answer：First, determine the time taken to complete one-quarter of the task. The printer starts at 9:00 AM and completes one-quarter by 12:00 PM, which is a duration of 3 hours. Therefore, the total time required to complete the entire task at the original speed would be (4 times 3 = 12) hours.Next, consider the increased speed after 12:00 PM. The printer's speed is increased by 25%, meaning it now operates at 125% of the original speed. The remaining three-quarters of the task will be completed at this increased speed. Since the original time for one-quarter of the task was 3 hours, the time for one-quarter at the increased speed is (frac{3}{1.25} = 2.4) hours. Therefore, the time to complete three-quarters of the task at the increased speed is (3 times 2.4 = 7.2) hours.Finally, add this duration to 12:00 PM to find the completion time. (7.2) hours is equivalent to 7 hours and 12 minutes. Adding this to 12:00 PM results in a completion time of 7:12 PM.

answer：Okay, so I have this problem about a rectangle ABCD where AB is 4 units and BC is 6 units. The task is to figure out the minimum area remaining after cutting out three isosceles right triangles. Hmm, okay. Let me try to visualize this.First, I know that an isosceles right triangle has two legs of equal length and a hypotenuse. The area of such a triangle is (1/2)*leg^2. So, if I can figure out the lengths of the legs for each triangle, I can calculate their areas and subtract them from the total area of the rectangle.The rectangle has sides AB=4 and BC=6, so its area is 4*6=24 square units. That's straightforward.Now, the tricky part is figuring out where to cut these three isosceles right triangles so that the remaining area is minimized. I need to maximize the total area of the triangles removed. So, I should try to make each triangle as large as possible without overlapping and without going outside the rectangle.Let me think about how to place these triangles. Since they are isosceles right triangles, their legs must be equal. So, maybe I can place them at the corners of the rectangle? Let me consider each corner.Starting with corner A: If I place a triangle at corner A, the legs would be along AB and AD. Since AB is 4, and AD is 6, the maximum leg length for the triangle at A would be 4, because beyond that, it would go beyond the rectangle. So, the area of this triangle would be (1/2)*4*4=8.Similarly, at corner B: The legs would be along BA and BC. BA is 4, BC is 6. So again, the maximum leg length is 4, giving an area of 8.At corner C: The legs would be along CB and CD. CB is 6, CD is 4. So, the maximum leg length here is 4, area is 8.At corner D: The legs would be along DC and DA. DC is 4, DA is 6. So, maximum leg length is 4, area is 8.Wait, but the problem says to remove three triangles, not four. So, maybe I can remove three of these corner triangles. If I remove three, each with area 8, the total area removed would be 24, which is the entire area of the rectangle. That can't be right because the remaining area would be zero, which doesn't make sense.So, perhaps I can't remove three triangles each of area 8. Maybe the triangles can't all be placed at the corners without overlapping or going outside the rectangle.Alternatively, maybe the triangles are placed along the sides but not necessarily at the corners. Let me think.Another approach: Maybe the triangles are placed such that their legs are along the sides of the rectangle, but not necessarily starting at the corners. For example, if I place a triangle on side AB, its legs could extend into the rectangle, but then another triangle on side BC might interfere.Wait, perhaps the triangles are placed in such a way that they share a common vertex or something. Let me try to sketch this mentally.Alternatively, maybe the triangles are placed along the diagonals of the rectangle. Since it's a rectangle, the diagonals are equal and bisect each other. The length of the diagonal can be calculated using the Pythagorean theorem: sqrt(4^2 + 6^2)=sqrt(16+36)=sqrt(52)=2*sqrt(13). Hmm, not sure if that helps.Wait, the problem mentions that the triangles are isosceles right triangles. So, their legs are equal, and the right angle is between the legs. So, perhaps each triangle is placed such that one leg is along a side of the rectangle, and the other leg extends into the rectangle.Let me consider placing one triangle on side AB. If I place a triangle with legs along AB and AD, starting at corner A, as I thought earlier. Similarly, another triangle on side BC, starting at corner B, and another on side CD, starting at corner C. But then, as I thought before, each would have area 8, but removing three would remove 24, which is the entire area.But that can't be, so maybe the triangles are smaller.Alternatively, maybe the triangles are placed in such a way that they overlap or share some space, but since we need to maximize the area removed, overlapping would not be efficient.Wait, perhaps the triangles are placed not at the corners but somewhere else on the sides, allowing for larger triangles without overlapping.Let me think