🧠:[<think>Okay, so I need to figure out how many lines pass through the point (0, 1) and intersect the parabola y² = 4x at only one point. The options are 0, 1, 2, or 3. Hmm, let's start by understanding the problem.First, the parabola y² = 4x. I remember that this is a standard right-opening parabola with vertex at the origin. The general form of a parabola that opens to the right is y² = 4ax, where 'a' is the focal length. Here, 4a = 4, so a = 1. So the focus is at (1, 0). But maybe that's not directly relevant here.We need lines passing through (0, 1) that are tangent to this parabola. Because if a line intersects the parabola at exactly one point, it must be a tangent line. So essentially, the question is asking how many tangent lines to the parabola y² = 4x pass through the point (0, 1).Alternatively, we can rephrase it: Find the number of tangent lines from the point (0, 1) to the parabola y² = 4x. Each tangent line from an external point should correspond to a certain number, depending on the position of the point relative to the parabola.Wait, so maybe we need to check whether (0,1) is inside the parabola, on the parabola, or outside. If it's outside, there should be two tangent lines. If it's on the parabola, there's exactly one tangent line. If it's inside, there are no tangent lines.So first, let's check if (0,1) is on the parabola. Plugging x=0 into y²=4x gives y²=0, so y=0. The point (0,1) is not on the parabola. Next, check if it's inside or outside.For a parabola, the "inside" is the region where the points are towards the opening direction. Since this parabola opens to the right, the region x ≥ 0 is the "outside," and x < 0 is the "inside." Wait, but (0,1) is at x=0, which is on the vertex. But since the parabola is at x=0 when y=0, but (0,1) is above that. Maybe we need another way to check.Alternatively, for any point (x1, y1), the condition to be inside the parabola y² = 4ax is y1² < 4ax1. Let's plug in (0,1). Here, y1² = 1, and 4ax1 = 4*1*0 = 0. So 1 < 0? No. So 1 > 0, so the point is outside the parabola. Therefore, there should be two tangent lines from (0,1) to the parabola. But the options include 2 (option C). But let's confirm.Wait, maybe I should do this more carefully. Let's derive the equation of the tangent lines to the parabola y² = 4x. The general equation of a tangent line to this parabola at a point (at², 2at) is ty = x + at². Since here, 4a = 4, so a=1. Therefore, the tangent line at point (t², 2t) is t y = x + t².So the tangent lines are of the form t y = x + t². Now, these lines must pass through the point (0,1). Let's substitute x=0 and y=1 into the equation:t*(1) = 0 + t² ⇒ t = t² ⇒ t² - t = 0 ⇒ t(t - 1) = 0 ⇒ t = 0 or t = 1.Therefore, there are two values of t: t=0 and t=1. So there are two tangent lines passing through (0,1).Wait, but let's check what these tangent lines are.For t=0: 0*y = x + 0 ⇒ 0 = x + 0 ⇒ x=0. So the line x=0. But does x=0 intersect the parabola y²=4x at only one point? Plugging x=0 into the parabola equation gives y²=0 ⇒ y=0. So the line x=0 intersects the parabola at (0,0). However, the line x=0 is the y-axis, but the point (0,1) is on this line. Wait, but x=0 passes through (0,1) and intersects the parabola at (0,0). So that's one point of intersection.But the problem says "a line passes through (0,1) and intersects the parabola at only one point." So in this case, x=0 passes through (0,1) and intersects the parabola at (0,0). So two different points? Wait, but (0,1) is not on the parabola. So the line x=0 passes through (0,1) and intersects the parabola at (0,0). So that's two points: (0,1) is not on the parabola, but the line passes through (0,1) and (0,0). So that would actually intersect the parabola at (0,0). So is (0,1) considered an intersection point? Wait, the