answer：Okay, so I have this problem here where I need to find the area of the region formed by the moving point ( P(a, b) ) given that in the plane region ( {(x, y) mid |x| leq 1, |y| leq 1} ), it's always true that ( a x - 2 b y leq 2 ). The options are A. 4, B. 8, C. 16, D. 32. Hmm, let me try to figure this out step by step.First, I need to understand what the problem is saying. We have a region in the ( xy )-plane which is a square with sides of length 2, centered at the origin, because ( |x| leq 1 ) and ( |y| leq 1 ). So, this square goes from ( x = -1 ) to ( x = 1 ) and ( y = -1 ) to ( y = 1 ).Now, the condition given is that for all points ( (x, y) ) in this square, the inequality ( a x - 2 b y leq 2 ) must hold. So, no matter where ( (x, y) ) is in that square, ( a x - 2 b y ) should not exceed 2. I need to find the area of the region where the point ( P(a, b) ) can move such that this condition is always satisfied.Hmm, so this seems like a problem involving inequalities and regions in the plane. Maybe I can think of ( a x - 2 b y leq 2 ) as a linear inequality in terms of ( x ) and ( y ), but since ( x ) and ( y ) are bounded within the square, this inequality must hold for all ( x ) and ( y ) in that square. So, perhaps I need to find the maximum value of ( a x - 2 b y ) over the square and set it to be less than or equal to 2.Yes, that makes sense. So, if I can find the maximum of ( a x - 2 b y ) over ( |x| leq 1 ) and ( |y| leq 1 ), and then set that maximum to be ( leq 2 ), that should give me the condition on ( a ) and ( b ).How do I find the maximum of a linear function over a rectangle? I remember that for linear functions, the maximum occurs at one of the corners of the feasible region. So, since ( |x| leq 1 ) and ( |y| leq 1 ) define a rectangle (a square in this case), the maximum of ( a x - 2 b y ) will occur at one of the four corners: ( (1, 1) ), ( (1, -1) ), ( (-1, 1) ), or ( (-1, -1) ).So, let's compute ( a x - 2 b y ) at each of these four points:1. At ( (1, 1) ): ( a(1) - 2b(1) = a - 2b )2. At ( (1, -1) ): ( a(1) - 2b(-1) = a + 2b )3. At ( (-1, 1) ): ( a(-1) - 2b(1) = -a - 2b )4. At ( (-1, -1) ): ( a(-1) - 2b(-1) = -a + 2b )So, the maximum of these four expressions must be less than or equal to 2. Therefore, we have four inequalities:1. ( a - 2b leq 2 )2. ( a + 2b leq 2 )3. ( -a - 2b leq 2 )4. ( -a + 2b leq 2 )These four inequalities define the region in the ( ab )-plane where ( P(a, b) ) can move. So, I need to find the area of the region defined by these four inequalities.Let me write them down again:1. ( a - 2b leq 2 )2. ( a + 2b leq 2 )3. ( -a - 2b leq 2 )4. ( -a + 2b leq 2 )Hmm, maybe I can rearrange these inequalities to make them easier to handle.Starting with the first inequality: ( a - 2b leq 2 ). Let's write this as ( a leq 2 + 2b ).Second inequality: ( a + 2b leq 2 ). So, ( a leq 2 - 2b ).Third inequality: ( -a - 2b leq 2 ). Multiply both sides by -1 (remembering to reverse the inequality sign): ( a + 2b geq -2 ).Fourth inequality: ( -a + 2b leq 2 ). Multiply both sides by -1: ( a - 2b geq -2 ).So now, the four inequalities are:1. ( a leq 2 + 2b )2. ( a leq 2 - 2b )3. ( a + 2b geq -2 )4. ( a - 2b geq -2 )Hmm, so these are four linear inequalities. Let me try to visualize the region defined by these inequalities in the ( ab )-plane.Alternatively, maybe I can express these inequalities in terms of ( a ) and ( b ) and see if they form a polygon, perhaps a quadrilateral or something else, and then compute its area.Let me write all four inequalities again:1. ( a leq 2 + 2b ) (Inequality 1)2. ( a leq 2 - 2b ) (Inequality 2)3. ( a + 2b geq -2 ) (Inequality 3)4. ( a - 2b geq -2 ) (Inequality 4)So, let's try to find the intersection points of these lines because the region defined by the inequalities is a polygon whose vertices are the intersection points of these lines.First, let's find the intersection of Inequality 1 and Inequality 2:Set ( 2 + 2b = 2 - 2b ). Solving for ( b ):( 2 + 2b = 2 - 2b )Subtract 2 from both sides: ( 2b = -2b )Add ( 2b ) to both sides: ( 4b = 0 )So, ( b = 0 )Plugging back into Inequality 1: ( a = 2 + 2(0) = 2 )So, the intersection point is ( (2, 0) )Next, find the intersection of Inequality 2 and Inequality 3:Set ( 2 - 2b = -2 - 2b ). Wait, let's write Inequality 3 as ( a = -2 - 2b ). So, set ( 2 - 2b = -2 - 2b ).But that would imply ( 2 = -2 ), which is not possible. So, these two lines are parallel? Wait, no, let me check.Wait, Inequality 2 is ( a leq 2 - 2b ), which is a line with slope -2.Inequality 3 is ( a + 2b geq -2 ), which can be rewritten as ( a geq -2 - 2b ), which is a line with slope -2 as well. So, they are parallel. Therefore, they don't intersect.Hmm, so perhaps the region is bounded by these four lines, but some of them are parallel. Let's see.Wait, maybe I should instead consider all four lines:1. ( a = 2 + 2b )2. ( a = 2 - 2b )3. ( a = -2 - 2b )4. ( a = -2 + 2b )Wait, hold on. Let me correct that. Inequality 3 is ( a + 2b geq -2 ), which is ( a geq -2 - 2b ). Similarly, Inequality 4 is ( a - 2b geq -2 ), which is ( a geq -2 + 2b ).So, actually, the four lines are:1. ( a = 2 + 2b ) (from Inequality 1)2. ( a = 2 - 2b ) (from Inequality 2)3. ( a = -2 - 2b ) (from Inequality 3)4. ( a = -2 + 2b ) (from Inequality 4)So, these are four straight lines with slopes 2 and -2. So, lines 1 and 3 have slope 2, while lines 2 and 4 have slope -2.Wait, no. Let me check:Wait, ( a = 2 + 2b ) can be written as ( a = 2b + 2 ), so slope is 2.Similarly, ( a = 2 - 2b ) is ( a = -2b + 2 ), slope is -2.( a = -2 - 2b ) is ( a = -2b - 2 ), slope is -2.( a = -2 + 2b ) is ( a = 2b - 2 ), slope is 2.So, lines 1 and 4 have slope 2, lines 2 and 3 have slope -2.Therefore, lines 1 and 4 are parallel, and lines 2 and 3 are parallel.So, we have two pairs of parallel lines. So, the region defined by these inequalities is a parallelogram.Wait, but actually, let me see. The four inequalities are:1. ( a leq 2 + 2b )2. ( a leq 2 - 2b )3. ( a geq -2 - 2b )4. ( a geq -2 + 2b )So, the region is bounded between these four lines. So, it's a quadrilateral bounded by these four lines.To find the vertices of this quadrilateral, I need to find the intersection points of these lines.So, let's find the intersection points:1. Intersection of ( a = 2 + 2b ) and ( a = 2 - 2b ): we already did this earlier, it's ( (2, 0) ).2. Intersection of ( a = 2 - 2b ) and ( a = -2 - 2b ):Set ( 2 - 2b = -2 - 2b )Again, subtracting ( -2b ) from both sides: ( 2 = -2 ), which is not possible. So, these lines are parallel and don't intersect.Wait, that can't be. Maybe I made a mistake.Wait, no, ( a = 2 - 2b ) and ( a = -2 - 2b ) are both lines with slope -2, so they are parallel and don't intersect.Similarly, ( a = 2 + 2b ) and ( a = -2 + 2b ) are both lines with slope 2, so they are parallel and don't intersect.Therefore, the region is actually a parallelogram, but since the lines are parallel, the region is unbounded? Wait, no, because we have four inequalities, two upper bounds and two lower bounds.Wait, perhaps it's a bounded region because the upper bounds and lower bounds intersect each other.Wait, let me try to find all intersection points between the lines.So, lines:1. ( a = 2 + 2b ) (Line 1)2. ( a = 2 - 2b ) (Line 2)3. ( a = -2 - 2b ) (Line 3)4. ( a = -2 + 2b ) (Line 4)So, let's find the intersection points between Line 1 and Line 3:Set ( 2 + 2b = -2 - 2b )So, ( 2 + 2b = -2 - 2b )Bring variables to one side: ( 2b + 2b = -2 - 2 )( 4b = -4 )( b = -1 )Then, plug back into Line 1: ( a = 2 + 2(-1) = 0 )So, intersection point is ( (0, -1) )Next, intersection of Line 1 and Line 4:Set ( 2 + 2b = -2 + 2b )Subtract ( 2b ) from both sides: ( 2 = -2 ), which is not possible. So, no intersection.Intersection of Line 2 and Line 3:Set ( 2 - 2b = -2 - 2b )Again, subtract ( -2b ): ( 2 = -2 ), which is not possible. So, no intersection.Intersection of Line 2 and Line 4:Set ( 2 - 2b = -2 + 2b )Bring variables to one side: ( -2b - 2b = -2 - 2 )( -4b = -4 )( b = 1 )Plug back into Line 2: ( a = 2 - 2(1) = 0 )So, intersection point is ( (0, 1) )Intersection of Line 3 and Line 4:Set ( -2 - 2b = -2 + 2b )Bring variables to one side: ( -2b - 2b = -2 + 2 )( -4b = 0 )( b = 0 )Plug back into Line 3: ( a = -2 - 2(0) = -2 )So, intersection point is ( (-2, 0) )Wait, so let's list all the intersection points we have:1. Line 1 and Line 2: ( (2, 0) )2. Line 1 and Line 3: ( (0, -1) )3. Line 2 and Line 4: ( (0, 1) )4. Line 3 and Line 4: ( (-2, 0) )Wait, but hold on, Line 1 and Line 4 don't intersect, same with Line 2 and Line 3. So, the four intersection points we have are ( (2, 0) ), ( (0, -1) ), ( (0, 1) ), and ( (-2, 0) ).So, connecting these points, we get a quadrilateral. Let me plot these points mentally:- ( (2, 0) ): on the right side- ( (0, -1) ): below the origin- ( (0, 1) ): above the origin- ( (-2, 0) ): on the left sideSo, connecting these points in order, we get a diamond-shaped figure, which is a quadrilateral symmetric about both axes.Wait, actually, connecting ( (2, 0) ) to ( (0, -1) ) to ( (-2, 0) ) to ( (0, 1) ) and back to ( (2, 0) ). So, it's a kite-shaped quadrilateral.But actually, since all sides are equal in a way, maybe it's a rhombus.Wait, let's compute the distances between consecutive points:1. From ( (2, 0) ) to ( (0, -1) ): distance is ( sqrt{(2-0)^2 + (0 - (-1))^2} = sqrt{4 + 1} = sqrt{5} )2. From ( (0, -1) ) to ( (-2, 0) ): distance is ( sqrt{(0 - (-2))^2 + (-1 - 0)^2} = sqrt{4 + 1} = sqrt{5} )3. From ( (-2, 0) ) to ( (0, 1) ): distance is ( sqrt{(-2 - 0)^2 + (0 - 1)^2} = sqrt{4 + 1} = sqrt{5} )4. From ( (0, 1) ) to ( (2, 0) ): distance is ( sqrt{(0 - 2)^2 + (1 - 0)^2} = sqrt{4 + 1} = sqrt{5} )So, all sides are equal, so it's a rhombus.Now, to find the area of this rhombus, I can use the formula:Area = (diagonal1 * diagonal2) / 2So, what are the lengths of the diagonals?Looking at the points:- One diagonal connects ( (2, 0) ) and ( (-2, 0) ), which is along the a-axis. The length is ( 2 - (-2) = 4 ).- The other diagonal connects ( (0, 1) ) and ( (0, -1) ), which is along the b-axis. The length is ( 1 - (-1) = 2 ).So, diagonals are 4 and 2.Therefore, the area is ( (4 * 2)/2 = 4 ).Wait, but hold on, the options are 4, 8, 16, 32. So, 4 is an option, which is A.But wait, let me double-check because sometimes the area can be miscalculated.Alternatively, maybe I can use the shoelace formula to compute the area of the quadrilateral with vertices at ( (2, 0) ), ( (0, -1) ), ( (-2, 0) ), ( (0, 1) ).Let me list the coordinates in order:1. ( (2, 0) )2. ( (0, -1) )3. ( (-2, 0) )4. ( (0, 1) )5. Back to ( (2, 0) )Shoelace formula:Area = ( frac{1}{2} |(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_5) - (y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_5)| )Plugging in the values:First part:( x_1 y_2 = 2 * (-1) = -2 )( x_2 y_3 = 0 * 0 = 0 )( x_3 y_4 = (-2) * 1 = -2 )( x_4 y_5 = 0 * 0 = 0 )Sum: ( -2 + 0 + (-2) + 0 = -4 )Second part:( y_1 x_2 = 0 * 0 = 0 )( y_2 x_3 = (-1) * (-2) = 2 )( y_3 x_4 = 0 * 0 = 0 )( y_4 x_5 = 1 * 2 = 2 )Sum: ( 0 + 2 + 0 + 2 = 4 )So, Area = ( frac{1}{2} | -4 - 4 | = frac{1}{2} | -8 | = frac{1}{2} * 8 = 4 )So, the area is indeed 4. So, the answer should be A. 4.Wait, but let me think again. The problem says "the area of the region formed by the moving point ( P(a, b) )". So, is this region indeed the rhombus with area 4? Or did I make a mistake in interpreting the inequalities?Wait, let me check the inequalities again.We had four inequalities:1. ( a leq 2 + 2b )2. ( a leq 2 - 2b )3. ( a geq -2 - 2b )4. ( a geq -2 + 2b )So, plotting these lines, the region is bounded between these four lines, forming a rhombus with vertices at ( (2, 0) ), ( (0, 1) ), ( (-2, 0) ), ( (0, -1) ). So, yes, that's correct.But wait, the shoelace formula gave me 4, but let me visualize the shape. It's a diamond with diagonals of length 4 and 2, so area is (4*2)/2=4. So, that's correct.But wait, the options are A.4, B.8, C.16, D.32. So, 4 is an option, but I have a feeling that maybe I missed something because 4 seems small.Wait, let me think about the original condition. The inequality ( a x - 2 b y leq 2 ) must hold for all ( |x| leq 1 ) and ( |y| leq 1 ). So, the maximum of ( a x - 2 b y ) is 2. So, the region of ( (a, b) ) is the set of points where the maximum of ( a x - 2 b y ) over the square is less than or equal to 2.But I found that the region is a rhombus with area 4. So, is that correct?Alternatively, maybe I should consider the dual problem. The condition ( a x - 2 b y leq 2 ) for all ( |x| leq 1 ) and ( |y| leq 1 ) can be rewritten as ( max_{|x| leq 1, |y| leq 1} (a x - 2 b y) leq 2 ).As I did before, the maximum occurs at the corners, so I set the maximum at each corner to be less than or equal to 2, leading to four inequalities. Then, solving those inequalities gives me the region for ( (a, b) ), which is a rhombus with area 4.But let me think about scaling. The original square in the ( xy )-plane is of size 2x2. The transformation ( a x - 2 b y ) can be thought of as a linear functional. The condition that this functional is bounded by 2 over the unit square leads to a certain region in the ( ab )-plane.Wait, another approach: the set of points ( (a, b) ) such that ( a x - 2 b y leq 2 ) for all ( |x| leq 1 ), ( |y| leq 1 ) is equivalent to the set of ( (a, b) ) such that the maximum of ( a x - 2 b y ) over the square is less than or equal to 2.As I found earlier, the maximum is achieved at the corners, so the four inequalities must hold.Alternatively, perhaps I can use the concept of dual norms or something, but that might be overcomplicating.Alternatively, maybe I can think of the inequality ( a x - 2 b y leq 2 ) as defining a half-plane for each ( (x, y) ), and the intersection of all these half-planes is the region where ( (a, b) ) must lie.But since ( x ) and ( y ) are bounded, the intersection of all these half-planes is the region defined by the four inequalities I found earlier.So, I think my initial approach is correct.But just to be thorough, let me consider another point in the region and see if it satisfies all inequalities.For example, take ( a = 0 ), ( b = 0 ). Then, ( a x - 2 b y = 0 leq 2 ), which is true. So, the origin is inside the region.What about ( a = 1 ), ( b = 0 ). Then, ( a x - 2 b y = x leq 1 leq 2 ), which is true. So, ( (1, 0) ) is inside.What about ( a = 2 ), ( b = 0 ). Then, ( a x - 2 b y = 2x ). The maximum over ( |x| leq 1 ) is 2, so ( 2 leq 2 ), which is true. So, ( (2, 0) ) is on the boundary.Similarly, ( a = 0 ), ( b = 1 ). Then, ( a x - 2 b y = -2 y ). The maximum over ( |y| leq 1 ) is 2, so ( 2 leq 2 ), which is true. So, ( (0, 1) ) is on the boundary.So, all these points are correctly included or on the boundary.Wait, but if I take ( a = 1 ), ( b = 1 ), then ( a x - 