about the maximum possible size of an isosceles right triangle that can fit into the rectangle without overlapping.Suppose I place a triangle on side AB, with legs along AB and AD. The maximum leg length is 4, as before, giving area 8.Similarly, on side BC, maximum leg length is 6, but wait, BC is 6, but the other side is AB=4, so actually, the maximum leg length is limited by the shorter side, which is 4. So, same as before, area 8.Wait, but if I place a triangle on BC, the legs would be along BC and BA. BC is 6, BA is 4, so the maximum leg length is 4, area 8.Similarly, on side CD, which is 4, the maximum leg length is 4, area 8.But again, if I remove three such triangles, each of area 8, total removed area is 24, which is the entire area. That can't be.So, perhaps the triangles are not placed at the corners but somewhere else.Alternatively, maybe the triangles are placed such that their legs are not along the sides but at some angle, allowing for larger triangles.Wait, but the problem says "isosceles right triangles," so the legs must be equal and at right angles. So, their legs must be aligned with the sides of the rectangle or at 45 degrees.Wait, if I place a triangle with legs at 45 degrees to the sides, maybe I can fit a larger triangle.Let me think about that.If I place a triangle with legs at 45 degrees, the legs would extend diagonally across the rectangle. The maximum length of such a leg would be limited by the shorter side of the rectangle.Wait, the rectangle is 4x6, so the shorter side is 4. If I place a triangle with legs of length x at 45 degrees, the projection along each side would be x*cos(45) and x*sin(45), which are both x/√2.So, to fit within the rectangle, x/√2 must be less than or equal to 4 and 6. Since 4 is the shorter side, x/√2 ≤4, so x ≤4√2≈5.656.But the hypotenuse of such a triangle would be x√2, which would be 4√2*√2=8, which is longer than the longer side of the rectangle, which is 6. So, that won't fit.Therefore, placing triangles at 45 degrees might not be feasible without exceeding the rectangle's boundaries.Hmm, maybe I need to think differently.Wait, perhaps the triangles are placed such that their legs are along the sides, but not starting at the corners. For example, on side AB, which is 4 units, I can place a triangle with legs of length x along AB and AD, starting somewhere along AB, not necessarily at corner A.Similarly, on side BC, which is 6 units, place another triangle with legs of length y along BC and BA, starting somewhere along BC.And on side CD, which is 4 units, place another triangle with legs of length z along CD and CB, starting somewhere along CD.But then, the total area removed would be (1/2)x^2 + (1/2)y^2 + (1/2)z^2, and I need to maximize this sum without overlapping.But this seems complicated. Maybe there's a smarter way.Wait, perhaps the triangles are placed in such a way that they form a sort of "staircase" along the rectangle, each triangle fitting into the space left by the previous one.Alternatively, maybe the triangles are placed at the midpoints of the sides.Wait, let me think about the figure mentioned in the problem. Since the user mentioned a figure, but I don't have it, I need to make some assumptions.Given that the original solution mentioned bisectors of angles B and C intersecting at E, and then extending BE to meet AD at F, and drawing a perpendicular from E to CD at G. Then, the triangles are EBC, ABF, and CEG.Wait, let me try to reconstruct that.So, rectangle ABCD, with AB=4, BC=6. Let me label the rectangle with A at the bottom-left corner, B at the bottom-right, C at the top-right, and D at the top-left.So, AB is the bottom side, BC is the right side, CD is the top, DA is the left.Now, the bisectors of angles B and C intersect at point E.Angle B is the right angle at corner B, so its bisector would be a line that splits the 90-degree angle into two 45-degree angles. Similarly, angle C is the right angle at corner C, so its bisector would also split the 90-degree angle into two 45-degree angles.So, the bisector of angle B would go from B towards the interior of the rectangle at a 45-degree angle from both BA and BC.Similarly, the bisector of angle C would go from C towards the interior of the rectangle at a 45-degree angle from both CB and CD.These two bisectors intersect at point E.Given that, let's try to find the coordinates of E.Let me assign coordinates to the rectangle. Let me place point A at (0,0), so point B is at (4,0), point C is at (4,6), and point D is at (0,6).Now, the bisector of angle B is the line from B(4,0) at a 45-degree angle upwards into the rectangle. Similarly, the bisector of angle C is the line from C(4,6) at a 45-degree angle downwards into the rectangle.The equation of the bisector from B: Since it's at 45 degrees, its slope is 1. So, the equation is y = x - 4, because it passes through (4,0).The equation of the bisector from C: It's also at 45 degrees, but going downwards, so its slope is -1. The equation is y = -x + 10, because it passes through (4,6).Now, to find point E, we solve the system:y = x - 4y = -x + 