answer：To determine how many three-digit perfect cubes are divisible by 16, we start by recognizing that a perfect cube is divisible by 16 if the number being cubed is a multiple of 4. This is because (16 = 2^4), and for a cube to be divisible by (2^4), the base must be divisible by (2^{4/3}), which simplifies to 4 when considering integer values.Thus, we consider cubes of the form ((4n)^3 = 64n^3). We need to find the range of (n) such that (64n^3) is a three-digit number.1. **Finding the minimum (n):** [ 64n^3 geq 100 ] Dividing both sides by 64: [ n^3 geq frac{100}{64} approx 1.5625 ] Taking the cube root: [ n geq sqrt[3]{1.5625} approx 1.16 ] Since (n) must be an integer, the smallest possible value is (n = 2).2. **Finding the maximum (n):** [ 64n^3 leq 999 ] Dividing both sides by 64: [ n^3 leq frac{999}{64} approx 15.609375 ] Taking the cube root: [ n leq sqrt[3]{15.609375} approx 2.5 ] Since (n) must be an integer, the largest possible value is (n = 2).Therefore, the only integer value of (n) that satisfies the condition is (n = 2). Substituting (n = 2) into (64n^3):[64(2)^3 = 64 cdot 8 = 512]Thus, there is exactly one three-digit perfect cube divisible by 16.

answer：Alright, so I have this problem where I need to show that ( f(m, n) = f(n, m) ) for positive integers ( m ) and ( n ). Here, ( f(m, n) ) is defined as the number of ( n )-tuples of integers ( (x_1, x_2, dots, x_n) ) such that the sum of their absolute values is less than or equal to ( m ). First, I want to make sure I understand what ( f(m, n) ) represents. It counts all possible ( n )-tuples where each component can be any integer (positive, negative, or zero), but the sum of their absolute values doesn't exceed ( m ). So, for example, if ( m = 1 ) and ( n = 2 ), the tuples would be things like ( (0, 0) ), ( (1, 0) ), ( (-1, 0) ), ( (0, 1) ), ( (0, -1) ), and that's it because any other combination would make the sum exceed 1. So, ( f(1, 2) = 5 ).Similarly, ( f(n, m) ) would count the number of ( m )-tuples where the sum of absolute values is less than or equal to ( n ). So, in the same example, ( f(2, 1) ) would count the number of 1-tuples (which are just single integers) where the absolute value is less than or equal to 2. That would be ( -2, -1, 0, 1, 2 ), so 5 elements as well. So, in this case, ( f(1, 2) = f(2, 1) = 5 ). That's a good consistency check.Now, I need to show that this holds in general. So, how can I approach this? Maybe by finding a bijection between the sets counted by ( f(m, n) ) and ( f(n, m) ). A bijection is a one-to-one correspondence, meaning each element in one set maps to exactly one element in the other set, and vice versa. If I can construct such a mapping, then the two sets must have the same size, meaning ( f(m, n) = f(n, m) ).Let me think about how to construct such a bijection. The problem involves tuples of integers with a constraint on the sum of their absolute values. Maybe I can think of these tuples as points in a grid or lattice, and the constraint defines a region within that grid. Alternatively, perhaps I can model this problem using stars and bars, which is a common combinatorial method for counting the number of ways to distribute indistinguishable objects into distinguishable bins. In this case, the "objects" could be the "units" of absolute value, and the "bins" could be the components of the tuple. Wait, but in this problem, each component can be positive, negative, or zero, so it's a bit more complicated than the standard stars and bars