2 b y = x - 2 y ). The maximum over ( |x| leq 1 ), ( |y| leq 1 ) would be when ( x = 1 ), ( y = -1 ), giving ( 1 - 2*(-1) = 1 + 2 = 3 ), which is greater than 2. So, ( (1, 1) ) is outside the region, which is consistent with our earlier result because ( (1, 1) ) is outside the rhombus.Similarly, ( a = 1 ), ( b = 0.5 ): ( a x - 2 b y = x - y ). The maximum over the square would be when ( x = 1 ), ( y = -1 ), giving ( 1 - (-1) = 2 ), which is equal to 2, so ( (1, 0.5) ) is on the boundary.Wait, but according to our earlier region, ( (1, 0.5) ) is inside the rhombus because the rhombus extends to ( a = 2 ) and ( b = 1 ). Wait, no, actually, the rhombus has vertices at ( (2, 0) ), ( (0, 1) ), ( (-2, 0) ), ( (0, -1) ). So, ( (1, 0.5) ) lies on the line connecting ( (2, 0) ) and ( (0, 1) ), so it's on the boundary.Wait, let me check if ( (1, 0.5) ) satisfies all four inequalities:1. ( a leq 2 + 2b ): ( 1 leq 2 + 2*(0.5) = 2 + 1 = 3 ). True.2. ( a leq 2 - 2b ): ( 1 leq 2 - 2*(0.5) = 2 - 1 = 1 ). So, ( 1 leq 1 ). True.3. ( a geq -2 - 2b ): ( 1 geq -2 - 2*(0.5) = -2 -1 = -3 ). True.4. ( a geq -2 + 2b ): ( 1 geq -2 + 2*(0.5) = -2 +1 = -1 ). True.So, it's on the boundary because it satisfies the second inequality as equality.So, seems consistent.Therefore, I think my initial conclusion is correct. The area is 4, which is option A.But wait, hold on. The problem says "the area of the region formed by the moving point ( P(a, b) )". So, is the region really the rhombus with area 4? Or is there a scaling factor I'm missing?Wait, another way to think about this is using the concept of the dual space. The condition ( a x - 2 b y leq 2 ) for all ( |x| leq 1 ), ( |y| leq 1 ) can be interpreted as the point ( (a, b) ) lying inside the dual of the unit square under the linear functional ( a x - 2 b y ).But perhaps that's overcomplicating. Alternatively, maybe I can use the concept of the maximum of a linear function over a convex set.Wait, the maximum of ( a x - 2 b y ) over ( |x| leq 1 ), ( |y| leq 1 ) is equal to ( max { a - 2b, a + 2b, -a - 2b, -a + 2b } ). So, setting each of these to be less than or equal to 2 gives the four inequalities.So, the region is indeed the intersection of these four half-planes, which is the rhombus with vertices at ( (2, 0) ), ( (0, 1) ), ( (-2, 0) ), ( (0, -1) ), and area 4.Therefore, the answer should be A. 4.But wait, another thought: the original square is 2x2, area 4. The dual region is also 4? That seems interesting. Maybe it's a coincidence or maybe it's because of the specific coefficients.Alternatively, maybe I can think of this as a linear transformation. The original square is transformed by the linear functional ( a x - 2 b y ), and the condition is that the maximum value is 2. So, perhaps the region is scaled accordingly.But regardless, through the process of finding the intersection points and computing the area, I arrived at 4, which is one of the options.Therefore, I think the correct answer is A. 4.**Final Answer**boxed{A}

answer：Okay, so I have this problem here: I need to consider two positive even integers less than 15. They don't have to be distinct, which means they can be the same number. Then, for each pair of these numbers, I have to calculate the sum of the two numbers plus their product. Finally, I need to figure out how many different possible values this calculation can result in. Hmm, let me break this down. First, let's list all the positive even integers less than 15. Since they have to be even, starting from 2, the numbers would be 2, 4, 6, 8, 10, 12, and 14. So that's seven numbers in total. Now, since the two numbers don't have to be distinct, I can pair each number with itself and with every other number. So, the total number of pairs would be 7 (for the same numbers) plus 7 choose 2 (for different numbers). Wait, 7 choose 2 is 21, so total pairs would be 28. But I don't think I need to list all 28 pairs because some of them might result in the same value when I compute the sum plus the product. So, my goal is to find how many unique results I can get from these 28 pairs.Let me denote the two numbers as x and y, both of which are even numbers from the list: 2, 4, 6, 8, 10, 12, 14. The expression I need to compute is x + y + xy. Hmm, that looks a bit like factoring. Let me see if I can rewrite this expression to make it easier to handle.If I factor the expression, x + y + xy, I can factor out an x from the first and third terms: x(1 + y) + y. Alternatively, I can factor out a y: y(1 + x) + x. Hmm, not sure if that helps directly. Wait, another approach: if I add 1 to both sides, maybe I can factor it as (x + 1)(y + 1) - 1. Let me check:(x + 1)(y + 1) = xy + x + y + 1. So, if I subtract 1 from both sides, I get (x + 1)(y + 1) - 1 = xy + x + y. Yes! So, the expression x + y + xy is equal to (x + 1)(y + 1) - 1. That seems useful because now I can think of it as the product of (x + 1) and (y + 1) minus 1.So, maybe instead of computing x + y + xy for each pair, I can compute (x + 1)(y + 1) - 1. That might make it easier to see if different pairs can result in the same value.Let me write down the possible values of (x + 1) and (y + 1). Since x and y are even numbers from 2 to 14, adding 1 to each gives me 3, 5, 7, 9, 11, 13, 15. So, (x + 1) and (y + 1) can be 3, 5, 7, 9, 11, 13, or 15.Therefore, the expression becomes (a)(b) - 1, where a and b are elements from the set {3, 5, 7, 9, 11, 13, 15}. So, now my problem reduces to finding how many unique products a*b there are, and then subtracting 1 from each to get the final result. But since subtraction is a linear operation, the number of unique values of a*b - 1 will be the same as the number of unique a*b, because subtracting 1 from each just shifts all the values down by 1, but doesn't affect uniqueness.Therefore, if I can find how many unique products a*b exist where a and b are in {3, 5, 7, 9, 11, 13, 15}, then that will be the number of unique values for the original expression.So, now my task is to compute all possible products a*b where a and b are in that set, and count the unique ones.Let me list all possible products:First, let me list the set again: {3, 5, 7, 9, 11, 13, 15}.I can create a multiplication table for these numbers:- 3*3 = 9- 3*5 = 15- 3*7 = 21- 3*9 = 27- 3*11 = 33- 3*13 = 39- 3*15 = 45- 5*3 = 15 (duplicate)- 5*5 = 25- 5*7 = 35- 5*9 = 45 (duplicate)- 5*11 = 55- 5*13 = 65- 5*15 = 75- 7*3 = 21 (duplicate)- 7*5 = 35 (duplicate)- 7*7 = 49- 7*9 = 63- 7*11 = 77- 7*13 = 91- 7*15 = 105- 9*3 = 27 (duplicate)- 9*5 = 45 (duplicate)- 9*7 = 63 (duplicate)- 9*9 = 81- 9*11 = 99- 9*13 = 117- 9*15 = 135- 11*3 = 33 (duplicate)- 11*5 = 55 (duplicate)- 11*7 = 77 (duplicate)- 11*9 = 99 (duplicate)- 11*11 = 121- 11*13 = 143- 11*15 = 165- 13*3 = 39 (duplicate)- 13*5 = 65 (duplicate)- 13*7 = 91 (duplicate)- 13*9 = 117 (duplicate)- 13*11 = 143 (duplicate)- 13*13 = 169- 13*15 = 195- 15*3 = 45 (duplicate)- 15*5 = 75 (duplicate)- 15*7 = 105 (duplicate)- 15*9 = 135 (duplicate)- 15*11 = 165 (duplicate)- 15*13 = 195 (duplicate)- 15*15 = 225Now, let me list all these products and note the duplicates:From 3:9, 15, 21, 27, 33, 39, 45From 5:15, 25, 35, 45, 55, 65, 75From 7:21, 35, 49, 63, 77, 91, 105From 9:27, 45, 63, 81, 99, 117, 135From 11:33, 55, 77, 99, 121, 143, 165From 13:39, 65, 91, 117, 143, 169, 195From 15:45, 75, 105, 135, 165, 195, 225Now, let me compile all these products into a single list and identify the unique ones.Starting from the smallest:9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 55, 63, 65, 75, 77, 81, 91, 99, 105, 117, 121, 135, 143, 165, 169, 195, 225.Wait, let me count how many unique products there are. I'll list them in order:1. 