10Setting them equal:x - 4 = -x + 102x = 14x = 7But wait, x=7 is beyond the rectangle, which only goes up to x=4. That can't be right. Did I make a mistake?Wait, no. Wait, the bisector from B is going into the rectangle, so it's from (4,0) with slope 1, so it's y = x - 4. But when x=4, y=0, and as x increases, y increases. But since the rectangle only goes up to x=4, this line would go outside the rectangle. That can't be.Wait, maybe I got the direction wrong. The bisector of angle B is inside the rectangle, so it should go from B(4,0) towards the interior, which would be towards decreasing x and increasing y. So, actually, the slope should be negative, not positive.Wait, no. If you bisect the right angle at B, which is between BA (along negative x-axis) and BC (along positive y-axis), the bisector would go into the rectangle at a 45-degree angle from BA and BC. So, from B(4,0), moving towards the interior, the direction would be towards decreasing x and increasing y, which would have a slope of -1, not 1.Similarly, the bisector from C(4,6) would go towards decreasing x and decreasing y, which would have a slope of -1 as well.Wait, let me recast this.From point B(4,0), the bisector of the right angle would go into the rectangle at 45 degrees from both BA and BC. So, from B, moving towards the interior, the direction is towards the upper left, which would have a slope of -1. So, the equation is y = -1*(x - 4) + 0, which simplifies to y = -x + 4.Similarly, from point C(4,6), the bisector of the right angle would go into the rectangle at 45 degrees from both CB and CD. So, from C, moving towards the interior, the direction is towards the lower left, which would also have a slope of -1. So, the equation is y = -1*(x - 4) + 6, which simplifies to y = -x + 10.Now, let's find the intersection point E of these two lines:y = -x + 4y = -x + 10Wait, these are parallel lines? No, wait, both have slope -1, so they are parallel and will never intersect. That can't be right. There must be a mistake.Wait, no, actually, the bisector from B is y = -x + 4, and the bisector from C is y = -x + 10. These are parallel lines, so they don't intersect. That contradicts the original problem statement which says they intersect at E. So, I must have made a mistake in determining the equations.Wait, perhaps the bisectors are not both with slope -1. Let me think again.At point B(4,0), the two sides are BA (from B to A, which is left along the x-axis) and BC (from B to C, which is up along the y-axis). The angle between BA and BC is 90 degrees. The bisector would be a line that splits this 90-degree angle into two 45-degree angles.So, from B, the bisector would go into the rectangle at a 45-degree angle from both BA and BC. So, from B(4,0), moving towards the interior, the direction is such that it makes equal angles with BA and BC.Similarly, at point C(4,6), the two sides are CB (from C to B, down along the y-axis) and CD (from C to D, left along the x-axis). The angle between CB and CD is 90 degrees, and the bisector would split this into two 45-degree angles.So, from C(4,6), the bisector would go into the rectangle at a 45-degree angle from both CB and CD.Wait, so from B(4,0), the bisector would have a slope of 1, not -1, because it's going up and to the left, making equal angles with BA (left along x-axis) and BC (up along y-axis). Similarly, from C(4,6), the bisector would have a slope of -1, going down and to the left, making equal angles with CB (down along y-axis) and CD (left along x-axis).Wait, let me confirm that.From B(4,0), the bisector should make a 45-degree angle with both BA and BC. BA is along the negative x-axis, and BC is along the positive y-axis. So, the bisector would go into the rectangle at a 45-degree angle from BA, which is towards the upper left, so the slope would be positive, because from B(4,0), going up and left would have a positive slope? Wait, no, going left is negative x direction, and up is positive y direction, so the slope is (Δy)/(Δx) = positive y / negative x, which is negative. So, slope is -1.Wait, no, actually, if you move from B(4,0) towards the interior at a 45-degree angle from BA (which is along negative x-axis), the direction would be towards the upper left, which is negative x and positive y, so the slope is (Δy)/(Δx) = positive / negative = negative. So, slope is -1.Similarly, from C(4,6), the bisector makes a 45-degree angle with CB (down along y-axis) and CD (left along x-axis). So, moving from C(4,6) towards the interior, it would go down and left, which is negative y and negative x, so slope is (Δy)/(Δx) = negative / negative = positive. So, slope is +1.Wait, that makes more sense. So, from B(4,0), the bisector has slope -1, equation y = -x + 4.From C(4,6), the bisector has slope +1, equation y = x + 2.Wait, let me check that.From C(4,6), moving down and left at 45 degrees, so slope is +1? Wait, no, moving down and left would be negative slope.Wait, I'm getting confused. Let me think carefully.From B(4,0), moving towards the interior at 45 degrees from BA (left along x-axis) and BC (up along