, let me consider that the equation is |x - 2| = |x - 1| + |x - 3| + |x - 4|Let me denote A = |x - 2|, B = |x - 1| + |x - 3| + |x - 4|We can think of B as the sum of distances from x to 1, 3, and 4, while A is the distance from x to 2.So, the equation is asking for points x where the distance to 2 is equal to the sum of distances to 1, 3, and 4.Intuitively, since 1, 3, and 4 are spread out, the sum of distances to them is likely to be larger than the distance to 2, except possibly at certain points.But from the earlier analysis, it seems that B is always greater than A.Wait, let me consider the function f(x) = B - A = |x - 1| + |x - 3| + |x - 4| - |x - 2|We can analyze f(x) to see if it ever equals zero.Let me consider the critical points at x = 1, 2, 3, 4.Compute f(x) at these points:At x = 1:f(1) = 0 + 2 + 3 - 1 = 4At x = 2:f(2) = 1 + 1 + 2 - 0 = 4At x = 3:f(3) = 2 + 0 + 1 - 1 = 2At x = 4:f(4) = 3 + 1 + 0 - 2 = 2So, f(x) is positive at these points.Now, let's check the behavior between these points.Between x = 1 and x = 2:f(x) = (x - 1) + (3 - x) + (4 - x) - (2 - x) = (x - 1 + 3 - x + 4 - x) - (2 - x) = (6 - x) - (2 - x) = 6 - x - 2 + x = 4So, f(x) = 4 in this interval.Between x = 2 and x = 3:f(x) = (x - 1) + (3 - x) + (4 - x) - (x - 2) = (x - 1 + 3 - x + 4 - x) - (x - 2) = (6 - x) - (x - 2) = 6 - x - x + 2 = 8 - 2xSo, f(x) = 8 - 2x in this interval.At x = 2: f(2) = 8 - 4 = 4At x = 3: f(3) = 8 - 6 = 2So, f(x) decreases from 4 to 2 in this interval.Between x = 3 and x = 4:f(x) = (x - 1) + (x - 3) + (4 - x) - (x - 2) = (x - 1 + x - 3 + 4 - x) - (x - 2) = (x - 0) - (x - 2) = x - x + 2 = 2So, f(x) = 2 in this interval.For x < 1:f(x) = (1 - x) + (3 - x) + (4 - x) - (2 - x) = (8 - 3x) - (2 - x) = 8 - 3x - 2 + x = 6 - 2xSo, as x approaches negative infinity, f(x) approaches positive infinity.For x > 4:f(x) = (x - 1) + (x - 3) + (x - 4) - (x - 2) = (3x - 8) - (x - 2) = 2x - 6As x approaches positive infinity, f(x) approaches positive infinity.So, f(x) is always positive. Therefore, the equation f(x) = 0 has no solution.Therefore, the original equation |x - 2| = |x - 1| + |x - 3| + |x - 4| has no real solutions.Wait, but earlier I thought x = 2 and x = 2.25 were solutions, but plugging them in showed they weren't. So, maybe the initial thought was incorrect.Therefore, the number of real solutions is zero.

answer：First, I need to determine the circumference of the circular field by dividing the total fencing cost by the cost per meter. This will give me the total length of the fence required.Next, I'll use the formula for the circumference of a circle, which is ( C = pi times text{diameter} ), to find the diameter. By rearranging the formula, I can solve for the diameter by dividing the circumference by ( pi ).Finally, I'll perform the calculations to find the exact diameter of the field.