answer：First, I need to calculate the total force generated by the Muay Thai fighter's kicks using the polynomial expression ( F(x) = 3x^3 - 5x^2 + 2x - 1 ) when ( x = 4 ). I'll substitute 4 into the equation and perform the arithmetic step by step to ensure accuracy.Next, I'll solve the inequality ( -2y + 15 > y + 3 ) introduced by the BJJ black belt. To find the optimal range of techniques ( y ), I'll isolate ( y ) by performing algebraic operations on both sides of the inequality, making sure to reverse the inequality sign when multiplying or dividing by a negative number.

answer：To determine the total number of symphonies Beethoven could have composed in 5 years, I'll substitute ( x = 5 ) into the polynomial ( S(x) = 2x^2 + 3x + 4 ). Calculating each term step by step will give me the final total.Next, to find the range of years where Mozart's average compositions per year are 60 or fewer, I'll solve the inequality ( frac{600}{x+2} leq 60 ). By manipulating the inequality, I'll isolate ( x ) and determine the valid range of years that satisfy this condition.

answer：Alright, so I have this problem about a geometric sequence. Let me try to parse it step by step. First, part (Ⅰ) says: In a geometric sequence {a_n} where each term is positive, a₁ = 2, and 2a₁, a₃, 3a₂ form an arithmetic sequence. I need to find the general formula for the geometric sequence {a_n}.Okay, so let's recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r. So, in general, a_n = a₁ * r^{n-1}.Given that a₁ = 2, so the first term is 2. Then, a₂ would be 2r, a₃ would be 2r², and so on.Now, the problem says that 2a₁, a₃, 3a₂ form an arithmetic sequence. Hmm, arithmetic sequence means that the difference between consecutive terms is constant. So, the difference between the second and first term should be equal to the difference between the third and second term.Let me write that down. Let me denote the three terms as follows:First term: 2a₁ = 2*2 = 4.Second term: a₃ = 2r².Third term: 3a₂ = 3*(2r) = 6r.So, the three terms are 4, 2r², 6r.Since they form an arithmetic sequence, the difference between the second and first term should be equal to the difference between the third and second term.So, 2r² - 4 = 6r - 2r².Let me write that equation:2r² - 4 = 6r - 2r².Now, let's solve for r.First, bring all terms to one side:2r² - 4 - 6r + 2r² = 0.Combine like terms:(2r² + 2r²) + (-6r) + (-4) = 04r² - 6r - 4 = 0.Hmm, that's a quadratic equation in terms of r. Let me write it as:4r² - 6r - 4 = 0.I can try to simplify this equation. Let's see if I can divide all terms by 2:2r² - 3r - 2 = 0.Okay, that's simpler. Now, let's solve for r using the quadratic formula.The quadratic formula is r = [ -b ± sqrt(b² - 4ac) ] / (2a).Here, a = 2, b = -3, c = -2.So, plugging in:r = [ 3 ± sqrt( (-3)² - 4*2*(-2) ) ] / (2*2)Compute discriminant:D = 9 - 4*2*(-2) = 9 + 16 = 25.So, sqrt(D) = 5.Thus,r = [3 ± 5]/4.So, two possible solutions:1) r = (3 + 5)/4 = 8/4 = 2.2) r = (3 - 5)/4 = (-2)/4 = -0.5.But wait, the problem says each term is positive. So, if r is negative, then the terms would alternate in sign. Since all terms are positive, r must be positive. So, r = 2 is the only valid solution.Therefore, the common ratio r is 2.So, the general formula for the geometric sequence is a_n = a₁ * r^{n-1} = 2 * 2^{n-1} = 2^{n}.Wait, 2 * 2^{n-1} is equal to 2^{1 + n -1} = 2^{n}. Yeah, that's correct.So, part (Ⅰ) is solved, and the general formula is a_n = 2^{n}.Wait, let me double-check. Let me compute 2a₁, a₃, 3a₂ with a₁=2 and r=2.2a₁ = 4.a₃ = 2*2² = 8.3a₂ = 3*(2*2) = 12.So, the three terms are 4, 8, 12.Is that an arithmetic sequence? Let's check the differences:8 - 4 = 4.12 - 8 = 4.Yes, both differences are 4, so it's indeed an arithmetic sequence. So, that checks out.Alright, so part (Ⅰ) is done.Now, moving on to part (Ⅱ): If the sequence {b_n} satisfies b_n = 11 - 2 log₂ a_n, find the maximum value of the sum of the first n terms of {b_n}, denoted as T_n.So, first, let me write down what b_n is.Given that a_n = 2^{n}, so log₂ a_n = log₂ (2^{n}) = n.Therefore, b_n = 11 - 2n.So, the sequence {b_n} is defined by b_n = 11 - 2n.So, now, we need to find T_n, which is the sum of the first n terms of {b_n}, and find its maximum value.First, let's write down T_n.Since {b_n} is an arithmetic sequence, because each term is linear in n.Wait, b_n = 11 - 2n, so each term decreases by 2 each time.So, {b_n} is an arithmetic sequence with first term b₁ = 11 - 2*1 = 9, and common difference d = -2.Therefore, the sum of the first n terms of an arithmetic sequence is given by:T_n = (n/2)*(2b₁ + (n - 1)d).Alternatively, T_n = (n/2)*(b₁ + b_n).Either formula can be used.Let me compute T_n using the first formula.Given b₁ = 9, d = -2.So,T_n = (n/2)*(2*9 + (n - 1)*(-2)).Compute inside the brackets:2*9 = 18.(n - 1)*(-2) = -2n + 2.So, 18 - 2n + 2 = 20 - 2n.Therefore,T_n = (n/2)*(20 - 2n) = (n/2)*(20 - 2n) = n*(10 - n) = 10n - n².So, T_n = -n² + 10n.Alternatively, T_n is a quadratic function in terms of n, which is a parabola opening downward, so it has a maximum at its vertex.Since n is a positive integer, we can find the value of n that maximizes T_n.But let's think about it.Since T_n is a quadratic function in n, T_n = -n² + 10n, which can be rewritten as T_n = -(n² - 10n).Completing the square:T_n = -(n² -10n +25 -25) = -(n -5)^2 +25.So, the maximum value is 25 when n =5.But wait, n has to be a positive integer, so n=5 gives the maximum T_n=25.But let me verify.Alternatively, since T_n is a quadratic function with a maximum at n = -b/(2a) where the quadratic is an^2 + bn + c.In our case, T_n = -n² +10n, so a = -1, b =10.Thus, the vertex is at n = -10/(2*(-1)) = -10/-2 =5.So, n=5 is where the maximum occurs.Since n must be an integer, n=5 is the point where T_n is maximized.Therefore, the maximum value of T_n is T_5.Compute T_5:T_5 = -5² +10*5 = -25 +50 =25.Alternatively, compute T_n as the sum of the first n terms:b₁=9, b₂=7, b₃=5, b₄=3, b₅=1, b₆=-1, etc.So, the terms are decreasing by 2 each time.So, the sum up to n=5 is 9 +7 +5 +3 +1 =25.If we go beyond n=5, the terms become negative, so adding them would decrease the total sum.Therefore, the maximum sum occurs at n=5, and the maximum value is25.Therefore, the answer to part (Ⅱ) is25.Wait, let me just double-check my steps.First, I found that b_n =11 -2n, which is correct because a_n=2^n, so log₂ a_n =n.Then, the sequence {b_n} is 9,7,5,3,1,-1,... So, it's an arithmetic sequence with first term 9 and common difference -2.Sum of first n terms is T_n = (n/2)*(2*9 + (n-1)*(-2)) = (n/2)*(18 -2n +2) = (n/2)*(20 -2n) =n*(10 -n) =10n -n².Which is correct.Then, to find the maximum, since it's a quadratic in n, opening downward, the maximum occurs at vertex n=5.Thus, T_5=25.Yes, that seems correct.Alternatively, since the terms are positive up to n=5 (since b_5=1, and b_6=-1), so the sum is maximized when we include all positive terms, which is up to n=5.So, that makes sense.Therefore, the maximum value of T_n is25.**Final Answer**Ⅰ. The general formula is boxed{a_n = 2^n}.Ⅱ. The maximum value of ( T_n ) is boxed{25}.