y-axis). So, direction is towards upper left, which is negative x and positive y. So, slope is (Δy)/(Δx) = positive / negative = negative. So, slope is -1. Equation: y - 0 = -1(x - 4) → y = -x + 4.From C(4,6), moving towards the interior at 45 degrees from CB (down along y-axis) and CD (left along x-axis). So, direction is towards lower left, which is negative x and negative y. So, slope is (Δy)/(Δx) = negative / negative = positive. So, slope is +1. Equation: y - 6 = 1*(x - 4) → y = x - 4 + 6 → y = x + 2.Now, let's find the intersection point E of these two lines:y = -x + 4y = x + 2Set them equal:-x + 4 = x + 2-2x = -2x = 1Then, y = 1 + 2 = 3So, point E is at (1,3).Okay, that makes sense. So, E is inside the rectangle at (1,3).Now, the next step is to extend BE to intersect AD at point F.BE is the line from B(4,0) to E(1,3). Let's find the equation of BE.The slope of BE is (3 - 0)/(1 - 4) = 3/(-3) = -1.So, equation of BE: y - 0 = -1(x - 4) → y = -x + 4.Wait, that's the same as the bisector from B, which makes sense because BE is part of that bisector.Now, AD is the left side of the rectangle, which is the line x=0.So, to find point F, we need to find where BE intersects AD, which is at x=0.From the equation y = -x + 4, when x=0, y=4.So, point F is at (0,4).Okay, so F is at (0,4).Next, through point E(1,3), draw a perpendicular line to CD, intersecting it at point G.CD is the top side of the rectangle, which is the line y=6.A perpendicular to CD would be a vertical line, since CD is horizontal. Wait, no, CD is horizontal, so a perpendicular would be vertical. But point E is at (1,3). So, a vertical line through E would be x=1, which intersects CD at (1,6). So, point G is at (1,6).Wait, but the problem says "draw a perpendicular line to CD", which is horizontal, so the perpendicular is vertical. So, yes, x=1 intersects CD at (1,6).So, point G is at (1,6).Now, the three triangles to be removed are:1. Triangle EBC: with vertices at E(1,3), B(4,0), and C(4,6).2. Triangle ABF: with vertices at A(0,0), B(4,0), and F(0,4).3. Triangle CEG: with vertices at C(4,6), E(1,3), and G(1,6).Wait, let me confirm that these are isosceles right triangles.First, triangle EBC: Let's check the lengths of EB, BC, and EC.EB is from E(1,3) to B(4,0). Distance: sqrt((4-1)^2 + (0-3)^2) = sqrt(9 + 9) = sqrt(18) = 3√2.BC is from B(4,0) to C(4,6). Distance: 6 units.EC is from E(1,3) to C(4,6). Distance: sqrt((4-1)^2 + (6-3)^2) = sqrt(9 + 9) = sqrt(18) = 3√2.So, triangle EBC has sides EB=3√2, BC=6, EC=3√2.Wait, but in a triangle, the sum of any two sides must be greater than the third side. Here, EB + EC = 3√2 + 3√2 = 6√2 ≈8.485, which is greater than BC=6. So, it's a valid triangle.Now, is it an isosceles right triangle?Let's check the angles. Since EB=EC=3√2, and BC=6.Using the Pythagorean theorem: (3√2)^2 + (3√2)^2 = 18 + 18 = 36, which equals BC^2=6^2=36.So, yes, triangle EBC is an isosceles right triangle with legs EB and EC, each of length 3√2, and hypotenuse BC=6.Okay, so area of triangle EBC is (1/2)*(3√2)^2 = (1/2)*18 = 9.Next, triangle ABF: vertices at A(0,0), B(4,0), F(0,4).Let's check the lengths:AB is from A(0,0) to B(4,0): 4 units.AF is from A(0,0) to F(0,4): 4 units.BF is from B(4,0) to F(0,4): sqrt((4-0)^2 + (0-4)^2) = sqrt(16 + 16) = sqrt(32) = 4√2.So, triangle ABF has sides AB=4, AF=4, BF=4√2.Again, checking if it's an isosceles right triangle.AB=AF=4, and BF=4√2.Check Pythagorean theorem: 4^2 + 4^2 = 16 + 16 = 32 = (4√2)^2.Yes, so triangle ABF is an isosceles right triangle with legs AB and AF, each of length 4, and hypotenuse BF=4√2.Area is (1/2)*4*4=8.Third triangle, CEG: vertices at C(4,6), E(1,3), G(1,6).Let's check the lengths:CE is from C(4,6) to E(1,3): sqrt((4-1)^2 + (6-3)^2)=sqrt(9+9)=sqrt(18)=3√2.EG is from E(1,3) to G(1,6): 3 units.CG is from C(4,6) to G(1,6): 3 units.Wait, so triangle CEG has sides CE=3√2, EG=3, CG=3.Wait, but in a triangle, the sum of any two sides must be greater than the third. Here, EG + CG = 3 + 3 = 6, which is greater than CE=3√2≈4.242. So, it's a valid triangle.Is it an isosceles right triangle?Let's check the angles. Since EG=CG=3, it's an isosceles triangle with legs EG and CG, each of length 3, and hypotenuse CE=3√2.Check Pythagorean theorem: 3^2 + 3^2 = 9 + 9 = 18 = (3√2)^2.Yes, so triangle CEG is an isosceles right triangle with legs EG and CG, each of length 3, and hypotenuse CE=3√2.Area is (1/2)*3*3=4.5.So, the three triangles have areas 9, 8, and 4.5, totaling 9+8+4.5=21.5.Subtracting this from the total area of the rectangle (24), the remaining area is 24 - 21.5=2.5.So, the minimum remaining area is 2.5 square units.Wait, but is this the minimum? Could there be a way to remove more area by placing the triangles differently?I think in this configuration, the triangles are placed such that they are as large as possible without overlapping, so this should give the minimum remaining area.Alternatively, maybe there's a way to place the triangles to remove more area, but I can't think of a way right now. Given the constraints of the problem and the specific construction, I think 2.5 is the correct answer.

