answer：Alright, so I have this problem here that I need to prove. It says that given a natural number ( n ), with ( x_0 = 0 ) and each ( x_i > 0 ) for ( i = 1, 2, ldots, n ), and the sum of all ( x_i ) from 1 to ( n ) equals 1, I need to show that the sum from ( i = 1 ) to ( n ) of ( frac{x_i}{sqrt{1 + x_0 + cdots + x_{i-1}} cdot sqrt{x_i + cdots + x_n}} ) is between 1 and ( frac{pi}{2} ). Okay, let me try to unpack this. First, since ( x_0 = 0 ), the term ( 1 + x_0 + cdots + x_{i-1} ) simplifies to ( 1 + sum_{k=1}^{i-1} x_k ). Similarly, ( x_i + cdots + x_n ) is just ( sum_{k=i}^n x_k ). So, each term in the sum is ( frac{x_i}{sqrt{1 + sum_{k=1}^{i-1} x_k} cdot sqrt{sum_{k=i}^n x_k}} ).Given that the total sum of all ( x_i ) is 1, this might be useful in simplifying or bounding the expression. Maybe I can use some inequality here, like Cauchy-Schwarz or AM-GM. Let me think about Cauchy-Schwarz first because it often helps with square roots and products.The Cauchy-Schwarz inequality states that for any non-negative real numbers ( a ) and ( b ), ( sqrt{a} cdot sqrt{b} leq frac{a + b}{2} ). Wait, actually, that's the AM-GM inequality. Hmm, maybe I should recall both:- **AM-GM Inequality**: ( sqrt{ab} leq frac{a + b}{2} )- **Cauchy-Schwarz Inequality**: ( (sum a_i b_i)^2 leq (sum a_i^2)(sum b_i^2) )I think the AM-GM might be more directly applicable here because of the product of square roots. Let me try that.So, if I take ( a = 1 + sum_{k=1}^{i-1} x_k ) and ( b = sum_{k=i}^n x_k ), then by AM-GM:( sqrt{a} cdot sqrt{b} leq frac{a + b}{2} )Which means:( sqrt{1 + sum_{k=1}^{i-1} x_k} cdot sqrt{sum_{k=i}^n x_k} leq frac{1 + sum_{k=1}^{i-1} x_k + sum_{k=i}^n x_k}{2} )Simplifying the numerator on the right:( 1 + sum_{k=1}^{i-1} x_k + sum_{k=i}^n x_k = 1 + sum_{k=1}^n x_k = 1 + 1 = 2 )So, the right-hand side becomes ( frac{2}{2} = 1 ). Therefore:( sqrt{1 + sum_{k=1}^{i-1} x_k} cdot sqrt{sum_{k=i}^n x_k} leq 1 )Taking reciprocals (since both sides are positive), we get:( frac{1}{sqrt{1 + sum_{k=1}^{i-1} x_k} cdot sqrt{sum_{k=i}^n x_k}} geq 1 )Multiplying both sides by ( x_i ):( frac{x_i}{sqrt{1 + sum_{k=1}^{i-1} x_k} cdot sqrt{sum_{k=i}^n x_k}} geq x_i )Summing this over all ( i ) from 1 to ( n ):( sum_{i=1}^n frac{x_i}{sqrt{1 + sum_{k=1}^{i-1} x_k} cdot sqrt{sum_{k=i}^n x_k}} geq sum_{i=1}^n x_i = 1 )Okay, that takes care of the lower bound. So, the sum is at least 1. That wasn't too bad. Now, the upper bound is trickier because it involves ( frac{pi}{2} ). Hmm, ( pi ) usually comes into play with trigonometric functions or integrals. Maybe I need to relate this sum to an integral somehow.Let me think about the terms in the sum. Each term is ( frac{x_i}{sqrt{1 + S_{i-1}} cdot sqrt{S_i}} ), where ( S_{i-1} = sum_{k=1}^{i-1} x_k ) and ( S_i = sum_{k=i}^n x_k ). Wait, actually, ( S_i = 1 - S_{i-1} ) because the total sum is 1.So, each term can be rewritten as ( frac{x_i}{sqrt{1 + S_{i-1}} cdot sqrt{1 - S_{i-1}}} ). That simplifies to ( frac{x_i}{sqrt{(1 + S_{i-1})(1 - S_{i-1})}} ).Simplifying