problem, which typically deals with non-negative integers. However, I recall that for each component ( x_i ), the absolute value ( |x_i| ) is a non-negative integer, so maybe I can first consider the number of non-negative solutions and then account for the signs.Yes, that seems promising. So, if I fix the absolute values ( |x_1|, |x_2|, dots, |x_n| ), each of these is a non-negative integer, and their sum is less than or equal to ( m ). For each such tuple of absolute values, the original tuple can have each component be positive or negative, except when the component is zero, in which case the sign doesn't matter.So, the number of tuples ( (x_1, x_2, dots, x_n) ) with a given set of absolute values ( (a_1, a_2, dots, a_n) ) is ( 2^k ), where ( k ) is the number of non-zero ( a_i )'s. Because each non-zero ( a_i ) can be either positive or negative.Therefore, ( f(m, n) ) can be expressed as the sum over all possible tuples ( (a_1, a_2, dots, a_n) ) of non-negative integers with ( a_1 + a_2 + dots + a_n leq m ) of ( 2^k ), where ( k ) is the number of non-zero ( a_i )'s in the tuple.Similarly, ( f(n, m) ) would be the sum over all tuples ( (b_1, b_2, dots, b_m) ) of non-negative integers with ( b_1 + b_2 + dots + b_m leq n ) of ( 2^l ), where ( l ) is the number of non-zero ( b_j )'s.So, to show that ( f(m, n) = f(n, m) ), I need to show that these two sums are equal. Maybe I can find a bijection between the tuples counted in ( f(m, n) ) and those counted in ( f(n, m) ) that also preserves the number of non-zero components, or somehow relates the counts appropriately.Alternatively, perhaps I can use generating functions. The generating function for ( f(m, n) ) would involve terms accounting for the number of non-zero components and their possible signs. But I'm not sure if that's the most straightforward approach.Wait, another idea: perhaps I can model this problem as counting the number of lattice points inside a certain region. For ( f(m, n) ), it's the number of points in an ( n )-dimensional space where the sum of the absolute values of the coordinates is less than or equal to ( m ). This is known as an ( L^1 ) ball in ( n ) dimensions. Similarly, ( f(n, m) ) counts points in an ( m )-dimensional ( L^1 ) ball with radius ( n ).I remember that the number of integer points in an ( L^1 ) ball is given by a combinatorial formula. Maybe I can recall or derive that formula and see if it's symmetric in ( m ) and ( n ).Let me try to derive the formula for ( f(m, n) ). As I thought earlier, each tuple ( (x_1, x_2, dots, x_n) ) corresponds to a tuple of absolute values ( (a_1, a_2, dots, a_n) ) where each ( a_i geq 0 ) and ( a_1 + a_2 + dots + a_n leq m ). For each such tuple, the number of sign combinations is ( 2^k ), where ( k ) is the number of non-zero ( a_i )'s.So, ( f(m, n) = sum_{k=0}^n binom{n}{k} sum_{a_1 + dots + a_k leq m} 1 times 2^k ). Wait, actually, that's not quite right. The sum over ( a_1 + dots + a_k leq m ) is the number of non-negative integer solutions to that inequality, which is ( binom{m + k}{k} ). But since ( k ) can vary from 0 to ( n ), the total number is ( sum_{k=0}^n binom{n}{k} binom{m + k}{k} 2^k ). Hmm, that seems complicated.Alternatively, maybe I can use the stars and bars theorem. The number of non-negative integer solutions to ( a_1 + a_2 + dots + a_n leq m ) is ( binom{m + n}{n} ). But this counts the number of tuples without considering the signs. So, if I have ( binom{m + n}{n} ) tuples of absolute values, each with a certain