answer：Okay, so I need to find the maximum value of the function f(x) = |x| - |x - 3|. Hmm, absolute value functions can sometimes be tricky because they have different expressions depending on whether the inside is positive or negative. Maybe I should break this down into different cases based on the value of x.First, let me recall that the absolute value function |a| is equal to a if a is non-negative, and -a if a is negative. So, for |x|, it will be x when x is greater than or equal to 0, and -x when x is less than 0. Similarly, |x - 3| will be x - 3 when x - 3 is non-negative (so x >= 3), and -(x - 3) when x - 3 is negative (so x < 3).Therefore, I can split the function f(x) into different cases based on the value of x. The critical points where the expressions inside the absolute values change their signs are at x = 0 and x = 3. So, I can divide the real number line into three intervals:1. x < 02. 0 <= x < 33. x >= 3I'll analyze each interval separately.**Case 1: x < 0**In this interval, x is negative, so |x| = -x. Also, since x < 0, x - 3 is definitely negative (because 3 is positive and x is negative, so subtracting 3 makes it even more negative). Therefore, |x - 3| = -(x - 3) = -x + 3.So, f(x) = |x| - |x - 3| becomes:f(x) = (-x) - (-x + 3) = -x + x - 3 = -3.So, in this interval, f(x) is constant at -3.**Case 2: 0 <= x < 3**In this interval, x is non-negative, so |x| = x. However, x - 3 is still negative because x is less than 3. Therefore, |x - 3| = -(x - 3) = -x + 3.So, f(x) = |x| - |x - 3| becomes:f(x) = x - (-x + 3) = x + x - 3 = 2x - 3.This is a linear function with a slope of 2, so it's increasing as x increases. Therefore, the minimum value in this interval occurs at x = 0, and the maximum occurs at x approaching 3.Let me compute the value at x = 0:f(0) = 2*0 - 3 = -3.And as x approaches 3 from the left, f(x) approaches 2*3 - 3 = 6 - 3 = 3.So, in this interval, f(x) ranges from -3 to 3.**Case 3: x >= 3**In this interval, x is non-negative, so |x| = x. Also, x - 3 is non-negative because x is at least 3. Therefore, |x - 3| = x - 3.So, f(x) = |x| - |x - 3| becomes:f(x) = x - (x - 3) = x - x + 3 = 3.Therefore, in this interval, f(x) is constant at 3.**Putting it all together:**- For x < 0, f(x) = -3.- For 0 <= x < 3, f(x) increases from -3 to 3.- For x >= 3, f(x) = 3.So, the function starts at -3 when x is negative, increases linearly to 3 as x approaches 3, and then remains at 3 for all x greater than or equal to 3.Therefore, the maximum value of f(x) is 3. It occurs for all x >= 3.Wait, just to make sure I didn't make a mistake, let me test a value in each interval.For x < 0, say x = -1:f(-1) = |-1| - |-1 - 3| = 1 - | -4 | = 1 - 4 = -3. Correct.For 0 <= x < 3, say x = 1:f(1) = |1| - |1 - 3| = 1 - 2 = -1. Wait, but according to my earlier calculation, f(x) = 2x - 3, so f(1) should be 2*1 - 3 = -1. That's correct.And at x = 2:f(2) = |2| - |2 - 3| = 2 - 1 = 1. According to the formula, 2*2 - 3 = 4 - 3 = 1. Correct.At x approaching 3 from the left, say x = 2.999:f(2.999) = 2*2.999 - 3 ≈ 5.998 - 3 = 2.998, which is approximately 3. Correct.For x >= 3, say x = 4:f(4) = |4| - |4 - 3| = 4 - 1 = 3. Correct.So, everything checks out. The function reaches a maximum value of 3 at x = 3 and beyond.I think that's thorough enough. I considered all intervals, tested the endpoints, and even checked intermediate points to ensure my calculations were correct. So, yeah, the maximum value is definitely 3.**Final Answer**The maximum value of the function is boxed{3}.

answer：To solve the equation (3 in {1, m + 2}), I need to determine the value of (m).First, I recognize that for the number 3 to be an element of the set ({1, m + 2}), it must be equal to one of the elements in the set. Since 3 is not equal to 1, it must be equal to (m + 2).Setting up the equation:[3 = m + 2]To solve for (m), I subtract 2 from both sides of the equation:[3 - 2 = m][1 = m]Therefore, the value of (m) is 1.

answer：First, I will apply the Euclidean algorithm to find the greatest common divisor of 8251 and 6105.I start by dividing 8251 by 6105. The quotient is 1 and the remainder is 2146.Next, I divide 6105 by 2146. The quotient is 2 and the remainder is 1813.Then, I divide 2146 by 1813. The quotient is 1 and the remainder is 333.After that, I divide 1813 by 333. The quotient is 5 and the remainder is 148.I continue by dividing 333 by 148. The quotient is 2 and the remainder is 37.Next, I divide 148 by 37. The quotient is 4 and the remainder is 0.Since the remainder is now 0, the last non-zero remainder is 37, which is the greatest common divisor of 8251 and 6105.