the denominator:( (1 + S_{i-1})(1 - S_{i-1}) = 1 - S_{i-1}^2 )So, each term is ( frac{x_i}{sqrt{1 - S_{i-1}^2}} ). Hmm, that looks a bit like the derivative of an inverse sine function. Because ( frac{d}{dtheta} sin^{-1}(theta) = frac{1}{sqrt{1 - theta^2}} ). Maybe I can make a substitution here.Let me define ( theta_i = sin^{-1}(S_i) ). Then, ( S_i = sin(theta_i) ), and ( frac{dS_i}{dtheta_i} = cos(theta_i) ). But I'm not sure if that's directly applicable here.Wait, maybe I can consider the sum as a Riemann sum. If I let ( S_{i-1} ) be a partition of the interval from 0 to 1, then the term ( x_i ) is like the width of the partition, and ( frac{1}{sqrt{1 - S_{i-1}^2}} ) is like the function evaluated at some point. So, the sum might approximate the integral of ( frac{1}{sqrt{1 - t^2}} ) from 0 to 1.Indeed, ( int_0^1 frac{1}{sqrt{1 - t^2}} dt = frac{pi}{2} ). So, if the sum is a Riemann sum for this integral, then the sum should be less than ( frac{pi}{2} ). But I need to make this precise.To see this, let me consider the function ( f(t) = frac{1}{sqrt{1 - t^2}} ). If I partition the interval [0,1] into points ( t_0 = 0, t_1, t_2, ldots, t_n = 1 ), where ( t_i = S_i = sum_{k=1}^i x_k ). Then, each ( x_i = t_i - t_{i-1} ), and the sum becomes:( sum_{i=1}^n frac{x_i}{sqrt{1 - t_{i-1}^2}} )Which is a left Riemann sum for the integral ( int_0^1 frac{1}{sqrt{1 - t^2}} dt ). Since ( f(t) ) is increasing on [0,1), the left Riemann sum underestimates the integral. Wait, but our sum is a left Riemann sum, so it should be less than the integral. But the integral is ( frac{pi}{2} ), so that would give us the upper bound.But wait, in our case, the sum is:( sum_{i=1}^n frac{x_i}{sqrt{1 - t_{i-1}^2}} )Which is indeed a left Riemann sum for ( int_0^1 frac{1}{sqrt{1 - t^2}} dt ). Since ( f(t) ) is increasing, the left Riemann sum is less than the integral. Therefore:( sum_{i=1}^n frac{x_i}{sqrt{1 - t_{i-1}^2}} < int_0^1 frac{1}{sqrt{1 - t^2}} dt = frac{pi}{2} )So, that gives us the upper bound. Therefore, combining both results, we have:( 1 leq sum_{i=1}^n frac{x_i}{sqrt{1 + x_0 + cdots + x_{i-1}} cdot sqrt{x_i + cdots + x_n}} < frac{pi}{2} )Wait, but I need to make sure that the Riemann sum interpretation is valid. Since ( t_i = S_i ) is a partition of [0,1], and ( x_i = t_i - t_{i-1} ), then yes, the sum is indeed a left Riemann sum. Therefore, the inequality holds.So, to summarize:1. For the lower bound, I used the AM-GM inequality to show that each term is at least ( x_i ), and summing up gives 1.2. For the upper bound, I recognized the sum as a left Riemann sum for the integral of ( frac{1}{sqrt{1 - t^2}} ), which equals ( frac{pi}{2} ), and since the function is increasing, the sum is less than the integral.I think that covers both bounds. I should double-check if there are any cases where the inequalities might not hold, but since all ( x_i ) are positive and sum to 1, the conditions for AM-GM and the Riemann sum interpretation are satisfied. So, I feel confident about this proof.