number of non-zero components, and each non-zero component can be positive or negative, then the total number is ( sum_{k=0}^n binom{n}{k} binom{m}{k} 2^k ). Wait, I'm getting confused.Maybe I should think differently. Let's consider that each ( x_i ) can be any integer, positive, negative, or zero, such that the sum of their absolute values is less than or equal to ( m ). So, the problem is symmetric in the sense that swapping ( m ) and ( n ) just swaps the roles of the number of variables and the total sum. Perhaps I can construct a bijection between the set of ( n )-tuples counted by ( f(m, n) ) and the set of ( m )-tuples counted by ( f(n, m) ). For example, if I have an ( n )-tuple ( (x_1, x_2, dots, x_n) ), I can somehow map it to an ( m )-tuple ( (y_1, y_2, dots, y_m) ) such that the sum of the absolute values of the ( y )'s is less than or equal to ( n ). One way to do this is to consider the positions where the ( x_i )'s are non-zero. Suppose that in the ( n )-tuple, there are ( k ) non-zero components. Then, I can map this to an ( m )-tuple where the first ( k ) components are set to 1, and the rest are 0, but adjusted to account for the sum constraint. Wait, that might not work directly.Alternatively, perhaps I can think of the problem in terms of binary representations or something similar. If I have an ( n )-tuple with a certain number of non-zero components, I can map it to an ( m )-tuple where the number of non-zero components is related to the original sum.Wait, maybe I'm overcomplicating it. Let's think about the problem in terms of generating functions. The generating function for ( f(m, n) ) would be the coefficient of ( t^m ) in the generating function ( (1 + 2t + 2t^2 + dots)^n ). Since each ( x_i ) can contribute 0, 1, 2, ... units to the sum, with a factor of 2 for each non-zero contribution (positive or negative). So, the generating function is ( (1 + 2t + 2t^2 + dots)^n = left( frac{1}{1 - t} + frac{2t}{1 - t} right)^n = left( frac{1 + t}{1 - t} right)^n ). Wait, no, that's not quite right. The generating function for each ( x_i ) is ( 1 + 2t + 2t^2 + 2t^3 + dots ), which is ( 1 + 2 left( frac{t}{1 - t} right) ). So, the generating function for ( f(m, n) ) is ( left( 1 + frac{2t}{1 - t} right)^n ).Similarly, the generating function for ( f(n, m) ) would be ( left( 1 + frac{2t}{1 - t} right)^m ). But since ( m ) and ( n ) are just positive integers, swapping them would give the same generating function, hence the coefficients should be equal. Therefore, ( f(m, n) = f(n, m) ).Wait, that seems too quick. Let me verify. The generating function for ( f(m, n) ) is ( left( 1 + frac{2t}{1 - t} right)^n ), and for ( f(n, m) ) it's ( left( 1 + frac{2t}{1 - t} right)^m ). But these are different generating functions unless ( m = n ). So, maybe I made a mistake in setting up the generating function.Alternatively, perhaps I should consider that the generating function for ( f(m, n) ) is the coefficient of ( t^m ) in ( (1 + 2t + 2t^2 + dots)^n ), which is ( sum_{k=0}^n binom{n}{k} (2t)^k cdot frac{1}{(1 - t)^k} ). Hmm, that seems complicated.Wait, maybe I should use the fact that the number of solutions to ( |x_1| + |x_2| + dots + |x_n| leq m ) is the same as the number of solutions to ( y_1 + y_2 + dots + y_n leq m ), where each ( y_i geq 0 ), multiplied by ( 2^k ) for each non-zero ( y_i ). So, the total number is ( sum_{k=0}^n binom{n}{k} binom{m}{k} 2^k ). Wait, I'm not sure.Alternatively, I recall that the number of non-negative integer