answer：Okay, so I need to figure out why the Dirichlet function is Lebesgue integrable but not Riemann integrable. Hmm, let me start by recalling what the Dirichlet function is. It's defined on the interval [0,1], and it's 1 when x is irrational and 0 when x is rational. That seems pretty straightforward, but I remember that this function has some interesting properties when it comes to integration.First, let me think about the Riemann integral. I know that for a function to be Riemann integrable, it needs to be continuous almost everywhere, or more formally, the set of discontinuities should have measure zero. But wait, the Dirichlet function is discontinuous everywhere, right? Because between any two points, no matter how close, there's both a rational and an irrational number. So, that means the function jumps between 0 and 1 infinitely often in any interval. That must mean it's not Riemann integrable because it's too discontinuous.But why exactly does that matter? Well, the Riemann integral relies on the function being "nice" enough so that the upper and lower sums converge to the same value as the partition gets finer. If the function is too discontinuous, those upper and lower sums don't converge. Let me try to visualize this. If I take any interval, no matter how small, there are both rationals and irrationals in it. So, if I choose sample points that are all rational, the Riemann sum would be zero, but if I choose all irrationals, the sum would be the length of the interval. Since these two possibilities give different results, the Riemann sums don't settle down to a single value. That means the function isn't Riemann integrable.Now, moving on to the Lebesgue integral. I remember that Lebesgue integration is more general and can handle functions that aren't Riemann integrable. The key here is that the Lebesgue integral doesn't care as much about the function's behavior on sets of measure zero. In this case, the rationals in [0,1] are countable, so they have measure zero. Since the function is 1 almost everywhere (except on a set of measure zero), the integral should just be the measure of the interval where it's 1, which is [0,1]. So, the integral should be 1.Wait, let me make sure I'm not mixing things up. The Lebesgue integral of the Dirichlet function would be the integral over [0,1] of d(x) dx. Since d(x) is 1 on the irrationals and 0 on the rationals, and the irrationals have measure 1 (since the rationals have measure 0), the integral should indeed be 1. That makes sense because the Lebesgue integral focuses on the "size" of the sets where the function takes different values, rather than the behavior at individual points.But why exactly does the Lebesgue integral handle this better? I think it's because Lebesgue integration partitions the range of the function instead of the domain. So, instead of looking at intervals and summing up function values times interval lengths, it looks at the sets where the function is a certain value and sums up those values times the measure of those sets. In this case, the function is 1 almost everywhere, so the integral is just 1 times the measure of [0,1], which is 1.Putting it all together, the Dirichlet function isn't Riemann integrable because it's too discontinuous—every interval has both rationals and irrationals, making the upper and lower sums not converge. But it is Lebesgue integrable because the Lebesgue integral can handle functions that are non-zero only on sets of measure zero, and in this case, the function is 1 almost everywhere, so the integral is straightforward.I think I've got it. The key difference is how Riemann and Lebesgue integrals handle discontinuities and the "size" of sets where the function behaves differently. Riemann integrals require the function to be continuous almost everywhere in a certain sense, while Lebesgue integrals are more forgiving and can integrate functions that are non-zero only on sets of measure zero.

answer：Okay, so I have this problem where I need to find positive integers p and q such that 3/5 is less than p/q, which is less than 5/8. And I need to make sure that q is as small as possible. Then, once I find p and q, I have to figure out what q minus p is.Alright, let's break this down. First, I know that 3/5 is 0.6 and 5/8 is 0.625. So, I need a fraction p/q that's somewhere between 0.6 and 0.625. And I want the smallest possible q. Hmm, okay.I remember something about Farey sequences or mediant fractions. Maybe that could help. The mediant of two fractions a/b and c/d is (a+c)/(b+d). So, if I take the mediant of 3/5 and 5/8, that would be (3+5)/(5+8) = 8/13. Let me check if 8/13 is between 3/5 and 5/8.Calculating 8 divided by 13, that's approximately 0.615. Yeah, that's between 0.6 and 0.625. So, 8/13 is a valid fraction between 3/5 and 5/8. Now, is 13 the smallest possible denominator?Let me think. Maybe I can find a smaller q. Let's try q=1,2,3,4,5,6,7,8,9,10,11,12 and see if any of these have a p such that 3/5 < p/q <5/8.Starting with q=1: The only possible fraction is 1/1, which is 1. That's way bigger than 5/8, so no.q=2: Possible fractions are 1/2=0.5 and 2/2=1. 0.5 is less than 0.6, so no good.q=3: Fractions are 1/3≈0.333, 2/3≈0.666, 3/3=1. 0.666 is bigger than 0.625, so no.q=4: Fractions are 1/4=0.25, 2/4=0.5, 3/4=0.75, 4/4=1. All either too small or too big.q=5: 3/5=0.6, which is the lower bound, but we need something greater than that. 4/5=0.8, which is too big.q=6: Let's see. 1/6≈0.166, 2/6≈0.333, 3/6=0.5, 4/6≈0.666, 5/6≈0.833, 6/6=1. Again, nothing between 0.6 and 0.625.q=7: Fractions are 1/7≈0.142, 2/7≈0.285, 3/7≈0.428, 4/7≈0.571, 5/7≈0.714, 6/7≈0.857, 7/7=1. 4/7 is about 0.571, which is less than 0.6, and 5/7 is 0.714, which is too big. So, nothing in between.q=8: 5/8 is the upper bound, which is 0.625. So, fractions are 1/8=0.125, 2/8=0.25, 3/8=0.375, 4/8=0.5, 5/8=0.625, 6/8=0.75, etc. So, nothing between 0.6 and 0.625 except 5/8 itself, which is the upper limit.q=9: Let's see. 1/9≈0.111, 2/9≈0.222, 3/9≈0.333, 4/9≈0.444, 5/9≈0.555, 6/9≈0.666, 7/9≈0.777, 8/9≈0.888, 9/9=1. 5/9≈0.555 is less than 0.6, and 6/9≈0.666 is too big. So, nothing in between.q=10: Fractions are 1/10=0.1, 2/10=0.2, 3/10=0.3, 4/10=0.4, 5/10=0.5, 6/10=0.6, 7/10=0.7, 8/10=0.8, 9/10=0.9, 10/10=1. So, 6/10=0.6 is the lower bound, and 7/10=0.7 is too big. So, nothing in between.q=11: Let's compute some fractions. 1/11≈0.0909, 2/11≈0.1818, 3/11≈0.2727, 4/11≈0.3636, 5/11≈0.4545, 6/11≈0.5455, 7/11≈0.6364, 8/11≈0.7273, etc. So, 6/11≈0.5455 is less than 0.6, and 7/11≈0.6364 is more than 0.625. So, nothing between 0.6 and 0.625.q=12: Fractions are 1/12≈0.0833, 2/12≈0.1667, 3/12=0.25, 4/12≈0.3333, 5/12≈0.4167, 6/12=0.5, 7/12≈0.5833, 8/12≈0.6667, etc. 7/12≈0.5833 is less than 0.6, and 8/12≈0.6667 is too big. So, nothing in between.So, after checking all q from 1 to 12, none of them have a fraction p/q that lies strictly between 3/5 and 5/8. The first q that works is 13, with p=8, giving 8/13≈0.615, which is between 0.6 and 0.625.Therefore, the smallest possible q is 13, and p is 8. So, q-p is 13-8=5.**Final Answer**The value of ( q - p ) is boxed{5}.