answer：Okay, so I need to figure out whether the statement "m > 0" is a sufficient, necessary, both, or neither condition for the function ( f(x) = |x(mx + 2)| ) to be increasing on the interval ((0, +infty)). Hmm, let's break this down step by step.First, let's understand the function ( f(x) = |x(mx + 2)| ). This is an absolute value function, which means it will always output non-negative values. The expression inside the absolute value is ( x(mx + 2) ), which simplifies to ( mx^2 + 2x ). So, ( f(x) = |mx^2 + 2x| ).Now, I need to determine when this function is increasing on the interval ((0, +infty)). To do this, I should consider the behavior of the function inside the absolute value and how the absolute value affects its increasing or decreasing nature.Let's first look at the expression inside the absolute value: ( mx^2 + 2x ). This is a quadratic function in terms of ( x ). The general shape of a quadratic function ( ax^2 + bx + c ) depends on the coefficient ( a ). If ( a > 0 ), the parabola opens upwards, and if ( a < 0 ), it opens downwards.In our case, the coefficient of ( x^2 ) is ( m ). So, if ( m > 0 ), the parabola opens upwards, and if ( m < 0 ), it opens downwards. The linear term here is ( 2x ), which affects the position of the parabola but not its direction.Since we're interested in the interval ((0, +infty)), let's analyze the behavior of ( mx^2 + 2x ) in this interval.1. **Case 1: ( m > 0 )** - The parabola opens upwards. - The function ( mx^2 + 2x ) is a quadratic that will eventually increase as ( x ) increases because the ( mx^2 ) term dominates for large ( x ). - However, near ( x = 0 ), the function might have a minimum point. Let's find the critical points by taking the derivative. The derivative of ( mx^2 + 2x ) with respect to ( x ) is ( 2mx + 2 ). Setting this equal to zero to find critical points: [ 2mx + 2 = 0 implies x = -frac{1}{m} ] Since ( m > 0 ), ( x = -frac{1}{m} ) is negative, which is outside our interval of interest ((0, +infty)). Therefore, on ((0, +infty)), the derivative ( 2mx + 2 ) is always positive because both ( m ) and ( x ) are positive. This means ( mx^2 + 2x ) is strictly increasing on ((0, +infty)). Since the inside function ( mx^2 + 2x ) is increasing and positive on ((0, +infty)) (because ( m > 0 ) and ( x > 0 )), the absolute value doesn't affect the increasing nature. Therefore, ( f(x) = |mx^2 + 2x| ) is also increasing on ((0, +infty)).2. **Case 2: ( m = 0 )** - The function simplifies to ( f(x) = |2x| = 2|x| ). - On ((0, +infty)), this is just ( 2x ), which is clearly increasing.3. **Case 3: ( m < 0 )** - The parabola opens downwards. - The function ( mx^2 + 2x ) will have a maximum point somewhere. Let's find the critical point again. The derivative is ( 2mx + 2 ). Setting this equal to zero: [ 2mx + 2 = 0 implies x = -frac{1}{m} ] Since ( m < 0 ), ( x = -frac{1}{m} ) is positive. This means the function ( mx^2 + 2x ) has a maximum at ( x = -frac{1}{m} ) within our interval ((0, +infty)). Therefore, the function ( mx^2 + 2x ) increases up to ( x = -frac{1}{m} ) and then decreases beyond that point. However, because we're taking the absolute value, the behavior changes. Let's analyze the sign of ( mx^2 + 2x ) around ( x = -frac{1}{m} ): - For ( x < -frac{1}{m} ), ( mx^2 + 2x ) is positive because the parabola is above the x-axis. - At ( x = -frac{1}{m} ), it reaches zero. - For ( x > -frac{1}{m} ), ( mx^2 + 2x ) becomes negative because