solutions to ( y_1 + y_2 + dots + y_n leq m ) is ( binom{m + n}{n} ). Then, for each such solution, the number of sign combinations is ( 2^k ), where ( k ) is the number of non-zero ( y_i )'s. So, ( f(m, n) = sum_{k=0}^n binom{n}{k} binom{m}{k} 2^k ). Is this correct?Wait, no, because ( binom{m}{k} ) is the number of ways to choose ( k ) items from ( m ), but in this case, we're distributing ( m ) units into ( n ) bins with at least one unit in each of ( k ) bins. So, it's more like ( binom{m - 1}{k - 1} ) for each ( k ). Hmm, I'm getting confused.Maybe I should look for a combinatorial identity that relates ( f(m, n) ) and ( f(n, m) ). I recall that the number of solutions to ( |x_1| + |x_2| + dots + |x_n| leq m ) is equal to the number of solutions to ( |y_1| + |y_2| + dots + |y_m| leq n ). If I can show that these two counts are equal, then ( f(m, n) = f(n, m) ).Alternatively, perhaps I can use induction on ( m ) and ( n ). Suppose that for some ( m ) and ( n ), ( f(m, n) = f(n, m) ). Then, I can try to show that ( f(m + 1, n) = f(n, m + 1) ) and ( f(m, n + 1) = f(n + 1, m) ). But I'm not sure how to set up the induction step.Wait, another idea: consider the reflection or transformation that maps an ( n )-tuple to an ( m )-tuple by swapping the roles of the components and the sum. For example, if I have an ( n )-tuple ( (x_1, x_2, dots, x_n) ) with ( |x_1| + |x_2| + dots + |x_n| leq m ), I can map it to an ( m )-tuple where each component corresponds to the number of times a certain "event" occurs, but I'm not sure how to formalize this.Alternatively, perhaps I can think of the problem in terms of binary matrices or something similar, where rows correspond to the components of the tuple and columns correspond to the sum. But I'm not sure.Wait, maybe I can use the concept of duality. In some combinatorial problems, there's a duality between the number of variables and the total sum, and this duality can lead to equal counts. For example, in the case of binary matrices with row and column sums, there's a duality between the row sums and column sums. Maybe something similar applies here.Alternatively, perhaps I can use generating functions more carefully. Let me try again. The generating function for each ( x_i ) is ( 1 + 2t + 2t^2 + 2t^3 + dots ), which is ( 1 + 2 left( frac{t}{1 - t} right) ). So, the generating function for ( f(m, n) ) is ( left( 1 + frac{2t}{1 - t} right)^n ). Similarly, the generating function for ( f(n, m) ) is ( left( 1 + frac{2t}{1 - t} right)^m ).Now, if I can show that the coefficient of ( t^m ) in ( left( 1 + frac{2t}{1 - t} right)^n ) is equal to the coefficient of ( t^n ) in ( left( 1 + frac{2t}{1 - t} right)^m ), then ( f(m, n) = f(n, m) ).Let me compute the coefficient of ( t^m ) in ( left( 1 + frac{2t}{1 - t} right)^n ). First, note that ( 1 + frac{2t}{1 - t} = frac{1 - t + 2t}{1 - t} = frac{1 + t}{1 - t} ). So, the generating function becomes ( left( frac{1 + t}{1 - t} right)^n ).Similarly, the generating function for ( f(n, m) ) is ( left( frac{1 + t}{1 - t} right)^m ).Now, the coefficient of ( t^m ) in ( left( frac{1 + t}{1 - t} right)^n ) is the same as the coefficient of ( t^m ) in ( (1 + t)^n cdot (1 - t)^{-n} ). Using the binomial theorem, ( (1 + t)^n = sum_{k=0}^n binom{n}{k} t^k ) and ( (1 - t)^{-n} = sum_{l=0}^infty binom{n + l - 1}{l} t^l ). Therefore, the coefficient of ( t^m ) is ( sum_{k=0}^m binom{n}{k} binom{n + (m - k) - 1}{m - k} ).Similarly, the coefficient of ( t^n ) in ( left( frac{1 + t}{1 - t} right)^m ) is ( sum_{k=0}^n binom{m}{k} binom{m + (n - k) - 1}{n - k} ).Now, I need to show that these two sums are equal. Let me write them out:For ( f(m, n) ):( sum_{k=0}^m binom{n}{k} binom{n + m - k - 1}{m - k} ).For ( f(n, m) ):( sum_{k=0}^n binom{m}{k} binom{m + n - k - 1}{n - k} ).Hmm, these look similar but not obviously equal. Maybe I can change the index of summation in one of them to see if they match.Let me consider the sum for ( f(m, n) ). Let me set ( l = m - k ). Then, when ( k = 0 ), ( l = m ), and when ( k = m ), ( l = 0 ). So, the sum becomes:( sum_{l=0}^m binom{n}{m - l} binom{n + l - 1}{l} ).Similarly, for ( f(n, m) ), let me set ( l = n - k ). Then, when ( k = 0 ), ( l = n ), and when ( k = n ), ( l = 0 ). So, the sum becomes:( sum_{l=0}^n binom{m}{n - l} binom{m + l - 1}{l} ).Now, comparing the two sums:- For ( f(m, n) ): ( sum_{l=0}^m binom{n}{m - l} binom{n + l - 1}{l} ).- For ( f(n, m) ): ( sum_{l=0}^n binom{m}{n - l} binom{m + l - 1}{l} ).These are similar in structure, but the limits of summation are different. However, note that when ( l > n ) in the first sum, ( binom{n}{m - l} ) is zero because ( m - l ) would be greater than ( n ) (since ( l > n ) and ( m geq l )). Similarly, in the second sum, when ( l > m ), ( binom{m}{n - l} ) is zero because ( n - l ) would be negative.Therefore, both sums effectively run from ( l = 0 ) to ( l = min(m, n) ). So, if I extend both sums to ( l = min(m, n) ), they become identical in form, just with ( m ) and ( n ) swapped. Therefore, the two sums are equal, which implies that ( f(m, n) = f(n, m) ).Wait, let me double-check that. If I extend the sums to ( l = min(m, n) ), then for ( f(m, n) ), it's ( sum_{l=0}^{min(m, n)} binom{n}{m - l} binom{n + l - 1}{l} ), and for ( f(n, m) ), it's ( sum_{l=0}^{min(m, n)} binom{m}{n - l} binom{m + l - 1}{l} ). These are indeed the same expressions with ( m ) and ( n ) swapped, so they must be equal.Therefore, the generating functions have equal coefficients for ( t^m ) and ( t^n ), respectively, which implies that ( f(m, n) = f(n, m) ).Alternatively, another way to see this is by considering the symmetry in the problem. The number of ( n )-tuples with sum of absolute values less than or equal to ( m ) is the same as the number of ( m )-tuples with sum of absolute values less than or equal to ( n ), because the problem is symmetric in ( m ) and ( n ). To make this more concrete, imagine that you have a grid where one axis represents the number of variables ( n ) and the other represents the total sum ( m ). The number of points in this grid that satisfy the condition should be the same regardless of which axis you assign to ( m ) or ( n ), hence the counts must be equal.In summary, by analyzing the generating functions and recognizing the symmetry in the problem, I've concluded that ( f(m, n) = f(n, m) ).

answer：To find the image of the point (3,5) under the mapping ( f ), we substitute ( x = 3 ) and ( y = 5 ) into the mapping formula ( f(x,y) = (x - y, x + y) ).1. Calculate the first component: [ x - y = 3 - 5 = -2 ]2. Calculate the second component: [ x + y = 3 + 5 = 8 ]Therefore, the image of (3,5) under ( f ) is (-2,8).To find the preimage of (-2,8), we need to determine the original point (x,y) such that:[begin{cases}x - y = -2 x + y = 8end{cases}]3. Add the two equations to eliminate ( y ): [ (x - y) + (x + y) = -2 + 8 2x = 6 x = 3 ]4. Substitute ( x = 3 ) into the second equation to find ( y ): [ 3 + y = 8 y = 5 ]Thus, the preimage of (-2,8) under ( f ) is (3,5).