92. 153. 214. 255. 276. 337. 358. 399. 4510. 4911. 5512. 6313. 6514. 7515. 7716. 8117. 9118. 9919. 10520. 11721. 12122. 13523. 14324. 16525. 16926. 19527. 225So, that's 27 unique products. Therefore, the number of unique values for (x + 1)(y + 1) is 27. Since the original expression is (x + 1)(y + 1) - 1, subtracting 1 from each of these 27 products will result in 27 unique values as well, because subtracting 1 doesn't cause any overlaps. For example, if two products were the same, subtracting 1 would still leave them the same, but since all products are unique, subtracting 1 will also result in unique values.Therefore, the number of different possible values resulting from x + y + xy is 27.Wait, hold on a second. Let me double-check because sometimes when you subtract 1, if two different products are consecutive numbers, subtracting 1 could potentially cause overlaps. But in this case, looking at the list of products, they are all unique and spaced apart by at least 1, so subtracting 1 won't cause any overlaps. For example, 9 becomes 8, 15 becomes 14, 21 becomes 20, and so on. Since all the products are unique and each is at least 1 apart, subtracting 1 will still keep them unique.Therefore, the number of unique results is indeed 27.But just to be thorough, let me check a few specific cases:Take x=2, y=2: (2+1)(2+1)-1=9-1=8x=2, y=4: (3)(5)-1=15-1=14x=2, y=6: 3*7-1=21-1=20x=2, y=8: 3*9-1=27-1=26x=2, y=10: 3*11-1=33-1=32x=2, y=12: 3*13-1=39-1=38x=2, y=14: 3*15-1=45-1=44Similarly, x=4, y=4: 5*5-1=25-1=24x=4, y=6: 5*7-1=35-1=34x=4, y=8:5*9-1=45-1=44 (Wait, 44 was already achieved by x=2, y=14)Wait, hold on, so 44 appears twice. So, does that mean 44 is achieved by both (2,14) and (4,8)? So, that would mean that the value 44 is achieved by two different pairs, so it's only counted once.But in my earlier count, I thought all products were unique, but in reality, when I subtract 1, some products might result in the same value. Wait, but no, because the products themselves are unique. Let me see:Wait, 3*15=45, and 5*9=45. So, both pairs (2,14) and (4,8) result in 45 when computing (x+1)(y+1). Therefore, when subtracting 1, both would result in 44. So, 44 is achieved by two different pairs, but it's still just one unique value.Wait, so in my earlier count, I considered the products (x+1)(y+1) as unique, but in reality, some products are duplicates, like 45. So, actually, the number of unique products is less than 27 because some products repeat.Wait, so my initial approach was wrong because I thought all products were unique, but actually, some products are duplicates, such as 15, 21, 27, etc., which are achieved by multiple pairs.So, I need to recount the unique products.Let me go back to the multiplication table and list all the products, noting duplicates.From 3:9, 15, 21, 27, 33, 39, 45From 5:15, 25, 35, 45, 55, 65, 75From 7:21, 35, 49, 63, 77, 91, 105From 9:27, 45, 63, 81, 99, 117, 135From 11:33, 55, 77, 99, 121, 143, 165From 13:39, 65, 91, 117, 143, 169, 195From 15:45, 75, 105, 135, 165, 195, 225Now, let's list all the products and count the unique ones:Starting from the smallest:9 (from 3*3)15 (from 3*5 and 5*3)21 (from 3*7 and 7*3)25 (from 5*5)27 (from 3*9 and 9*3)33 (from 3*11 and 11*3)35 (from 5*7 and 7*5)39 (from 3*13 and 13*3)45 (from 3*15, 5*9, 9*5, 15*3)49 (from 7*7)55 (from 5*11 and 11*5)63 (from 7*9 and 9*7)65 (from 5*13 and 13*5)75 (from 5*15 and 15*5)77 (from 7*11 and 11*7)81 (from 9*9)91 (from 7*13 and 13*7)99 (from 9*11 and 11*9)105 (from 7*15 and 15*7)117 (from 9*13 and 13*9)121 (from 11*11)135 (from 9*15 and 15*9)143 (from 11*13 and 13*11)165 (from 11*15 and 15*11)169 (from 13*13)195 (from 13*15 and 15*13)225 (from 15*15)So, compiling all these, the unique products are:9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 55, 63, 65, 75, 77, 81, 91, 99, 105, 117, 121, 135, 143, 165, 169, 195, 225.Wait, that's 27 unique products. But earlier, I thought that 45 was achieved by multiple pairs, but in the unique products list, it's only counted once. So, the number of unique products is 27, which means the number of unique (x + 1)(y + 1) - 1 is also 27, because each product is unique, so subtracting 1 won't cause any overlaps.Wait, but earlier, I saw that both (2,14) and (4,8) give 45, which becomes 44. So, 44 is achieved by two different pairs, but it's only one unique value. So, in the unique products list, 45 is only counted once, so 44 is only counted once in the final result.Therefore, the number of unique values is indeed 27.But let me double-check by actually computing some of these expressions:Take x=2, y=2: 2 + 2 + (2*2) = 4 + 4 = 8x=2, y=4: 2 + 4 + 8 = 14x=2, y=6: 2 + 6 + 12 = 20x=2, y=8: 2 + 8 + 16 = 26x=2, y=10: 2 + 10 + 20 = 32x=2, y=12: 2 + 12 + 24 = 38x=2, y=14: 2 + 14 + 28 = 44x=4, y=4: 4 + 4 + 16 = 24x=4, y=6: 4 + 6 + 24 = 34x=4, y=8: 4 + 8 + 32 = 44x=4, y=10: 4 + 10 + 40 = 54x=4, y=12: 4 + 12 + 48 = 64x=4, y=14: 4 + 14 + 56 = 74x=6, y=6: 6 + 6 + 36 = 48x=6, y=8: 6 + 8 + 48 = 62x=6, y=10: 6 + 10 + 60 = 76x=6, y=12: 6 + 12 + 72 = 90x=6, y=14: 6 + 14 + 84 = 104x=8, y=8: 8 + 8 + 64 = 80x=8, y=10: 8 + 10 + 80 = 98x=8, y=12: 8 + 12 + 96 = 116x=8, y=14: 8 + 14 + 112 = 134x=10, y=10: 10 + 10 + 100 = 120x=10, y=12: 10 + 12 + 120 = 142x=10, y=14: 10 + 14 + 140 = 164x=12, y=12: 12 + 12 + 144 = 168x=12, y=14: 12 + 14 + 168 = 194x=14, y=14: 14 + 14 + 196 = 224Wait, hold on, when I compute x=4, y=8, I get 44, which is the same as x=2, y=14. So, 44 is achieved twice, but it's only one unique value. Similarly, let's see if any other duplicates occur.Looking at the computed values:8, 14, 20, 26, 32, 38, 44, 24, 34, 44, 54, 64, 74, 48, 62, 76, 90, 104, 80, 98, 116, 134, 120, 142, 164, 168, 194, 224.Wait, so let's list them all:8, 14, 20, 26, 32, 38, 44, 24, 34, 54, 64, 74, 48, 62, 76, 90, 104, 80, 98, 116, 134, 120, 142, 164, 168, 194, 224.Now, let's sort them:8, 14, 20, 24, 26, 32, 34, 38, 44, 48, 54, 62, 64, 74, 76, 80, 90, 98, 104, 116, 120, 134, 142, 164, 168, 194, 224.Now, let's count them:1. 82. 143. 204. 245. 266. 327. 348. 389. 4410. 4811. 5412. 6213. 6414. 7415. 7616. 8017. 9018. 9819. 10420. 11621. 12022. 13423. 14224. 16425. 16826. 19427. 224So, that's 27 unique values. Therefore, despite some products being achieved by multiple pairs, the number of unique results is indeed 27.Wait, but earlier, when I thought that (x + 1)(y + 1) had 27 unique products, I thought that the number of unique results would be 27, and now, when I actually compute the expressions, I also get 27 unique results. So, that confirms it.Therefore, the answer is 27.But just to be absolutely sure, let me check if any other duplicates exist in the computed values. For example, is 44 the only duplicate? Let me see:Looking at the list:8, 14, 20, 24, 26, 32, 34, 38, 44, 48, 54, 62, 64, 74, 76, 80, 90, 98, 104, 116, 120, 134, 142, 164, 168, 194, 224.Each number appears only once, except for 44, which was achieved twice but is only listed once. So, yes, 27 unique values.Therefore, the number of different possible values is 27.**Final Answer**The number of different possible values is boxed{27}.