answer：Okay, so I have this problem where I need to find the minimum length of a tangent drawn from a point on the line y = x + 2 to the circle (x - 3)^2 + (y + 1)^2 = 2. The options are A: 4, B: 3, C: 2√2, D: 1. Hmm, let me think about how to approach this.First, I remember that the length of a tangent from a point to a circle can be found using the formula: length = √[(distance from point to center)^2 - radius^2]. So, if I can find the point on the line y = x + 2 that is closest to the center of the circle, then the tangent length from that point will be the minimum.The circle given is (x - 3)^2 + (y + 1)^2 = 2. So, the center of the circle is at (3, -1), and the radius is √2. Got that.Now, the line is y = x + 2. I need to find the point on this line that is closest to the center (3, -1). The closest point on a line to a given point is the perpendicular projection of that point onto the line. So, I should find the equation of the line perpendicular to y = x + 2 that passes through (3, -1).The slope of y = x + 2 is 1, so the slope of the perpendicular line will be -1 (since perpendicular slopes are negative reciprocals). So, the equation of the perpendicular line is y - (-1) = -1(x - 3), which simplifies to y + 1 = -x + 3, so y = -x + 2.Now, to find the intersection point of the two lines y = x + 2 and y = -x + 2. Let me set them equal:x + 2 = -x + 2Adding x to both sides:2x + 2 = 2Subtracting 2:2x = 0So, x = 0. Plugging back into y = x + 2, y = 0 + 2 = 2. So, the closest point on the line y = x + 2 to the center (3, -1) is (0, 2).Now, let me compute the distance between (3, -1) and (0, 2). Using the distance formula:Distance = √[(3 - 0)^2 + (-1 - 2)^2] = √[9 + 9] = √18 = 3√2.Okay, so the distance from the center to the point (0, 2) is 3√2. Now, the length of the tangent from (0, 2) to the circle is √[(distance)^2 - radius^2] = √[(3√2)^2 - (√2)^2] = √[18 - 2] = √16 = 4.Wait, so the minimum length of the tangent is 4? That's option A. Hmm, let me double-check my steps to make sure I didn't make a mistake.1. Found the center of the circle: (3, -1). Correct.2. Radius is √2. Correct.3. The line is y = x + 2. Correct.4. To find the closest point on the line to the center, I found the perpendicular line with slope -1 passing through (3, -1). Equation: y = -x + 2. Correct.5. Solved for intersection: x = 0, y = 2. Correct.6. Distance between (3, -1) and (0, 2): √[(3)^2 + (-3)^2] = √(9 + 9) = √18 = 3√2. Correct.7. Length of tangent: √[(3√2)^2 - (√2)^2] = √(18 - 2) = √16 = 4. Correct.Seems like all steps are correct. So, the minimum length of the tangent is 4, which is option A.But wait, let me think again. Is there another way to approach this problem? Maybe using calculus or optimization? Let me try that to confirm.Suppose I take a general point (x, y) on the line y = x + 2. Then, the length of the tangent from (x, y) to the circle is √[(x - 3)^2 + (y + 1)^2 - (√2)^2]. Since y = x + 2, substitute that in:Length = √[(x - 3)^2 + (x + 2 + 1)^2 - 2] = √[(x - 3)^2 + (x + 3)^2 - 2].Let me compute that:(x - 3)^2 = x^2 - 6x + 9(x + 3)^2 = x^2 + 6x + 9So, adding them: x^2 - 6x + 9 + x^2 + 6x + 9 = 2x^2 + 18Subtract 2: 2x^2 + 16So, length = √(2x^2 + 16)To minimize this, I can minimize the expression inside the square root, which is 2x^2 + 16. Since 2x^2 is always non-negative, the minimum occurs when x = 0, giving 16. So, the minimum length is √16 = 4. That's the same result as before.Okay, so both methods give me 4. That solidifies my confidence that the answer is A: 4.Just to make sure, let me visualize the situation. The circle is centered at (3, -1) with a small radius √2. The line y = x + 2 is a straight line with a slope of 1, crossing the y-axis at (0, 2). The closest point on this line to the circle's center is (0, 2), and the tangent from there is 4 units long. Since the radius is only √2, which is about 1.414, and the distance from center to the point is 3√2, which is about 4.242, the tangent length is 4, which is a bit less than the distance, as expected.I think I've covered all bases here. Calculated using two methods, both leading to the same answer, and the reasoning makes sense. So, I'm pretty sure the answer is A: 4.**Final Answer**boxed{A}