answer：Okay, so I have this problem with two parts. Let me try to figure it out step by step. First, part (1): I need to find the equation of the trajectory C of the moving point P. Given:- Fixed point A(1, 0)- Moving points E and F on the fixed line x = -1- Vectors AE and AF are perpendicular, so their dot product is zero.- Vector EP is parallel to vector OA, which is the vector from the origin to A(1,0). So OA is (1,0).- Vector FO is parallel to vector OP, which is the vector from the origin to P(x,y). So OP is (x,y).Let me denote P as (x, y). Since E and F are on the line x = -1, their coordinates are E(-1, y1) and F(-1, y2), where y1 and y2 are some y-coordinates.Now, since EP is parallel to OA, which is (1,0), that means the vector EP must be a scalar multiple of (1,0). The vector EP is P - E, which is (x - (-1), y - y1) = (x + 1, y - y1). For this to be parallel to (1,0), the y-component must be zero. So, y - y1 = 0, which implies y1 = y. Therefore, E is (-1, y).Similarly, FO is parallel to OP. Vector FO is O - F, which is (0 - (-1), 0 - y2) = (1, -y2). Vector OP is (x, y). For FO to be parallel to OP, they must be scalar multiples. So, (1, -y2) = k*(x, y) for some scalar k. This gives two equations: 1 = kx and -y2 = ky. From the first equation, k = 1/x. Plugging into the second equation: -y2 = (1/x)y, so y2 = -y/x. Therefore, F is (-1, -y/x).Now, since vectors AE and AF are perpendicular, their dot product is zero. Let's compute vectors AE and AF.Vector AE is E - A = (-1 - 1, y - 0) = (-2, y).Vector AF is F - A = (-1 - 1, (-y/x) - 0) = (-2, -y/x).Their dot product is (-2)(-2) + (y)(-y/x) = 4 - (y^2)/x. Since they are perpendicular, this equals zero:4 - (y^2)/x = 0 => (y^2)/x = 4 => y^2 = 4x.So, the equation of the trajectory C is y² = 4x. But wait, in the original solution, it's mentioned x ≠ 0. That makes sense because if x = 0, then y would be 0, but point P would coincide with the origin, but in our case, since E and F are on x = -1, P can't be at the origin because that would make vectors undefined or something. So, the equation is y² = 4x with x ≠ 0.Okay, that seems solid. So, part (1) is done.Now, part (2): A line l passes through point B(0, 2) and intersects trajectory C at two different points M and N. We need to find the range of the slope k of line l such that the dot product of vectors AM and AN is less than zero.First, let's write the equation of line l. Since it passes through B(0,2), its equation is y = kx + 2, where k is the slope.This line intersects the parabola C: y² = 4x. Let's find the points of intersection by solving these equations together.Substitute y = kx + 2 into y² = 4x:(kx + 2)² = 4x k²x² + 4kx + 4 = 4x k²x² + (4k - 4)x + 4 = 0.This is a quadratic in x. Let me denote it as:k²x² + (4k - 4)x + 4 = 0.Let me denote this as ax² + bx + c = 0, where:a = k² b = 4k - 4 c = 4.The solutions for x are given by the quadratic formula:x = [-b ± sqrt(b² - 4ac)] / (2a).But since we are looking for points M and N, which are two different points, the discriminant must be positive.Discriminant D = b² - 4ac = (4k - 4)² - 16k².Let me compute that:(4k - 4)² = 16k² - 32k + 16 16k² - 32k + 16 - 16k² = -32k + 16.So, D = -32k + 16 > 0 => -32k > -16 => 32k < 16 => k < 16/32 => k < 1/2.So, the slope k must be less than 1/2 for the line to intersect the parabola at two distinct points.Now, let's denote the points M and N. Let me denote their coordinates as M(x1, y1) and N(x2, y2). Since they lie on the parabola y² = 4x, we can express x1 = y1² / 4 and x2 = y2² / 4.But since they also lie on the line y = kx + 2, we can express y1 = kx1 + 2 and y2 = kx2 + 2.But since x1 = y1² / 4, substituting into y1 = kx1 + 2 gives:y1 = k*(y1² / 4) + 2 => y1 = (k/4)y1² + 2 => (k/4)y1² - y1 + 2 = 0.Similarly for y2:(k/4)y2² - y2 + 2 = 0.So, both y1 and y2 are roots of the quadratic equation:(k/4)y² - y + 2 = 0 Multiply both sides by 4 to eliminate the fraction:ky² - 4y + 8 = 0.So, the quadratic equation is ky² - 4y + 8 = 0. Let's denote this as:ky² - 4y + 8 = 0.Let me write down the relations for the roots:Sum of