answer：Alright, so I have this problem about determining which of the given functions are "conditionally constrained functions." The definition provided is that a function ( f(x) ) with domain ( mathbf{R} ) is conditionally constrained if there exists a constant ( omega > 0 ) such that ( |f(x)| leqslant omega|x| ) for all real numbers ( x ). There are four functions given, and I need to check each one to see if it satisfies this condition. Let me go through them one by one.**1. ( f(x) = 4x )**Okay, starting with the first function. Let's see. The function is linear, and it's straightforward. The absolute value of ( f(x) ) is ( |4x| ), which simplifies to ( 4|x| ). So, comparing this to the condition ( |f(x)| leqslant omega|x| ), we can see that if we choose ( omega = 4 ), the inequality holds for all real numbers ( x ). Therefore, function 1 is a conditionally constrained function.**2. ( f(x) = x^2 + 2 )**Moving on to the second function. This is a quadratic function. Let's compute ( |f(x)| ). Since ( x^2 ) is always non-negative, ( x^2 + 2 ) is always at least 2. Therefore, ( |f(x)| = x^2 + 2 ).Now, we need to see if there's a constant ( omega > 0 ) such that ( x^2 + 2 leqslant omega|x| ) for all real ( x ). Hmm, let's analyze this inequality.Divide both sides by ( |x| ) (assuming ( x neq 0 )):( |x| + frac{2}{|x|} leqslant omega ).Now, let's denote ( t = |x| ), so the inequality becomes:( t + frac{2}{t} leqslant omega ).We need this to hold for all ( t > 0 ). Let's analyze the function ( g(t) = t + frac{2}{t} ).To find the minimum of ( g(t) ), we can take the derivative:( g'(t) = 1 - frac{2}{t^2} ).Setting ( g'(t) = 0 ):( 1 - frac{2}{t^2} = 0 ) ( frac{2}{t^2} = 1 ) ( t^2 = 2 ) ( t = sqrt{2} ).So, the minimum value of ( g(t) ) is at ( t = sqrt{2} ):( g(sqrt{2}) = sqrt{2} + frac{2}{sqrt{2}} = sqrt{2} + sqrt{2} = 2sqrt{2} approx 2.828 ).But as ( t ) approaches 0, ( g(t) ) tends to infinity because ( frac{2}{t} ) becomes very large. Similarly, as ( t ) approaches infinity, ( g(t) ) also tends to infinity because ( t ) becomes very large. This means that ( g(t) ) doesn't have an upper bound; it can be made arbitrarily large by choosing ( t ) close to 0 or very large. Therefore, there's no constant ( omega ) such that ( x^2 + 2 leqslant omega|x| ) for all real ( x ). Hence, function 2 is not a conditionally constrained function.**3. ( f(x) = frac{2x}{x^2 - 2x + 5} )**Third function is a rational function. Let's compute ( |f(x)| ):( |f(x)| = left| frac{2x}{x^2 - 2x + 5} right| ).We need to find if there exists a constant ( omega > 0 ) such that:( left| frac{2x}{x^2 - 2x + 5} right| leqslant omega |x| ) for all real ( x ).Let me rearrange this inequality:( frac{2|x|}{|x^2 - 2x + 5|} leqslant omega |x| ).Assuming ( x neq 0 ), we can divide both sides by ( |x| ):( frac{2}{|x^2 - 2x + 5|} leqslant omega ).So, we need to find the maximum value of ( frac{2}{|x^2 - 2x + 5|} ) over all real ( x ). The maximum of this expression will give us the minimal possible ( omega ).First, let's analyze the denominator ( |x^2 - 2x + 5| ). Since ( x^2 - 2x + 5 ) is a quadratic, let's find its minimum value.The quadratic ( x^2 - 2x + 5 ) can be rewritten by completing the square:( x^2 - 2x + 1 + 4 = (x - 1)^2 + 4 ).Since ( (x - 1)^2 ) is always non-negative, the minimum value of the quadratic is 4, achieved at ( x = 1 ). Therefore, ( |x^2 - 2x + 5| = x^2 - 2x + 5 geq 4 ) for all real ( x ).Therefore, ( frac{2}{|x^2 - 2x + 5|} leq frac{2}{4} = frac{1}{2} ).Thus, the maximum value of ( frac{2}{|x^2 - 2x + 5|} ) is ( frac{1}{2} ), so choosing ( omega = frac{1}{2} ) satisfies the inequality for all real ( x ).Wait, hold on. Let me verify this because I think I might have made a mistake.If ( |x^2 - 2x + 5| geq 4 ), then ( frac{2}{|x^2 - 2x + 5|} leq frac{2}{4} = frac{1}{2} ). So, yes, the maximum value is ( frac{1}{2} ).Therefore, ( |f(x)| leq frac{1}{2}|x| ) for all real ( x ), which means function 3 is a conditionally constrained function.Wait, but let me check for ( x = 0 ). Plugging in ( x = 0 ), ( f(0) = 0 ), so ( |f(0)| = 0 leq omega cdot 0 = 0 ), which holds. For ( x neq 0 ), as we saw, ( |f(x)| leq frac{1}{2}|x| ). So, yes, function 3 is conditionally constrained.