answer：Alright, so I have this system of inequalities to solve over the set of integer pairs (a, b). The inequalities are:1. (11 > 2a - b)2. (25 > 2b - a)3. (42 < 3b - a)4. (46 < 2a + b)Hmm, okay. Let me try to figure this out step by step. I think the best approach is to rewrite each inequality to express one variable in terms of the other, and then find the overlapping integer solutions.Starting with the first inequality: (11 > 2a - b). Let me rearrange this to solve for b.Adding b to both sides: (11 + b > 2a)Then, subtracting 11: (b > 2a - 11)So, that's one boundary condition: b must be greater than (2a - 11).Moving on to the second inequality: (25 > 2b - a). Let's solve for a.Adding a to both sides: (25 + a > 2b)Subtracting 25: (a > 2b - 25)So, a must be greater than (2b - 25).Third inequality: (42 < 3b - a). Let's solve for a.Adding a to both sides: (42 + a < 3b)Subtracting 42: (a < 3b - 42)So, a must be less than (3b - 42).Fourth inequality: (46 < 2a + b). Let's solve for b.Subtracting 2a: (46 - 2a < b)So, b must be greater than (46 - 2a).Alright, so now I have four inequalities:1. (b > 2a - 11)2. (a > 2b - 25)3. (a < 3b - 42)4. (b > 46 - 2a)Hmm, so I need to find integer pairs (a, b) that satisfy all four inequalities.Maybe I can combine these inequalities to find bounds on a and b.Looking at inequalities 2 and 3: (2b - 25 < a < 3b - 42)So, (2b - 25 < 3b - 42)Let me solve this:Subtract 2b from both sides: (-25 < b - 42)Add 42 to both sides: (17 < b)So, b must be greater than 17.Similarly, from inequality 4: (b > 46 - 2a)But since b > 17, let's see what that implies for a.From inequality 2: (a > 2b - 25)Since b > 17, let's plug in b = 18 as a starting point.If b = 18, then a > 2*18 -25 = 36 -25 = 11. So, a >11.From inequality 3: a < 3*18 -42 = 54 -42 = 12. So, a <12.But a must be an integer, so a must be 12? Wait, but a <12, so a must be less than 12. But a >11, so a must be 12? Wait, that's conflicting.Wait, a >11 and a <12. So, a must be 12? But 12 is not less than 12. So, no solution for b=18.Wait, maybe I made a mistake.Wait, if b=18, then a must be greater than 11 and less than 12. Since a is an integer, there's no integer a that satisfies 11 < a <12. So, no solution for b=18.Hmm, okay. Let's try b=19.From inequality 2: a > 2*19 -25 = 38 -25 =13From inequality 3: a < 3*19 -42 =57 -42=15So, a must be greater than 13 and less than15. So, a can be 14.So, a=14, b=19. Let's check all inequalities.1. (11 > 2a - b = 28 -19=9). 11>9, which is true.2. (25 > 2b -a=38 -14=24). 25>24, true.3. (42 < 3b -a=57 -14=43). 42<43, true.4. (46 < 2a +b=28 +19=47). 46<47, true.So, (14,19) is a solution.Let me check if there are more solutions.Next, b=20.From inequality 2: a >2*20 -25=40 -25=15From inequality 3: a <3*20 -42=60 -42=18So, a must be greater than15 and less than18. So, a=16,17.Check a=16, b=20.1. (11 >32 -20=12). 11>12? No, that's false.So, (16,20) is invalid.a=17, b=20.1. (11 >34 -20=14). 11>14? No, false.So, no solutions for b=20.Wait, maybe I need to check the other inequalities as well.Wait, for a=16, b=20:Inequality 1: 11 >32 -20=12? 11>12? No.So, invalid.Similarly, a=17, b=20: 11 >34 -20=14? 11>14? No.So, no solutions for b=20.Let's try b=21.From inequality 2: a >2*21 -25=42 -25=17From inequality 3: a <3*21 -42=63 -42=21So, a must be greater than17 and less than21. So, a=18,19,20.Check each:a=18, b=21:1. 11 >36 -21=15? 11>15? No.a=19, b=21:1. 11 >38 -21=17? 11>17? No.a=20, b=21:1. 11 >40 -21=19? 11>19? No.So, no solutions for b=21.Hmm, seems like as b increases, 2a - b is increasing, so the first inequality is getting harder to satisfy.Wait, maybe I should check lower b.Wait, earlier I found b>17, so b must be at least 18. But when b=18, no solution. b=19, one solution. b=20,21: no solutions.Wait, maybe check b=17? But earlier, from inequality 2 and 3, we had b>17, so b must be at least 18.Wait, but let me double-check.From inequalities 2 and 3: 2b -25 < a <3b -42So, 2b -25 <3b -42Which simplifies to -25 <b -42 => 17 <b.So, b must be greater than17, so b>=18.So, b=18,19,20,...But for b=18, no solution.b=19: solution at a=14.Wait, but a=14, b=19.Wait, let me check the other inequalities for b=19.Inequality 4: b >46 -2a.So, 19 >46 -2a.So, 2a >46 -19=27.So, a>13.5. Since a is integer, a>=14.Which matches our earlier result.So, a=14 is the only solution for b=19.Wait, let me check b=19, a=14.Inequality 1: 11 >28 -19=9. True.Inequality 2:25 >38 -14=24. True.Inequality 3:42 <57 -14=43. True.Inequality 4:46 <28 +19=47. True.So, that works.Now, let's check b=22.From inequality 2: a >2*22 -25=44 -25=19From inequality 3: a <3*22 -42=66 -42=24So, a must be greater than19 and less than24. So, a=20,21,22,23.Check each:a=20, b=22:1. 11 >40 -22=18? 11>18? No.a=21, b=22:1. 11 >42 -22=20? 11>20? No.a=22, b=22:1. 11 >44 -22=22? 11>22? No.a=23, b=22:1. 11 >46 -22=24? 11>24? No.So, no solutions for b=22.Hmm, seems like as b increases beyond 19, the first inequality becomes harder to satisfy.Wait, maybe I should check b=16? But earlier, from inequalities 2 and 3, b>17, so b=16 is invalid.Wait, but let me check b=17 just in case.From inequality 2: a >2*17 -25=34 -25=9From inequality 3: a <3*17 -42=51 -42=9So, a must be greater than9 and less than9. Which is impossible. So, no solution for b=17.So, b must be at least18.Wait, so only b=19 gives a solution.But let me check b=19, a=14.Wait, is there another a for b=19?From inequality 2: a>13, so a=14,15,...From inequality 3: a<15, so a=14.So, only a=14.Wait, let me check if a=15, b=19.Inequality 1:11 >30 -19=11. 11>11? No, it's equal. So, not greater. So, invalid.So, only a=14.Wait, let me check if there are other b's.Wait, maybe I should check b=19, a=14.Wait, let me see if there are any other a's for b=19.Wait, from inequality 2: a>13, from inequality 3: a<15. So, a=14 is the only integer.So, only one solution: (14,19).Wait, but let me check b=19, a=14.Inequality 4:46 <2a +b=28 +19=47. 46<47, which is true.Okay, so that works.Wait, let me see if there are any other solutions.Wait, maybe I should check b=19, a=14.Wait, is there a way to have a higher a?Wait, from inequality 3: a <3b -42=57 -42=15. So, a<15.So, a=14 is the maximum.Wait, let me check if a=14, b=19 is the only solution.Wait, let me check b=19, a=14.Yes, that's the only one.Wait, let me check b=19, a=14.Wait, let me see if there are any other possible a's.Wait, from inequality 2: a>2b -25=38 -25=13.So, a>13, so a=14,15,...But from inequality 3: a<15, so a=14.So, only a=14.Wait, so that's the only solution.But let me check if there are any other b's.Wait, let me try b=20 again.From inequality 2: a>25, from inequality 3: a<18.Wait, that can't be, because 25>18, so no solution.Wait, that's not possible.Wait, no, wait, for b=20:From inequality 2: a>2*20 -25=40 -25=15From inequality 3: a<3*20 -42=60 -42=18So, a must be greater than15 and less than18. So, a=16,17.But when a=16, b=20:Inequality 1:11 >32 -20=12? 11>12? No.Similarly, a=17, b=20:11 >34 -20=14? 11>14? No.So, no solutions.Wait, maybe I should check b=23.From inequality 2: a>2*23 -25=46 -25=21From inequality 3: a<3*23 -42=69 -42=27So, a must be greater than21 and less than27. So, a=22,23,24,25,26.Check each:a=22, b=23:1. 11 >44 -23=21? 11>21? No.a=23, b=23:1. 11 >46 -23=23? 11>23? No.a=24, b=23:1. 11 >48 -23=25? 11>25? No.a=25, b=23:1. 11 >50 -23=27? 11>27? No.a=26, b=23:1. 11 >52 -23=29? 11>29? No.So, no solutions for b=23.Hmm, seems like as b increases, the first inequality is not satisfied.Wait, maybe I should check lower b's, but earlier we saw that b must be greater than17.Wait, let me check b=19, a=14.Is there a way to have a lower a?Wait, from inequality 2: a>13, so a=14 is the minimum.So, that's the only solution.Wait, let me check if a=14, b=19 is the only solution.Wait, let me check b=19, a=14.Yes, that's the only one.Wait, let me check if there are any other solutions with a lower b.Wait, b must be greater than17, so b=18,19,20,...But for b=18, no solution.b=19: solution.b=20: no.b=21: no.b=22: no.b=23: no.So, only b=19 gives a solution.Wait, let me check b=19, a=14.Yes, that's the only solution.Wait, let me see if I can find any other solutions.Wait, maybe I should check if a=14, b=19 is the only solution.Wait, let me see.From inequality 4: b >46 -2a.So, b >46 -28=18.So, b>18, which is consistent with b=19.So, that's the only solution.Wait, let me check if a=14, b=19 is the only solution.Yes, I think so.So, the only integer pair that satisfies all four inequalities is (14,19).**Final Answer**The integer pair satisfying all the inequalities is boxed{(14, 19)}.