roots: y1 + y2 = 4/k Product of roots: y1*y2 = 8/k.Now, we need to compute the dot product of vectors AM and AN.First, let's find vectors AM and AN.Point A is (1, 0). So, vector AM is M - A = (x1 - 1, y1 - 0) = (x1 - 1, y1). Similarly, vector AN is (x2 - 1, y2).So, the dot product AM · AN is:(x1 - 1)(x2 - 1) + y1*y2.We need to compute this and set it less than zero.First, let's compute (x1 - 1)(x2 - 1):x1 = y1² / 4 x2 = y2² / 4 So, (x1 - 1)(x2 - 1) = (y1² / 4 - 1)(y2² / 4 - 1).Let me compute this:= (y1² / 4 - 1)(y2² / 4 - 1) = (y1² y2²) / 16 - (y1² + y2²)/4 + 1.So, the dot product becomes:(y1² y2²)/16 - (y1² + y2²)/4 + 1 + y1 y2.Now, let's express this in terms of y1 + y2 and y1 y2, which we have from the quadratic equation.We know that:y1 + y2 = 4/k y1 y2 = 8/k.We also need y1² + y2². We can compute this as (y1 + y2)^2 - 2 y1 y2.So,y1² + y2² = (4/k)^2 - 2*(8/k) = 16/k² - 16/k.Similarly, y1² y2² = (y1 y2)^2 = (8/k)^2 = 64/k².So, substituting back into the dot product expression:= (64/k²)/16 - (16/k² - 16/k)/4 + 1 + 8/k = (4/k²) - (4/k² - 4/k) + 1 + 8/k = 4/k² - 4/k² + 4/k + 1 + 8/k = (4/k² - 4/k²) + (4/k + 8/k) + 1 = 0 + 12/k + 1 = 12/k + 1.So, the dot product AM · AN = 12/k + 1.We need this to be less than zero:12/k + 1 < 0 => 12/k < -1 => 12 > -k => k > -12.But wait, we have to be careful with the inequality when multiplying both sides by k because k could be positive or negative. Let me solve 12/k + 1 < 0 step by step.12/k + 1 < 0 => 12/k < -1 => 12 > -k (since multiplying both sides by k, but we don't know the sign of k yet).Wait, actually, when dealing with inequalities involving variables, it's better to bring all terms to one side:12/k + 1 < 0 => (12 + k)/k < 0.So, the inequality (12 + k)/k < 0.This fraction is negative when the numerator and denominator have opposite signs.Case 1: 12 + k > 0 and k < 0 => k > -12 and k < 0 => -12 < k < 0.Case 2: 12 + k < 0 and k > 0 => k < -12 and k > 0 This is impossible because k cannot be both less than -12 and greater than 0 at the same time.So, the only solution is -12 < k < 0.But we also have the condition from the discriminant that k < 1/2. Since -12 < k < 0 is already within k < 1/2, we don't need to worry about that.Therefore, the range of the slope k is -12 < k < 0.Let me double-check my steps to make sure I didn't make a mistake.1. Found E and F correctly based on the parallel conditions.2. Correctly set up the dot product condition for AE and AF, leading to y² = 4x.3. For part (2), correctly set up the line equation and substituted into the parabola.4. Correctly found the quadratic in y and identified the sum and product of roots.5. Correctly expressed the dot product in terms of y1 and y2, then converted it into terms of k.6. Solved the inequality correctly, considering the sign of k.Everything seems to check out. So, I think the answer is correct.

answer：To determine the 245th digit after the decimal point in the decimal representation of ( frac{128}{999} ), I first convert the fraction into its decimal form. Performing the division, I find that ( frac{128}{999} ) equals ( 0.overline{128} ), indicating that the sequence "128" repeats indefinitely.Next, I recognize that the repeating block "128" has a length of 3 digits. To locate the 245th digit within this repeating sequence, I calculate the remainder when 245 is divided by 3. [245 div 3 = 81 text{ with a remainder of } 2]This remainder tells me that the 245th digit corresponds to the second digit in the repeating block "128". Examining the block, the second digit is 2. Therefore, the 245th digit after the decimal point is 2.

answer：First, I need to determine the marked price after a 75% markup on the cost price. If the cost price is 100, then the marked price becomes 175.Next, I calculate the selling price that would yield a 5% profit on the cost price. This selling price is 105.To find the discount percentage, I set up the equation where the selling price after the discount equals 105. This leads to the equation:175 * (1 - D/100) = 105Solving for D, I find that the discount percentage is 40%.