**4. ( f(x) ) is an odd function defined on ( mathbf{R} ) and for all ( x_1, x_2 ), ( f(x_1) - f(x_2) leqslant 4 |x_1 - x_2| ).**Fourth function is a bit more abstract. It's an odd function, meaning ( f(-x) = -f(x) ) for all ( x ), and it satisfies the Lipschitz condition with constant 4. That is, ( |f(x_1) - f(x_2)| leq 4 |x_1 - x_2| ) for all ( x_1, x_2 ).We need to determine if ( |f(x)| leq omega |x| ) for some ( omega > 0 ).Since ( f ) is Lipschitz with constant 4, it implies that ( f ) is differentiable almost everywhere with derivative bounded by 4. But more directly, using the Lipschitz condition, we can set ( x_2 = 0 ). Let's see:Let ( x_2 = 0 ), then:( |f(x_1) - f(0)| leq 4 |x_1 - 0| ).But since ( f ) is odd, ( f(0) = 0 ). Therefore,( |f(x_1)| leq 4 |x_1| ).So, setting ( omega = 4 ), we have ( |f(x)| leq 4|x| ) for all real ( x ).Therefore, function 4 is a conditionally constrained function.Wait, let me make sure I didn't skip any steps. The Lipschitz condition with constant 4 ensures that the function doesn't grow faster than linearly with slope 4. Since it's odd, it passes through the origin, so the bound ( |f(x)| leq 4|x| ) is satisfied. Yes, that seems correct.**Summary:**1. Function 1: Yes, ( omega = 4 ).2. Function 2: No, because ( x^2 + 2 ) grows faster than linear.3. Function 3: Yes, ( omega = 1/2 ).4. Function 4: Yes, ( omega = 4 ).So, functions 1, 3, and 4 satisfy the condition.**Final Answer**The conditionally constrained functions are boxed{1}, boxed{3}, and boxed{4}.

answer：First, I need to define the variables for the problem. Let ( x ) represent the number of hours spent tutoring basic algebra, and ( y ) represent the number of hours spent tutoring advanced algebra.The teacher works a total of 20 hours each week, so the sum of ( x ) and ( y ) should equal 20. This gives me the equation:[ x + y = 20 ]She earns 40 per hour for basic algebra and 50 per hour for advanced algebra. She wants to earn at least 900 per week, which leads to the inequality:[ 40x + 50y geq 900 ]To simplify this inequality, I can divide both sides by 10:[ 4x + 5y geq 90 ]Now, I can express ( y ) in terms of ( x ) using the total hours equation:[ y = 20 - x ]Substituting this into the inequality:[ 4x + 5(20 - x) geq 90 ][ 4x + 100 - 5x geq 90 ][ -x + 100 geq 90 ][ -x geq -10 ][ x leq 10 ]This means she can spend up to 10 hours tutoring basic algebra. Consequently, she should spend at least 10 hours tutoring advanced algebra to meet her income goal.For the second part, if she decides to spend 12 hours on basic algebra, I can use the total hours equation to find the required hours for advanced algebra:[ 12 + y = 20 ][ y = 8 ]However, I need to check if this meets the income requirement:[ 40(12) + 50(8) = 480 + 400 = 880 ]Since 880 is less than 900, she needs to adjust her hours. Increasing the advanced algebra hours to 10:[ 40(10) + 50(10) = 400 + 500 = 900 ]This meets the goal, so she should spend 10 hours on each type of tutoring.

answer：First, I need to determine how many weeks the writer can add pages without exceeding the 350-page limit. The manuscript starts with 200 pages, and the writer adds 5 pages each week. I'll set up an inequality where the initial pages plus the added pages are less than or equal to 350.Next, I'll solve this inequality for ( w ) to find the maximum number of weeks the writer can continue revising.Then, after the revisions, the editors need to cut down 10% of the revised manuscript to improve readability. If the revised manuscript has ( x ) pages, I'll express the number of pages to be cut as a polynomial expression in terms of ( x ).