answer：Alright, so I have this function f(x) = x + a/(x - 2) where x is greater than 2. The graph of this function passes through the point A(3,7). I need to find the minimum value of this function. Hmm, okay, let me think about how to approach this step by step.First, since the graph passes through point A(3,7), that means when x is 3, f(x) is 7. So I can plug these values into the function to find the value of 'a'. Let me write that down:f(3) = 3 + a/(3 - 2) = 7.Simplifying the denominator, 3 - 2 is 1, so this becomes:3 + a/1 = 7.Which simplifies to:3 + a = 7.Subtracting 3 from both sides gives:a = 4.Okay, so now I know that a is 4. Therefore, the function becomes:f(x) = x + 4/(x - 2).Now, I need to find the minimum value of this function for x > 2. Since it's a function of a single variable, I can use calculus to find its minimum. Specifically, I can take the derivative of f(x) with respect to x, set it equal to zero, and solve for x. That should give me the critical points, and then I can determine which one gives the minimum value.Let's compute the derivative f'(x). The function is f(x) = x + 4/(x - 2). The derivative of x is 1, and the derivative of 4/(x - 2) can be found using the power rule or the quotient rule. Let me use the power rule for simplicity.First, rewrite 4/(x - 2) as 4*(x - 2)^(-1). Then, the derivative is:d/dx [4*(x - 2)^(-1)] = 4*(-1)*(x - 2)^(-2)*1 = -4/(x - 2)^2.So, putting it all together, the derivative f'(x) is:f'(x) = 1 - 4/(x - 2)^2.Now, to find the critical points, set f'(x) equal to zero:1 - 4/(x - 2)^2 = 0.Let me solve for x. First, move the second term to the other side:1 = 4/(x - 2)^2.Multiply both sides by (x - 2)^2 to eliminate the denominator:(x - 2)^2 = 4.Take the square root of both sides:x - 2 = ±2.So, solving for x:x = 2 ± 2.But wait, the original function is defined for x > 2, so x must be greater than 2. Therefore, x = 2 + 2 = 4 is the valid solution. The other solution, x = 2 - 2 = 0, is less than 2 and thus not in the domain we're considering.So, the critical point is at x = 4. Now, I need to check whether this critical point is a minimum. Since the function is defined for x > 2, and we have only one critical point, we can analyze the behavior of the derivative around x = 4 to determine if it's a minimum.Alternatively, we can use the second derivative test. Let me compute the second derivative f''(x) to check the concavity at x = 4.First, f'(x) = 1 - 4/(x - 2)^2. Let's find f''(x):The derivative of 1 is 0, and the derivative of -4/(x - 2)^2 is:-4 * d/dx [(x - 2)^(-2)] = -4*(-2)*(x - 2)^(-3)*1 = 8/(x - 2)^3.So, f''(x) = 8/(x - 2)^3.Now, evaluate f''(x) at x = 4:f''(4) = 8/(4 - 2)^3 = 8/(2)^3 = 8/8 = 1.Since f''(4) is positive (1 > 0), the function is concave upward at x = 4, which means that this critical point is indeed a local minimum. Given that the function tends to infinity as x approaches 2 from the right and as x approaches infinity, this local minimum is also the global minimum on the domain x > 2.Therefore, the minimum value of the function occurs at x = 4. Now, let's compute f(4) to find the minimum value.f(4) = 4 + 4/(4 - 2) = 4 + 4/2 = 4 + 2 = 6.So, the minimum value of the function is 6.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, plugging a = 4 into f(x) gives f(x) = x + 4/(x - 2). Then, taking the derivative, f'(x) = 1 - 4/(x - 2)^2. Setting that equal to zero gives 1 = 4/(x - 2)^2, so (x - 2)^2 = 4, leading to x = 4. Then, f''(x) = 8/(x - 2)^3, which at x = 4 is 8/(2)^3 = 1, which is positive, confirming a minimum. Finally, f(4) = 4 + 4/(2) = 4 + 2 = 6. All steps seem correct.Alternatively, I can also verify by testing points around x = 4 to see if the function indeed has a minimum there. Let's pick x = 3 and x = 5.f(3) = 3 + 4/(3 - 2) = 3 + 4 = 7.f(5) = 5 + 4/(5 - 2) = 5 + 4/3 ≈ 5 + 1.333 ≈ 6.333.So, f(3) is 7, f(4) is 6, and f(5) is approximately 6.333. This suggests that at x = 4, the function reaches a lower value than at x = 3 and x = 5, which supports that x = 4 is indeed the point of minimum.Another way to think about this function is that it's a combination of a linear term x and a reciprocal term 4/(x - 2). The linear term increases as x increases, while the reciprocal term decreases as x increases. There's a balance point where the rate of increase of the linear term equals the rate of decrease of the reciprocal term, which is exactly where the derivative is zero. That balance point is at x = 4, resulting in the minimum value.I can also consider the function in terms of transformations. The function f(x) = x + 4/(x - 2) can be seen as a transformation of the basic reciprocal function 1/x. However, since it's combined with a linear function, the behavior is a bit more complex, but the calculus approach still applies.Just to ensure thoroughness, let me consider the limits of the function as x approaches 2 from the right and as x approaches infinity.As x approaches 2 from the right, the term 4/(x - 2) approaches positive infinity, so f(x) approaches infinity.As x approaches infinity, the term x dominates, so f(x) approaches infinity as well.Therefore, the function tends to infinity at both ends of its domain, which means that the critical point we found at x = 4 must indeed be the global minimum.In summary, by substituting the given point into the function, we found the value of 'a' to be 4. Then, by taking the derivative and analyzing the critical points, we determined that the function has a minimum at x = 4, and evaluating the function at this point gave us the minimum value of 6.**Final Answer**The minimum value of the function is boxed{6}.