the parabola dips below the x-axis. Therefore, ( f(x) = |mx^2 + 2x| ) will be: - Increasing on ( (0, -frac{1}{m}) ) because ( mx^2 + 2x ) is positive and increasing. - Decreasing on ( (-frac{1}{m}, +infty) ) because ( mx^2 + 2x ) is negative and decreasing in magnitude. However, since ( f(x) ) is the absolute value, the decreasing part of ( mx^2 + 2x ) translates to an increasing part of ( f(x) ) because the function is reflecting above the x-axis. Wait, is that correct? Let me think again. If ( mx^2 + 2x ) is decreasing and negative beyond ( x = -frac{1}{m} ), then ( f(x) = |mx^2 + 2x| ) would actually be increasing because the magnitude is increasing as ( x ) increases beyond that point. So, in reality, ( f(x) ) would increase up to ( x = -frac{1}{m} ) and then continue to increase beyond that point because the absolute value flips the decreasing part to increasing. Hmm, that seems contradictory. Let me verify by considering specific values. Suppose ( m = -1 ). Then ( f(x) = |x(-x + 2)| = | -x^2 + 2x | ). Let's analyze this function: - For ( x < 2 ), ( -x^2 + 2x ) is positive. - At ( x = 2 ), it is zero. - For ( x > 2 ), it is negative. So, ( f(x) = | -x^2 + 2x | ) is: - ( -x^2 + 2x ) for ( x in (0, 2) ) - ( x^2 - 2x ) for ( x in (2, +infty) ) Now, let's find the derivative in each interval. For ( x in (0, 2) ): [ f'(x) = frac{d}{dx}(-x^2 + 2x) = -2x + 2 ] Setting this equal to zero: [ -2x + 2 = 0 implies x = 1 ] So, at ( x = 1 ), there is a critical point. Let's check the sign of the derivative around ( x = 1 ): - For ( x < 1 ), ( f'(x) > 0 ) (increasing) - For ( x > 1 ), ( f'(x) < 0 ) (decreasing) Therefore, ( f(x) ) increases up to ( x = 1 ) and then decreases until ( x = 2 ). For ( x in (2, +infty) ): [ f'(x) = frac{d}{dx}(x^2 - 2x) = 2x - 2 ] Setting this equal to zero: [ 2x - 2 = 0 implies x = 1 ] But ( x = 1 ) is not in this interval. So, for ( x > 2 ), ( f'(x) = 2x - 2 ) is always positive because ( x > 2 ) implies ( 2x - 2 > 2*2 - 2 = 2 > 0 ). Therefore, ( f(x) ) is increasing on ( (2, +infty) ). Putting it all together: - ( f(x) ) increases on ( (0, 1) ) - ( f(x) ) decreases on ( (1, 2) ) - ( f(x) ) increases on ( (2, +infty) ) So, overall, ( f(x) ) is not strictly increasing on the entire interval ( (0, +infty) ) when ( m < 0 ). It has a decreasing segment in the middle. Therefore, for ( m < 0 ), ( f(x) ) is not increasing on the entire interval ( (0, +infty) ).4. **Case 4: ( m = 0 )** - As previously mentioned, ( f(x) = 2|x| ), which simplifies to ( 2x ) on ( (0, +infty) ), which is strictly increasing.So, summarizing:- If ( m > 0 ), ( f(x) ) is increasing on ( (0, +infty) ).- If ( m = 0 ), ( f(x) ) is increasing on ( (0, +infty) ).- If ( m < 0 ), ( f(x) ) is not increasing on the entire interval ( (0, +infty) ); it has a decreasing segment.Therefore, the condition ( m > 0 ) is sufficient for ( f(x) ) to be increasing on ( (0, +infty) ), but it is not necessary because ( m = 0 ) also satisfies the condition.However, wait a minute. When ( m = 0 ), the function is ( f(x) = 2|x| ), which is indeed increasing on ( (0, +infty) ). So, ( m geq 0 ) would be the necessary and sufficient condition. But the original statement is about ( m > 0 ).So, if we consider ( m > 0 ), it is sufficient because it guarantees the function is increasing. But it's not necessary because ( m = 0 ) also works. Therefore, ( m > 0 ) is a sufficient but not necessary condition.But hold on, in the case of ( m = 