answer：Okay, so I have this problem here where I need to find the range of the real number ( a ) such that the function ( f(x) = x^3 + x^2 - ax - 4 ) has exactly one extremum point in the interval ( (-1, 1) ). The options are given as A: ( (1,5) ), B: ( [1,5) ), C: ( (1,5] ), and D: ( (-infty, 1) cup (5, +infty) ).First, I remember that extremum points of a function occur where the first derivative is zero or undefined. Since ( f(x) ) is a polynomial, its derivative will also be a polynomial, which is defined everywhere. So, I just need to find where the derivative equals zero.Let me compute the derivative of ( f(x) ). The derivative of ( x^3 ) is ( 3x^2 ), the derivative of ( x^2 ) is ( 2x ), the derivative of ( -ax ) is ( -a ), and the derivative of a constant like ( -4 ) is zero. So, putting it all together:( f'(x) = 3x^2 + 2x - a )Now, I need to find the values of ( x ) where ( f'(x) = 0 ). That is, solve the equation:( 3x^2 + 2x - a = 0 )This is a quadratic equation in terms of ( x ). The general form of a quadratic equation is ( ax^2 + bx + c = 0 ), so in this case, ( a = 3 ), ( b = 2 ), and ( c = -a ) (wait, that's confusing because the constant term is also denoted by ( a ). Maybe I should use a different variable to avoid confusion. Let me rewrite the quadratic equation as:( 3x^2 + 2x - k = 0 )where ( k = a ). That might make it clearer.So, solving for ( x ):The quadratic formula is ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). Plugging in the values:( x = frac{-2 pm sqrt{(2)^2 - 4(3)(-k)}}{2(3)} )Simplify inside the square root:( (2)^2 = 4 )( 4(3)(-k) = -12k )So, the discriminant becomes:( 4 - (-12k) = 4 + 12k )Therefore, the solutions are:( x = frac{-2 pm sqrt{4 + 12k}}{6} )Simplify the square root:( sqrt{4 + 12k} = sqrt{4(1 + 3k)} = 2sqrt{1 + 3k} )So, substituting back:( x = frac{-2 pm 2sqrt{1 + 3k}}{6} )Factor out a 2 in the numerator:( x = frac{2(-1 pm sqrt{1 + 3k})}{6} = frac{-1 pm sqrt{1 + 3k}}{3} )So, the critical points are at:( x = frac{-1 + sqrt{1 + 3k}}{3} ) and ( x = frac{-1 - sqrt{1 + 3k}}{3} )But remember that ( k = a ), so substituting back:( x = frac{-1 + sqrt{1 + 3a}}{3} ) and ( x = frac{-1 - sqrt{1 + 3a}}{3} )Now, these are the critical points where the function could have extrema. The problem states that there should be exactly one extremum in the interval ( (-1, 1) ). So, I need to ensure that only one of these critical points lies within ( (-1, 1) ).Let me analyze the two critical points:1. ( x_1 = frac{-1 + sqrt{1 + 3a}}{3} )2. ( x_2 = frac{-1 - sqrt{1 + 3a}}{3} )First, note that ( sqrt{1 + 3a} ) must be a real number, so the expression inside the square root must be non-negative:( 1 + 3a geq 0 implies a geq -frac{1}{3} )But since the options given start from 1, I think this is just a preliminary condition, and the main constraints will come from the interval ( (-1, 1) ).Now, let's analyze each critical point:Starting with ( x_1 = frac{-1 + sqrt{1 + 3a}}{3} ):I want to find when ( x_1 ) is in ( (-1, 1) ). Similarly, for ( x_2 ).But before that, maybe I can use a different approach. Since the function ( f'(x) = 3x^2 + 2x - a ) is a quadratic, it can have 0, 1, or 2 real roots. The number of real roots depends on the discriminant.The discriminant ( D ) of ( f'(x) ) is:( D = (2)^2 - 4(3)(-a) = 4 + 12a )For real roots, ( D geq 0 implies 4 + 12a geq 0 implies a geq -frac{1}{3} ), which is consistent with earlier.But since we are looking for exactly one extremum in ( (-1, 1) ), perhaps we can use the Intermediate Value Theorem on the derivative. That is, if ( f'(-1) ) and ( f'(1) ) have opposite signs, then by the Intermediate Value Theorem, there is at least one root in ( (-1, 1) ). But since it's a quadratic, if only one root is in the interval, the other root must be