answer：First, I'll distribute the multiplication across the terms inside the parentheses. This means I'll multiply (sqrt{3}) by both (sqrt{3}) and (-2).Next, I'll simplify each term. Multiplying (sqrt{3}) by itself gives (3), and multiplying (-2) by (sqrt{3}) results in (-2sqrt{3}).Then, I'll simplify (sqrt{12}) by recognizing that it can be expressed as (sqrt{4 times 3}), which simplifies to (2sqrt{3}).Finally, I'll combine all the simplified terms: (3 - 2sqrt{3} + 2sqrt{3}). The (-2sqrt{3}) and (2sqrt{3}) cancel each other out, leaving the final result as (3).

answer：First, I need to determine the total number of parts in the given ratio. The ratio of students in the first, second, and third grades is 3:3:4. Adding these together gives 3 + 3 + 4 = 10 parts.Next, I'll calculate the number of students to be sampled from the second grade. Since the ratio for the second grade is 3 parts out of the total 10 parts, the proportion of the sample from the second grade is 3/10. Multiplying this proportion by the total sample size of 50 students gives (3/10) * 50 = 15 students.Therefore, 15 students should be drawn from the second grade.

answer：Alright, so I have this problem where I need to find the value of ( c ), which is the sum of ( f(n) ) from ( n = 1 ) to ( n = 2003 ). The function given is ( f(n) = sin frac{n pi}{4} ). Hmm, okay. Let me break this down step by step.First, I know that the sine function is periodic, so maybe there's a repeating pattern in the values of ( f(n) ). Since the argument is ( frac{n pi}{4} ), the period of this sine function should be ( frac{2pi}{pi/4} = 8 ). So, every 8 terms, the values of ( f(n) ) should repeat. That means the sequence of ( f(n) ) is periodic with period 8.Let me write out the values of ( f(n) ) for ( n = 1 ) to ( n = 8 ) to see the pattern.- ( f(1) = sin frac{pi}{4} = frac{sqrt{2}}{2} )- ( f(2) = sin frac{2pi}{4} = sin frac{pi}{2} = 1 )- ( f(3) = sin frac{3pi}{4} = frac{sqrt{2}}{2} )- ( f(4) = sin pi = 0 )- ( f(5) = sin frac{5pi}{4} = -frac{sqrt{2}}{2} )- ( f(6) = sin frac{6pi}{4} = sin frac{3pi}{2} = -1 )- ( f(7) = sin frac{7pi}{4} = -frac{sqrt{2}}{2} )- ( f(8) = sin 2pi = 0 )Okay, so the sequence from ( n = 1 ) to ( n = 8 ) is: ( frac{sqrt{2}}{2}, 1, frac{sqrt{2}}{2}, 0, -frac{sqrt{2}}{2}, -1, -frac{sqrt{2}}{2}, 0 ).Now, let's compute the sum of one period, which is 8 terms.Sum = ( frac{sqrt{2}}{2} + 1 + frac{sqrt{2}}{2} + 0 - frac{sqrt{2}}{2} - 1 - frac{sqrt{2}}{2} + 0 )Let me compute term by term:1. ( frac{sqrt{2}}{2} ) ≈ 0.70712. ( +1 ) ≈ 1.70713. ( + frac{sqrt{2}}{2} ) ≈ 1.7071 + 0.7071 ≈ 2.41424. ( +0 ) remains 2.41425. ( - frac{sqrt{2}}{2} ) ≈ 2.4142 - 0.7071 ≈ 1.70716. ( -1 ) ≈ 1.7071 - 1 ≈ 0.70717. ( - frac{sqrt{2}}{2} ) ≈ 0.7071 - 0.7071 ≈ 08. ( +0 ) remains 0So, the sum over one period is 0. Interesting! That means every 8 terms, the sum cancels out.Therefore, if I can figure out how many complete periods are in 2003 terms, and then add the sum of the remaining terms, I can find the total sum ( c ).Let me compute how many complete periods are in 2003 terms. Since each period is 8 terms, I can divide 2003 by 8.2003 ÷ 8 = 250 with a remainder. Let me compute 250 × 8 = 2000. So, 2003 - 2000 = 3. So, there are 250 complete periods and 3 additional terms.Since each complete period sums to 0, the total sum ( c ) is just the sum of the first 3 terms of the next period.So, I need to compute ( f(2001) + f(2002) + f(2003) ).But wait, since the function is periodic with period 8, ( f(n) = f(n + 8k) ) for any integer ( k ). Therefore, ( f(2001) = f(2001 mod 8) ), ( f(2002) = f(2002 mod 8) ), and ( f(2003) = f(2003 mod 8) ).Let me compute 2001 mod 8:2000 is divisible by 8, so 2000 mod 8 = 0. Therefore, 2001 mod 8 = 1, 2002 mod 8 = 2, 2003 mod 8 = 3.So, ( f(2001) = f(1) = frac{sqrt{2}}{2} ), ( f(2002) = f(2) = 1 ), ( f(2003) = f(3) = frac{sqrt{2}}{2} ).Therefore, the sum of the remaining 3 terms is ( frac{sqrt{2}}{2} + 1 + frac{sqrt{2}}{2} ).Let me compute that:( frac{sqrt{2}}{2} + frac{sqrt{2}}{2} = sqrt{2} ), so total sum is ( sqrt{2} + 1 ).Wait, but earlier when I summed one period, the sum was 0. So, the total sum is just the sum of the first 3 terms, which is ( sqrt{2} + 1 ).But let me double-check my calculations.First, 2003 divided by 8: 8 × 250 = 2000, so 2003 = 8 × 250 + 3. So, 250 complete periods, each summing to 0, plus 3 terms.Yes, so the sum is ( f(1) + f(2) + f(3) ). Which is ( frac{sqrt{2}}{2} + 1 + frac{sqrt{2}}{2} ). That adds up to ( 1 + sqrt{2} ).Wait, but let me verify the individual terms again:( f(1) = sin frac{pi}{4} = frac{sqrt{2}}{2} )( f(2) = sin frac{pi}{2} = 1 )( f(3) = sin frac{3pi}{4} = frac{sqrt{2}}{2} )So, yes, adding them together: ( frac{sqrt{2}}{2} + 1 + frac{sqrt{2}}{2} = 1 + sqrt{2} ).Therefore, the total sum ( c = 1 + sqrt{2} ).But wait, let me think again. Is that correct? Because sometimes when dealing with periodic functions, especially sine, the sum over a period is zero, but sometimes there are phase shifts or other considerations.But in this case, since the function is purely sinusoidal with no phase shift, and we're summing over integer multiples, the periodicity should hold as I calculated.Alternatively, maybe I can think of it as a telescoping series or use complex exponentials, but that might complicate things.Alternatively, perhaps I can use the formula for the sum of sines with arithmetic sequence arguments.I recall that the sum ( sum_{k=1}^{n} sin(k theta) ) can be expressed using the formula:( sum_{k=1}^{n} sin(k theta) = frac{sinleft(frac{n theta}{2}right) cdot sinleft(frac{(n + 1) theta}{2}right)}{sinleft(frac{theta}{2}right)} )Let me verify this formula.Yes, the formula is correct. It's derived from the imaginary part of the sum of complex exponentials.So, in this case, ( theta = frac{pi}{4} ), and ( n = 2003 ).Therefore, the sum ( c = sum_{k=1}^{2003} sinleft( frac{k pi}{4} right) ) can be written as:( c = frac{sinleft( frac{2003 cdot frac{pi}{4}}{2} right) cdot sinleft( frac{(2003 + 1) cdot frac{pi}{4}}{2} right)}{sinleft( frac{frac{pi}{4}}{2} right)} )Simplify this expression step by step.First, compute the denominator:( sinleft( frac{pi}{8} right) ). That's a known value, approximately 0.38268, but we can keep it as ( sin(pi/8) ).Now, compute the arguments of the numerator sines:First sine term:( frac{2003 cdot frac{pi}{4}}{2} = frac{2003 pi}{8} )Second sine term:( frac{2004 cdot frac{pi}{4}}{2} = frac{2004 pi}{8} = frac{501 pi}{2} )So, the numerator is ( sinleft( frac{2003 pi}{8} right) cdot sinleft( frac{501 pi}{2} right) ).Let me simplify these sine terms.First, ( sinleft( frac{501 pi}{2} right) ).Note that ( frac{501}{2} = 250.5 ), so ( 250.5 pi ). Since sine has a period of ( 2pi ), we can subtract multiples of ( 2pi ) to find an equivalent angle between 0 and ( 2pi ).Compute ( 250.5 pi div 2pi = 125.25 ). So, 125 full periods, and 0.25 of a period remaining.0.25 of ( 2pi ) is ( frac{pi}{2} ). So, ( sin(250.5 pi) = sinleft( frac{pi}{2} right) = 1 ).Wait, but actually, ( sin(250.5 pi) ). Let me think again.Wait, ( 250.5 pi = 250pi + 0.5pi ). Since ( sin(250pi + 0.5pi) ). But ( 250pi ) is an integer multiple of ( pi ), so ( sin(250pi + x) = sin(x) ) if 250 is even, or ( sin(pi + x) = -sin(x) ) if 250 is odd.Wait, 250 is even, so ( sin(250pi + 0.5pi) = sin(0.5pi) = 1 ). So, yes, ( sinleft( frac{501 pi}{2} right) = 1 ).Now, the first sine term: ( sinleft( frac{2003 pi}{8} right) ).Let me compute ( frac{2003}{8} ). 