0 ), the function is still increasing. So, does that mean that ( m > 0 ) is not necessary? Yes, because the function can be increasing even when ( m = 0 ). Therefore, ( m > 0 ) is sufficient but not necessary.But wait, let's double-check. If ( m = 0 ), ( f(x) = 2x ), which is increasing. So, the function can be increasing even when ( m ) is not greater than 0. Therefore, ( m > 0 ) is not a necessary condition.But is ( m > 0 ) sufficient? Yes, because if ( m > 0 ), the function is increasing.Therefore, the correct answer should be A: Sufficient but not necessary condition.However, I initially thought that ( m > 0 ) might not be sufficient because of the absolute value, but upon further analysis, it seems that when ( m > 0 ), the function inside the absolute value is always increasing and positive on ( (0, +infty) ), so the absolute value doesn't affect the increasing nature.Wait, but in the case of ( m < 0 ), the function inside the absolute value can be positive and then negative, leading to a situation where the absolute value function first increases, then decreases, and then increases again. So, it's not strictly increasing over the entire interval.Therefore, ( m > 0 ) ensures that the function inside the absolute value is always positive and increasing, making the absolute value function also increasing. But ( m = 0 ) also results in an increasing function, so ( m > 0 ) is not necessary.Hence, the conclusion is that ( m > 0 ) is a sufficient but not necessary condition for ( f(x) ) to be increasing on ( (0, +infty) ).

answer：First, I need to determine the number of ways to select 2 balls out of the 5 that will be placed into their corresponding numbered boxes. This can be calculated using the combination formula ( C_5^2 ), which equals 10.Next, for the remaining 3 balls, they must be placed into boxes that do not correspond to their numbers. This is a derangement problem for 3 items. The number of derangements for 3 items is 2.Finally, to find the total number of ways to place the balls, I multiply the number of ways to select the 2 balls by the number of derangements for the remaining 3 balls: ( 10 times 2 = 20 ).

answer：Okay, so I've got these two equations to solve. Let's take them one at a time. The first one is ( x^2 - 6x + 1 = 0 ). Hmm, this looks like a quadratic equation. I remember that quadratic equations can be solved using factoring, completing the square, or the quadratic formula. This one doesn't look easily factorable to me because the coefficients don't seem to multiply nicely to give the constant term. So maybe I should try completing the square or use the quadratic formula.Alright, let's try completing the square first. To complete the square for ( x^2 - 6x + 1 = 0 ), I need to get it into the form ( (x - h)^2 = k ). So, I'll move the constant term to the other side: ( x^2 - 6x = -1 ). Now, to complete the square, I take the coefficient of x, which is -6, divide it by 2 to get -3, and then square it to get 9. So, I add 9 to both sides: ( x^2 - 6x + 9 = -1 + 9 ), which simplifies to ( (x - 3)^2 = 8 ). Taking the square root of both sides gives ( x - 3 = pm sqrt{8} ). Simplifying ( sqrt{8} ) to ( 2sqrt{2} ), I get ( x = 3 pm 2sqrt{2} ). So, the solutions are ( x = 3 + 2sqrt{2} ) and ( x = 3 - 2sqrt{2} ).Wait, let me double-check that. If I plug ( x = 3 + 2sqrt{2} ) back into the original equation, does it satisfy? Let's see: ( (3 + 2sqrt{2})^2 - 6(3 + 2sqrt{2}) + 1 ). Expanding ( (3 + 2sqrt{2})^2 ) gives ( 9 + 12sqrt{2} + 8 ), which is ( 17 + 12sqrt{2} ). Then subtracting ( 6(3 + 2sqrt{2}) ) gives ( 17 + 12sqrt{2} - 18 - 12sqrt{2} ), which simplifies to ( -1 ). Adding 1 gives 0. Okay, that checks out. Similarly, ( x = 3 - 2sqrt{2} ) should also satisfy the equation. So, I think that's correct.Now, moving on to the second equation: ( (2x - 3)^2 = 5(2x - 3) ). This looks a bit more complicated, but maybe I can simplify it. First, I can expand the left side: ( (2x - 3)^2 ) is ( 4x^2 - 12x + 9 ). So, the equation becomes ( 4x^2 - 12x + 9 = 5(2x - 3) ). Let's distribute the 5 on the right side: ( 4x^2 - 12x + 9 = 10x - 15 ). Now, let's move all terms to one side to set the equation to zero: ( 4x^2 - 12x + 9 - 10x + 15 = 0 ). Combining like terms, ( 4x^2 - 22x + 24 = 0 ).Hmm, this is a quadratic equation again. Maybe I can factor it. Let's see if there are two numbers that multiply to ( 4 times 24 = 96 ) and add up to -22. Hmm, 96 is a big number. Let's think: -6 and -16 multiply to 96 and add to -22. So, I can rewrite the middle term using these numbers: ( 4x^2 - 6x - 16x + 24 = 0 ). Now, factor by grouping: group the first two terms and the last two terms: ( (4x^2 - 6x) + (-16x + 24) = 0 ). Factor out 2x from the first group and -8 from the second group: ( 2x(2x - 3) - 8(2x - 3) = 0 ). Now, factor out the common term ( (2x - 3) ): ( (2x - 3)(2x - 8) = 0 ).So, setting each factor equal to zero: ( 2x - 3 = 0 ) gives ( x = frac{3}{2} ), and ( 2x - 8 = 0 ) gives ( x = 4 ). Let me verify these solutions. Plugging ( x = frac{3}{2} ) into the original equation: ( (2(frac{3}{2}) - 3)^2 = 5(2(frac{3}{2}) - 3) ). Simplifying inside the parentheses: ( (3 - 3)^2 = 5(3 - 3) ), which is ( 0 = 0 ). That works. Now, ( x = 4 ): ( (2(4) - 3)^2 = 5(2(4) - 3) ). Simplifying: ( (8 - 3)^2 = 5(8 - 3) ), which is ( 25 = 25 ). That also works. So, both solutions are correct.Wait, but I remember that sometimes when we square both sides or manipulate equations, we might introduce extraneous solutions. Did that happen here? Let me check. For ( x = frac{3}{2} ), plugging back into the original equation: ( (2x - 3)^2 = 5(2x - 3) ). As I did before, it simplifies to ( 0 = 0 ), which is valid. For ( x = 4 ), it also simplifies correctly. So, no extraneous solutions here.Alternatively, I could have approached the second equation differently. Instead of expanding, I could have noticed that both sides have a common factor of ( (2x - 3) ). So, I could have subtracted ( 5(2x - 3) ) from both sides to get ( (2x - 3)^2 - 5(2x - 3) = 0 ). Then factor out ( (2x - 3) ): ( (2x - 3)(2x - 3 - 5) = 0 ), which simplifies to ( (2x - 3)(2x - 8) = 0 ). This is the same as before, leading to ( x = frac{3}{2} ) and ( x = 4 ). So, that's another way to look at it without expanding, which might be quicker.I think I've covered both methods for the second equation, and both lead to the same solutions. It's good to know multiple approaches in case one seems too complicated or time-consuming.So, summarizing my solutions:For the first equation ( x^2 - 6x + 1 = 0 ), the solutions are ( x = 3 + 2sqrt{2} ) and ( x = 3 - 2sqrt{2} ).For the second equation ( (2x - 3)^2 = 5(2x - 3) ), the solutions are ( x = frac{3}{2} ) and ( x = 4 ).I think that's all. I've checked both solutions by plugging them back into the original equations, and they work. I didn't encounter any issues or extraneous solutions, so I'm confident in my answers.