outside.Alternatively, another method is to consider the number of roots of ( f'(x) = 0 ) in ( (-1, 1) ). Since it's a quadratic, it can have 0, 1, or 2 roots. We need exactly one root in ( (-1, 1) ).To ensure exactly one root in ( (-1, 1) ), we can use the condition that ( f'(-1) ) and ( f'(1) ) have opposite signs. This is because if the derivative changes sign over the interval, there must be a root in between. However, since it's a quadratic, if both roots are in the interval, then the derivative would have two roots, which we don't want. So, to have exactly one root, the function ( f'(x) ) must cross the x-axis once in ( (-1, 1) ), which would happen if ( f'(-1) ) and ( f'(1) ) have opposite signs.Let me compute ( f'(-1) ) and ( f'(1) ):First, ( f'(x) = 3x^2 + 2x - a )Compute ( f'(-1) ):( f'(-1) = 3(-1)^2 + 2(-1) - a = 3(1) - 2 - a = 3 - 2 - a = 1 - a )Compute ( f'(1) ):( f'(1) = 3(1)^2 + 2(1) - a = 3 + 2 - a = 5 - a )So, ( f'(-1) = 1 - a ) and ( f'(1) = 5 - a )We need ( f'(-1) ) and ( f'(1) ) to have opposite signs. So, their product should be negative:( (1 - a)(5 - a) < 0 )Let me solve this inequality:First, find the critical points where each factor is zero:( 1 - a = 0 implies a = 1 )( 5 - a = 0 implies a = 5 )So, the critical points are at ( a = 1 ) and ( a = 5 ). These divide the real line into three intervals:1. ( a < 1 )2. ( 1 < a < 5 )3. ( a > 5 )Now, test each interval:1. For ( a < 1 ), say ( a = 0 ):( (1 - 0)(5 - 0) = 1 * 5 = 5 > 0 ). So, positive.2. For ( 1 < a < 5 ), say ( a = 3 ):( (1 - 3)(5 - 3) = (-2)(2) = -4 < 0 ). So, negative.3. For ( a > 5 ), say ( a = 6 ):( (1 - 6)(5 - 6) = (-5)(-1) = 5 > 0 ). So, positive.So, the inequality ( (1 - a)(5 - a) < 0 ) holds when ( 1 < a < 5 ).Therefore, the range of ( a ) is ( (1, 5) ).But wait, let me check the endpoints ( a = 1 ) and ( a = 5 ) to see if they are included or not.When ( a = 1 ):( f'(x) = 3x^2 + 2x - 1 )Let's find its roots:( 3x^2 + 2x - 1 = 0 )Using quadratic formula:( x = frac{-2 pm sqrt{4 + 12}}{6} = frac{-2 pm sqrt{16}}{6} = frac{-2 pm 4}{6} )So, roots are:( x = frac{-2 + 4}{6} = frac{2}{6} = frac{1}{3} )and( x = frac{-2 - 4}{6} = frac{-6}{6} = -1 )So, one root is at ( x = -1 ) and the other at ( x = frac{1}{3} ).But the interval is ( (-1, 1) ), which is open. So, ( x = -1 ) is not included. Therefore, only ( x = frac{1}{3} ) is in ( (-1, 1) ). So, at ( a = 1 ), there is exactly one extremum in ( (-1, 1) ). So, ( a = 1 ) should be included.Similarly, when ( a = 5 ):( f'(x) = 3x^2 + 2x - 5 )Find its roots:( 3x^2 + 2x - 5 = 0 )Quadratic formula:( x = frac{-2 pm sqrt{4 + 60}}{6} = frac{-2 pm sqrt{64}}{6} = frac{-2 pm 8}{6} )So, roots are:( x = frac{-2 + 8}{6} = frac{6}{6} = 1 )and( x = frac{-2 - 8}{6} = frac{-10}{6} = -frac{5}{3} )So, one root is at ( x = 1 ) and the other at ( x = -frac{5}{3} ). Again, since the interval is ( (-1, 1) ), ( x = 1 ) is not included. So, only ( x = -frac{5}{3} ) is outside the interval. Therefore, at ( a = 5 ), there are no extrema in ( (-1, 1) ). So, ( a = 5 ) should not be included.Therefore, combining these results, the range of ( a ) is ( [1, 5) ).Wait, but earlier, using the product of ( f'(-1) ) and ( f'(1) ), I found that ( a ) must be in ( (1, 5) ). But when checking the endpoints, ( a = 1 ) is acceptable because it gives exactly one extremum in the interval, while ( a = 5 ) does not. So, the correct interval should include 1 but exclude 5, which is ( [1, 5) ).Therefore, the answer should be option B: ( [1,5) ).But let me double-check by considering the behavior of the function ( f'(x) ) at ( a = 1 ) and ( a = 5 ).At ( a = 1 ), as we saw, ( f'(x) ) has a root at ( x = -1 ) and ( x = 1/3 ). Since ( x = -1 ) is not in the open interval ( (-1, 1) ), only ( x = 1/3 ) is in the interval, so exactly one extremum.At ( a = 5 ), ( f'(x) ) has roots at ( x = 1 ) and ( x = -5/3 ). Again, ( x = 1 ) is not in ( (-1, 1) ), so no extrema in the interval.Therefore, the correct range is ( [1, 5) ), which is option B.**Final Answer**boxed{B}