answer：Okay, so I need to find the length of the chord that's cut off by a given circle from a given line. Hmm, let me first understand what the problem is asking.The circle is given in parametric form:[begin{cases}x = 2 + 5costheta y = 1 + 5sinthetaend{cases}]where (theta) is the parameter. I remember that parametric equations of a circle are usually in the form (x = h + rcostheta) and (y = k + rsintheta), where ((h, k)) is the center and (r) is the radius. So, in this case, the center of the circle must be at ((2, 1)) and the radius is 5. Got that.The line is also given in parametric form:[begin{cases}x = -2 + 4t y = -1 - 3tend{cases}]where (t) is the parameter. I need to find the points where this line intersects the circle, and then find the distance between those two points, which will be the length of the chord.Alright, so the plan is:1. Convert the parametric equations of the line into a standard linear equation (maybe slope-intercept form or something else).2. Substitute this into the equation of the circle to find the points of intersection.3. Solve for the parameter (t), which will give me the points.4. Once I have the two points, calculate the distance between them to get the chord length.Let me start by converting the parametric equations of the line into a standard form. The parametric equations are:[x = -2 + 4t quad text{and} quad y = -1 - 3t]I can solve for (t) in the first equation:[x = -2 + 4t implies x + 2 = 4t implies t = frac{x + 2}{4}]Now, substitute this expression for (t) into the equation for (y):[y = -1 - 3left(frac{x + 2}{4}right)]Let me simplify this:[y = -1 - frac{3x + 6}{4} = -1 - frac{3x}{4} - frac{6}{4} = -1 - frac{3x}{4} - frac{3}{2}]Combine the constants:[-1 - frac{3}{2} = -frac{5}{2}]So, the equation becomes:[y = -frac{3x}{4} - frac{5}{2}]Alternatively, I can write this as:[y = -frac{3}{4}x - frac{5}{2}]Alright, so that's the equation of the line in slope-intercept form. Now, I need to find where this line intersects the circle. The circle has center ((2, 1)) and radius 5, so its equation is:[(x - 2)^2 + (y - 1)^2 = 25]Now, substitute the expression for (y) from the line into the circle's equation:[(x - 2)^2 + left(-frac{3}{4}x - frac{5}{2} - 1right)^2 = 25]Simplify the (y) term:[-frac{3}{4}x - frac{5}{2} - 1 = -frac{3}{4}x - frac{7}{2}]So, the equation becomes:[(x - 2)^2 + left(-frac{3}{4}x - frac{7}{2}right)^2 = 25]Now, let's compute each term.First, expand ((x - 2)^2):[(x - 2)^2 = x^2 - 4x + 4]Next, expand (left(-frac{3}{4}x - frac{7}{2}right)^2). Let me compute this step by step.First, square each term:[left(-frac{3}{4}xright)^2 = frac{9}{16}x^2][left(-frac{7}{2}right)^2 = frac{49}{4}]Now, the cross term:[2 times left(-frac{3}{4}xright) times left(-frac{7}{2}right) = 2 times frac{21}{8}x = frac{21}{4}x]So, putting it all together:[left(-frac{3}{4}x - frac{7}{2}right)^2 = frac{9}{16}x^2 + frac{21}{4}x + frac{49}{4}]Now, substitute back into the circle equation:[(x^2 - 4x + 4) + left(frac{9}{16}x^2 + frac{21}{4}x + frac{49}{4}right) = 25]Combine like terms. Let's convert all terms to have a denominator of 16 to make it easier.First term: (x^2 = frac{16}{16}x^2)Second term: (-4x = -frac{64}{16}x)Third term: (4 = frac{64}{16})Fourth term: (frac{9}{16}x^2)Fifth term: (frac{21}{4}x = frac{84}{16}x)Sixth term: (frac{49}{4} = frac{196}{16})So, substituting:[frac{16}{16}x^2 - frac{64}{16}x + frac{64}{16} + frac{9}{16}x^2 + frac{84}{16}x + frac{196}{16} = 25]Combine like terms:- (x^2) terms: (frac{16}{16} + frac{9}{16} = frac{25}{16}x^2)- (x) terms: (-frac{64}{16} + frac{84}{16} = frac{20}{16}x = frac{5}{4}x)- Constant terms: (frac{64}{16} + frac{196}{16} = frac{260}{16} = frac{65}{4})So, the equation becomes:[frac{25}{16}x^2 + frac{5}{4}x + frac{65}{4} = 25]Multiply both sides by 16 to eliminate denominators:[25x^2 + 20x + 260 = 400]Subtract 400 from both sides:[25x^2 + 20x + 260 - 400 = 0]Simplify:[25x^2 + 20x - 140 = 0]Let me check my calculations so far to make sure I didn't make a mistake. Hmm, when I multiplied through by 16, let's verify:Original equation after substitution:[frac{25}{16}x^2 + frac{5}{4}x + frac{65}{4} = 25]Multiply each term by 16:- (25x^2)- (20x) (since (5/4 * 16 = 20))- (260) (since (65/4 * 16 = 260))- (25 * 16 = 400)Yes, so 25x² + 20x + 260 = 400, which leads to 25x² + 20x - 140 = 0.Now, let's simplify this quadratic equation. Maybe divide all terms by 5 to make it simpler:[5x^2 + 4x - 28 = 0]Wait, 25 divided by 5 is 5, 20 divided by 5 is 4, 140 divided by 5 is 28. So, yes, correct.Now, let's solve for x using the quadratic formula. The quadratic is:[5x^2 + 4x - 28 = 0]Quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Where (a = 5), (b = 4), (c = -28).Compute discriminant:[b^2 - 4ac = 16 - 4(5)(-28) = 16 + 560 = 576]Square root of discriminant:[sqrt{576} = 24]So, solutions:[x = frac{-4 pm 24}{10}]Compute both solutions:1. (x = frac{-4 + 24}{10} = frac{20}{10} = 2)2. (x = frac{-4 - 24}{10} = frac{-28}{10} = -2.8)So, the x-coordinates where the line intersects the circle are (x = 2) and (x = -2.8). Now, let's find the corresponding y-coordinates using the equation of the line (y = -frac{3}{4}x - frac{5}{2}).First, for (x = 2):[y = -frac{3}{4}(2) - frac{5}{2} = -frac{6}{4} - frac{5}{2} = -frac{3}{2} - frac{5}{2} = -frac{8}{2} = -4]So, one point is ((2, -4)).Next, for (x = -2.8). Hmm, let me convert -2.8 into a fraction to make calculation easier. Since 0.8 is 4/5, so -2.8 is -14/5.So, (x = -14/5). Plugging into the line equation:[y = -frac{3}{4}left(-frac{14}{5}right) - frac{5}{2}]Compute each term:First term:[-frac{3}{4} times -frac{14}{5} = frac{42}{20} = frac{21}{10} = 2.1]Second term:[-frac{5}{2} = -2.5]So, adding them together:[2.1 - 2.5 = -0.4]Convert -0.4 to a fraction: that's -2/5.So, the other point is ((-14/5, -2/5)). Alternatively, in decimal, that's (-2.8, -0.4).So, the two points of intersection are ((2, -4)) and ((-2.8, -0.4)). Now, I need to find the distance between these two points to get the length of the chord.The distance formula is:[d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]Let me assign:Point A: ((2, -4))Point B: ((-2.8, -0.4))Compute differences:[x_2 - x_1 = -2.8 - 2 = -4.8][y_2 - y_1 = -0.4 - (-4) = -0.4 + 4 = 3.6]Now, square these differences:[(-4.8)^2 = 23.04][(3.6)^2 = 12.96]Add them:[23.04 + 12.96 = 36]So, the distance is:[d = sqrt{36} = 6]Wait, that's a nice number. So, the length of the chord is 6 units.But let me double-check my calculations to make sure I didn't make any mistakes.First, the points of intersection: (2, -4) and (-2.8, -0.4). Let me verify these points lie on both the line and the circle.For point (2, -4):Check the line equation: (y = -frac{3}{4}x - frac{5}{2})Plug in x=2: (y = -frac{3}{4}*2 - frac{5}{2} = -frac{3}{2} - frac{5}{2} = -4). Correct.Check the circle equation: ((2 - 2)^2 + (-4 - 1)^2 = 0 + 25 = 25). Correct.For point (-2.8, -0.4):Check the line equation: (y = -frac{3}{4}*(-2.8) - frac{5}{2})Compute: (2.1 - 2.5 = -0.4). Correct.Check the circle equation: ((-2.8 - 2)^2 + (-0.4 - 1)^2 = (-4.8)^2 + (-1.4)^2 = 23.04 + 1.96 = 25). Correct.So, both points are indeed on the circle and the line. Now, the distance between them:Difference in x: -2.8 - 2 = -4.8Difference in y: -0.4 - (-4) = 3.6Squares: (-4.8)^2 = 23.04, (3.6)^2 = 12.96Sum: 36Square root: 6So, the chord length is 6. That seems correct.Alternatively, another way to find the chord length is by using the formula involving the distance from the center to the line and the radius. Maybe I can verify using that method.The formula is:[text{Length of chord} = 2sqrt{r^2 - d^2}]where (r) is the radius and (d) is the distance from the center to the line.Given that the radius (r = 5), I need to find (d), the distance from the center (2,1) to the line.First, let me write the equation of the line in standard form (Ax + By + C = 0).The line equation is:[y = -frac{3}{4}x - frac{5}{2}]Multiply both sides by 4 to eliminate fractions:[4y = -3x - 10]Bring all terms to one side:[3x + 4y + 10 = 0]So, (A = 3), (B = 4), (C = 10).The distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is:[d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]Plugging in the center (2,1):[d = frac{|3*2 + 4*1 + 10|}{sqrt{3^2 + 4^2}} = frac{|6 + 4 + 10|}{sqrt{9 + 16}} = frac{|20|}{5} = 4]So, the distance from the center to the line is 4.Now, using the chord length formula:[text{Length} = 2sqrt{5^2 - 4^2} = 2sqrt{25 - 16} = 2sqrt{9} = 2*3 = 6]So, that's another way to get the chord length as 6. This confirms my earlier result.Therefore, the length of the chord is 6 units.**Final Answer**The length of the chord is boxed{6}.