answer：Okay, so I need to find the maximum distance from a point on the ellipse (frac{x^2}{16} + frac{y^2}{4} = 1) to the line (x + 2y - sqrt{2} = 0). Hmm, let me think about how to approach this.First, I remember that the distance from a point ((x_0, y_0)) to a line (Ax + By + C = 0) is given by the formula:[d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]In this case, the line is (x + 2y - sqrt{2} = 0), so (A = 1), (B = 2), and (C = -sqrt{2}). So, the distance from any point ((x, y)) on the ellipse to this line would be:[d = frac{|x + 2y - sqrt{2}|}{sqrt{1^2 + 2^2}} = frac{|x + 2y - sqrt{2}|}{sqrt{5}}]Now, since the point ((x, y)) lies on the ellipse (frac{x^2}{16} + frac{y^2}{4} = 1), I can parameterize the ellipse to express (x) and (y) in terms of a parameter. I recall that a standard parametrization for an ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) is (x = a cos theta) and (y = b sin theta), where (theta) is the parameter.In this case, (a^2 = 16) so (a = 4), and (b^2 = 4) so (b = 2). Therefore, the parametrization is:[x = 4 cos theta][y = 2 sin theta]Substituting these into the distance formula, we get:[d(theta) = frac{|4 cos theta + 2 times 2 sin theta - sqrt{2}|}{sqrt{5}} = frac{|4 cos theta + 4 sin theta - sqrt{2}|}{sqrt{5}}]Hmm, so (d(theta)) is a function of (theta), and I need to find its maximum value. To do this, I can simplify the expression inside the absolute value.Let me factor out the 4 from the first two terms:[4 cos theta + 4 sin theta = 4(cos theta + sin theta)]So, the distance becomes:[d(theta) = frac{|4(cos theta + sin theta) - sqrt{2}|}{sqrt{5}}]I remember that expressions like (A cos theta + B sin theta) can be rewritten using a single sine or cosine function with a phase shift. Specifically, (A cos theta + B sin theta = C cos(theta - phi)), where (C = sqrt{A^2 + B^2}) and (phi = arctanleft(frac{B}{A}right)).In this case, (A = 1) and (B = 1), so:[cos theta + sin theta = sqrt{2} cosleft(theta - frac{pi}{4}right)]Wait, let me verify that. If I have (A = 1) and (B = 1), then (C = sqrt{1^2 + 1^2} = sqrt{2}), and (phi = arctan(1/1) = pi/4). So yes, that's correct.Therefore, substituting back in:[4(cos theta + sin theta) = 4 sqrt{2} cosleft(theta - frac{pi}{4}right)]So, the distance becomes:[d(theta) = frac{|4 sqrt{2} cosleft(theta - frac{pi}{4}right) - sqrt{2}|}{sqrt{5}} = frac{sqrt{2} |4 cosleft(theta - frac{pi}{4}right) - 1|}{sqrt{5}}]Wait, let me make sure I did that correctly. If I factor out (sqrt{2}) from (4 sqrt{2} cos(theta - pi/4)), it would be:[4 sqrt{2} cos(theta - pi/4) = sqrt{2} times 4 cos(theta - pi/4)]So, yes, that's correct. So, the expression inside the absolute value is:[4 sqrt{2} cosleft(theta - frac{pi}{4}right) - sqrt{2} = sqrt{2} left(4 cosleft(theta - frac{pi}{4}right) - 1right)]Therefore, the distance is:[d(theta) = frac{sqrt{2} |4 cosleft(theta - frac{pi}{4}right) - 1|}{sqrt{5}} = frac{sqrt{2}}{sqrt{5}} |4 cosleft(theta - frac{pi}{4}right) - 1|]Simplifying (frac{sqrt{2}}{sqrt{5}}) gives (sqrt{frac{2}{5}}), but maybe it's better to keep it as (frac{sqrt{2}}{sqrt{5}}) for now.So, to find the maximum distance, I need to maximize (|4 cosleft(theta - frac{pi}{4}right) - 1|). Since the absolute value function is involved, the maximum will occur when the expression inside is either maximized or minimized.The term (4 cosleft(theta - frac{pi}{4}right)) has a maximum value of 4 and a minimum value of -4 because the cosine function oscillates between -1 and 1. Therefore, (4 cos(phi)) oscillates between -4 and 4.So, let's consider (4 cos(phi) - 1), where (phi = theta - frac{pi}{4}). The maximum value of this expression is when (4 cos(phi)) is maximum, which is 4. So, (4 - 1 = 3). The minimum value is when (4 cos(phi)) is minimum, which is -4. So, (-4 - 1 = -5).Therefore, the expression inside the absolute value, (4 cos(phi) - 1), ranges from -5 to 3. Therefore, the absolute value (|4 cos(phi) - 1|) will range from 0 to 5, because the maximum absolute value occurs at the minimum of the expression, which is -5, giving an absolute value of 5.Wait, let me think again. If the expression inside the absolute value can be as low as -5 and as high as 3, then the absolute value will have a maximum of 5 (when the expression is -5) and a minimum of 0 (when the expression is 0). So, the maximum of (|4 cos(phi) - 1|) is 5.Therefore, the maximum value of (d(theta)) is:[d_{text{max}} = frac{sqrt{2}}{sqrt{5}} times 5 = frac{5 sqrt{2}}{sqrt{5}} = sqrt{5} times sqrt{2} = sqrt{10}]Wait, let me verify that step. (frac{5 sqrt{2}}{sqrt{5}}) can be simplified as:[frac{5 sqrt{2}}{sqrt{5}} = 5 times frac{sqrt{2}}{sqrt{5}} = 5 times sqrt{frac{2}{5}} = sqrt{frac{25 times 2}{5}} = sqrt{frac{50}{5}} = sqrt{10}]Yes, that's correct. So, the maximum distance is (sqrt{10}).But just to make sure I didn't make a mistake earlier, let me recap:1. I parameterized the ellipse as (x = 4 cos theta), (y = 2 sin theta).2. Substituted into the distance formula, got (d(theta) = frac{|4 cos theta + 4 sin theta - sqrt{2}|}{sqrt{5}}).3. Recognized that (4 cos theta + 4 sin theta = 4 sqrt{2} cos(theta - pi/4)).4. Rewrote the distance as (frac{sqrt{2}}{sqrt{5}} |4 cos(theta - pi/4) - 1|).5. Noted that (4 cos(theta - pi/4)) ranges from -4 to 4, so (4 cos(theta - pi/4) - 1) ranges from -5 to 3.6. Therefore, the absolute value (|4 cos(theta - pi/4) - 1|) ranges from 0 to 5.7. Thus, the maximum distance is (frac{sqrt{2}}{sqrt{5}} times 5 = sqrt{10}).Seems solid. I don't see any errors in the steps. So, I think the maximum distance is indeed (sqrt{10}).**Final Answer**The maximum distance is (boxed{sqrt{10}}).

answer：First, I need to understand what it means to shorten the horizontal coordinates of the graph of ( y = sin x + cos x ) to (frac{1}{2}) of their original length. This transformation affects the period of the function.I recall that shortening the horizontal coordinates by a factor of (frac{1}{2}) is equivalent to horizontally compressing the graph. In terms of function transformations, this is achieved by replacing ( x ) with ( 2x ) in the original function. So, the transformed function becomes ( y = sin(2x) + cos(2x) ).Next, I want to express this new function in a simpler form to easily identify its period. I can use the trigonometric identity for the sum of sine and cosine functions:[sin A + cos A = sqrt{2} sinleft(A + frac{pi}{4}right)]Applying this identity to ( sin(2x) + cos(2x) ), I get:[sin(2x) + cos(2x) = sqrt{2} sinleft(2x + frac{pi}{4}right)]Now, the function is in the form ( y = sqrt{2} sinleft(2x + frac{pi}{4}right) ).To find the period of this sine function, I use the formula for the period of ( sin(Bx + C) ), which is ( frac{2pi}{|B|} ). In this case, ( B = 2 ), so the period is:[frac{2pi}{2} = pi]Therefore, the smallest positive period of the transformed function ( y = f(x) ) is ( pi ).