8 × 250 = 2000, so ( frac{2003}{8} = 250.375 ). So, ( 250.375 pi ).Again, subtract multiples of ( 2pi ) to find the equivalent angle.Compute ( 250.375 pi div 2pi = 125.1875 ). So, 125 full periods, and 0.1875 of a period remaining.0.1875 of ( 2pi ) is ( 0.1875 × 2pi = 0.375 pi = frac{3pi}{8} ).Therefore, ( sinleft( 250.375 pi right) = sinleft( frac{3pi}{8} right) ).Since ( sinleft( frac{3pi}{8} right) ) is a positive value, approximately 0.9239, but we can keep it as ( sinleft( frac{3pi}{8} right) ).So, putting it all together, the numerator is ( sinleft( frac{3pi}{8} right) times 1 = sinleft( frac{3pi}{8} right) ).Therefore, the entire sum ( c ) is:( c = frac{sinleft( frac{3pi}{8} right)}{sinleft( frac{pi}{8} right)} )Hmm, so now I need to compute ( frac{sinleft( frac{3pi}{8} right)}{sinleft( frac{pi}{8} right)} ).I remember that ( sin(3theta) = 3sintheta - 4sin^3theta ), but that might not be directly helpful here.Alternatively, perhaps using the identity for sine of multiple angles or complementary angles.Wait, ( frac{3pi}{8} = frac{pi}{2} - frac{pi}{8} ). So, ( sinleft( frac{3pi}{8} right) = sinleft( frac{pi}{2} - frac{pi}{8} right) = cosleft( frac{pi}{8} right) ).Yes, that's correct. So, ( sinleft( frac{3pi}{8} right) = cosleft( frac{pi}{8} right) ).Therefore, ( c = frac{cosleft( frac{pi}{8} right)}{sinleft( frac{pi}{8} right)} = cotleft( frac{pi}{8} right) ).Hmm, ( cot(pi/8) ) is a known value. Let me recall that ( cot(pi/8) = sqrt{2} + 1 ). Let me verify that.Yes, because ( tan(pi/8) = sqrt{2} - 1 ), so ( cot(pi/8) = 1 / (sqrt{2} - 1) ). Rationalizing the denominator:( frac{1}{sqrt{2} - 1} times frac{sqrt{2} + 1}{sqrt{2} + 1} = frac{sqrt{2} + 1}{2 - 1} = sqrt{2} + 1 ).Therefore, ( cot(pi/8) = sqrt{2} + 1 ).So, the sum ( c = sqrt{2} + 1 ).Wait, that's the same result as before when I considered the periodicity and added the first three terms. So, both methods give the same answer, which is reassuring.Therefore, I can be confident that the value of ( c ) is ( 1 + sqrt{2} ).But just to make sure, let me recap:1. The function ( f(n) = sin(npi/4) ) has a period of 8.2. Summing over one period gives 0.3. 2003 divided by 8 is 250 with a remainder of 3.4. Therefore, the sum is equal to the sum of the first 3 terms of the period.5. The first three terms are ( frac{sqrt{2}}{2}, 1, frac{sqrt{2}}{2} ), which add up to ( 1 + sqrt{2} ).6. Alternatively, using the sum formula for sine series, I arrived at the same result.So, both approaches confirm that ( c = 1 + sqrt{2} ).Just to be thorough, let me compute the sum numerically to check.Compute ( 1 + sqrt{2} approx 1 + 1.4142 = 2.4142 ).Now, compute the sum of the first 3 terms:( f(1) + f(2) + f(3) = frac{sqrt{2}}{2} + 1 + frac{sqrt{2}}{2} approx 0.7071 + 1 + 0.7071 = 2.4142 ). So, that's consistent.Alternatively, compute the sum using the formula:( c = frac{sin(2003 pi /8) cdot sin(501 pi /2)}{sin(pi /8)} ).We found that ( sin(501 pi /2) = 1 ), ( sin(2003 pi /8) = sin(3pi /8) approx 0.9239 ), and ( sin(pi /8) approx 0.3827 ).So, ( c approx 0.9239 / 0.3827 ≈ 2.4142 ), which is approximately ( 1 + sqrt{2} ).Therefore, all methods confirm that the sum is ( 1 + sqrt{2} ).I think that's solid. I don't see any mistakes in my reasoning, and both approaches lead to the same result. So, I'm confident that ( c = 1 + sqrt{2} ).**Final Answer**The value of ( c ) is boxed{1 + sqrt{2}}.