answer：Okay, so I have this limit to calculate:[lim _{n rightarrow infty} frac{n sqrt[4]{11 n}+sqrt{25 n^{4}-81}}{(n-7 sqrt{n}) sqrt{n^{2}-n+1}}]Hmm, limits involving sequences as n approaches infinity. I remember that for such limits, especially with polynomials and roots, we can often factor out the highest power of n in the numerator and denominator to simplify the expression.Let me start by looking at the numerator and the denominator separately.First, the numerator is:[n sqrt[4]{11 n} + sqrt{25 n^{4} - 81}]And the denominator is:[(n - 7 sqrt{n}) sqrt{n^{2} - n + 1}]I think I should handle each part step by step.Starting with the numerator:1. Let's look at the first term: ( n sqrt[4]{11 n} ).The fourth root of ( 11n ) can be written as ( (11n)^{1/4} ). So, this term is ( n times (11n)^{1/4} ).Let me simplify that:( n times (11n)^{1/4} = n times 11^{1/4} times n^{1/4} = 11^{1/4} times n^{1 + 1/4} = 11^{1/4} times n^{5/4} ).So, the first term is ( 11^{1/4} n^{5/4} ).2. Now, the second term in the numerator is ( sqrt{25n^4 - 81} ).That's the square root of ( 25n^4 - 81 ). Let me factor out the highest power inside the square root, which is ( n^4 ):( sqrt{25n^4 - 81} = sqrt{n^4 (25 - 81/n^4)} = n^2 sqrt{25 - 81/n^4} ).As n approaches infinity, ( 81/n^4 ) approaches 0, so this term behaves like ( n^2 times sqrt{25} = 5n^2 ).So, the numerator is approximately ( 11^{1/4} n^{5/4} + 5n^2 ).Now, let's look at the denominator:( (n - 7 sqrt{n}) sqrt{n^2 - n + 1} ).I can try to factor out the highest power of n from each part.First, let's handle ( n - 7 sqrt{n} ):Factor out n from the first term:( n - 7 sqrt{n} = n (1 - 7/n^{1/2}) ).Similarly, for the square root part ( sqrt{n^2 - n + 1} ):Factor out ( n^2 ) inside the square root:( sqrt{n^2 - n + 1} = sqrt{n^2 (1 - 1/n + 1/n^2)} = n sqrt{1 - 1/n + 1/n^2} ).So, putting it together, the denominator becomes:( n (1 - 7/n^{1/2}) times n sqrt{1 - 1/n + 1/n^2} = n^2 (1 - 7/n^{1/2}) sqrt{1 - 1/n + 1/n^2} ).So, the denominator is approximately ( n^2 times 1 times 1 = n^2 ) as n approaches infinity, since the terms with 1/n and 1/n^{1/2} go to zero.Now, let's rewrite the entire expression with these approximations:Numerator: ( 11^{1/4} n^{5/4} + 5n^2 ).Denominator: ( n^2 ).So, the expression becomes approximately:( frac{11^{1/4} n^{5/4} + 5n^2}{n^2} = frac{11^{1/4} n^{5/4}}{n^2} + frac{5n^2}{n^2} = 11^{1/4} n^{-3/4} + 5 ).As n approaches infinity, ( n^{-3/4} ) approaches 0, so the entire expression approaches 5.Wait, but let me double-check. Maybe I oversimplified.In the numerator, the dominant term is ( 5n^2 ), because ( n^2 ) grows faster than ( n^{5/4} ). So, actually, the numerator is dominated by ( 5n^2 ), and the denominator is ( n^2 ), so the limit should be 5.But let me verify step by step.First, numerator:( n sqrt[4]{11n} = n times (11n)^{1/4} = 11^{1/4} n^{1 + 1/4} = 11^{1/4} n^{5/4} ).( sqrt{25n^4 - 81} = 5n^2 sqrt{1 - 81/(25n^4)} approx 5n^2 (1 - (81)/(2 times 25n^4)) ) using the binomial approximation for large n.But for the leading term, it's 5n^2.So, numerator is approximately ( 11^{1/4} n^{5/4} + 5n^2 ).Denominator:( (n - 7 sqrt{n}) sqrt{n^2 - n + 1} ).As n becomes large, ( n - 7 sqrt{n} approx n ) and ( sqrt{n^2 - n + 1} approx n sqrt{1 - 1/n + 1/n^2} approx n (1 - 1/(2n) + 1/(2n^2)) approx n - 1/2 + 1/(2n) ).So, multiplying these approximations:( (n) times (n - 1/2 + 1/(2n)) approx n^2 - n/2 + 1/2 ).But for the leading term, it's n^2.So, denominator is approximately n^2.Therefore, the expression is approximately:( (5n^2) / (n^2) = 5 ).So, the limit should be 5.Wait, but earlier I thought the numerator had two terms, but the dominant term is 5n^2, so the other term is negligible.But just to be thorough, let's consider the exact expression.Numerator: ( n sqrt[4]{11n} + sqrt{25n^4 - 81} ).Let me write both terms with n^2 factored out.First term: ( n sqrt[4]{11n} = n times (11n)^{1/4} = 11^{1/4} n^{5/4} ).Second term: ( sqrt{25n^4 - 81} = 5n^2 sqrt{1 - 81/(25n^4)} ).So, numerator: ( 11^{1/4} n^{5/4} + 5n^2 sqrt{1 - 81/(25n^4)} ).Denominator: ( (n - 7 sqrt{n}) sqrt{n^2 - n + 1} ).Let me factor n from the first part and n^2 from the square root:( n(1 - 7/n^{1/2}) times n sqrt{1 - 1/n + 1/n^2} = n^2 (1 - 7/n^{1/2}) sqrt{1 - 1/n + 1/n^2} ).So, denominator is ( n^2 times [ (1 - 7/n^{1/2}) sqrt{1 - 1/n + 1/n^2} ] ).Now, let's write the entire expression:( frac{11^{1/4} n^{5/4} + 5n^2 sqrt{1 - 81/(25n^4)}}{n^2 (1 - 7/n^{1/2}) sqrt{1 - 1/n + 1/n^2}} ).Divide numerator and denominator by n^2:( frac{11^{1/4} n^{5/4 - 2} + 5 sqrt{1 - 81/(25n^4)}}{(1 - 7/n^{1/2}) sqrt{1 - 1/n + 1/n^2}} ).Simplify exponents:( 5/4 - 2 = 5/4 - 8/4 = -3/4 ).So, numerator becomes:( 11^{1/4} n^{-3/4} + 5 sqrt{1 - 81/(25n^4)} ).As n approaches infinity, ( n^{-3/4} ) approaches 0, and ( sqrt{1 - 81/(25n^4)} ) approaches 1.So, numerator approaches 5.Denominator:( (1 - 7/n^{1/2}) sqrt{1 - 1/n + 1/n^2} ).As n approaches infinity, ( 7/n^{1/2} ) approaches 0, ( 1/n ) approaches 0, and ( 1/n^2 ) approaches 0. So, the denominator approaches ( 1 times 1 = 1 ).Therefore, the entire expression approaches ( 5 / 1 = 5 ).So, the limit is 5.Wait, but let me check if I made any mistake in handling the terms.In the numerator, the first term is ( 11^{1/4} n^{5/4} ), which is much smaller than ( 5n^2 ) because ( n^{5/4} ) grows slower than ( n^2 ). So, indeed, the dominant term is ( 5n^2 ).In the denominator, after factoring, it's ( n^2 times ) something that approaches 1. So, the denominator is approximately ( n^2 ).Thus, the ratio is approximately ( 5n^2 / n^2 = 5 ).Therefore, the limit is 5.I think that's correct. So, the final answer is 5.**Final Answer**boxed{5}