line passes through (0,1), but that's not a point on the parabola. So the intersection points with the parabola are only where the line and the parabola both lie. So since (0,1) is not on the parabola, the line x=0 intersects the parabola only at (0,0). So that is a single point. So x=0 is a vertical line tangent to the parabola at the vertex, passing through (0,1). So that counts as one tangent line.Then, for t=1: The tangent line is 1*y = x + 1² ⇒ y = x + 1. Let's check if this line intersects the parabola only once. Substitute y = x + 1 into y² = 4x:(x + 1)^2 = 4x ⇒ x² + 2x + 1 = 4x ⇒ x² - 2x + 1 = 0 ⇒ (x - 1)^2 = 0 ⇒ x=1. So x=1, then y = 1 + 1 = 2. So the line intersects the parabola at (1,2), a single point. So that's another tangent line.Therefore, there are two tangent lines: x=0 and y = x + 1. So the answer is 2, option C.But wait, the options are A:0, B:1, C:2, D:3. So answer is C:2. But let me check again.Alternatively, perhaps there's another approach. Let me think. Suppose we consider a general line passing through (0,1). Let the equation be y = mx + 1. We can substitute this into the parabola equation y² = 4x and find when there's only one solution.So substituting y = mx + 1 into y² = 4x gives:(mx + 1)^2 = 4x ⇒ m²x² + 2mx + 1 = 4x ⇒ m²x² + (2m - 4)x + 1 = 0.For this quadratic in x to have exactly one solution, discriminant must be zero.Discriminant D = [2m - 4]^2 - 4 * m² * 1 = 0.Compute D:(2m - 4)^2 - 4m² = 0Expand (2m - 4)^2: 4m² - 16m + 16So D = 4m² -16m +16 -4m² = -16m +16 =0Thus, -16m +16 =0 ⇒ -16m = -16 ⇒ m=1.So m=1. So the line is y = x +1. That's one tangent line. But wait, but we also considered x=0 earlier, which is a vertical line. But in this approach, we assumed the line has a slope m, which would exclude vertical lines. So the vertical line x=0 is another tangent line. Let's check that.The line x=0 is vertical, passing through (0,1). As before, substituting x=0 into y²=4x gives y²=0 ⇒ y=0. So intersection at (0,0). So that's a single point. So x=0 is also a valid tangent line. But in the previous approach with y=mx +1, we found only m=1. So we need to consider vertical lines separately.So in total, there are two lines: one with slope m=1 and the vertical line x=0. Hence, two lines. Therefore, answer is C:2.Wait, but when I derived the tangent lines using the parametric form, I found t=0 and t=1. For t=0, the tangent line is x=0 (since t y = x + t² becomes 0*y = x +0 ⇒ x=0). For t=1, the tangent line is y = x +1. So that's two lines. So yes, the answer is 2.But let me check again for any possible mistakes. For example, is the line x=0 actually a tangent line? The parabola y²=4x at t=0 is the point (0,0). The tangent line there is x=0, which is the vertical line. So yes, it's a tangent line. So x=0 is a tangent line at the vertex (0,0). So that line passes through (0,1) and is tangent at (0,0). So that's valid.Then the other tangent line is y = x +1, tangent at (1,2). So two lines. Therefore, the answer should be 2. So option C.But let me confirm once more. Suppose I use the condition for a line to be tangent to a parabola. The general equation of the parabola is y²=4ax. The condition for a line y = mx + c to be tangent to this parabola is c = a/m. Wait, in this case, the parabola is y²=4x, so a=1. Therefore, the condition is c = 1/m.But in our case, the line passes through (0,1), so c=1. Therefore, if the line is y = mx +1, then the condition for tangency is 1 = 1/m ⇒ m=1. So that gives m=1, which is the line y = x +1. But that only accounts for one line. The vertical line x=0 is not covered by this condition because vertical lines have undefined slope. So vertical lines must be considered separately.Thus, combining both, we get two lines: x=0 and y=x+1. Therefore, answer is C:2.Alternatively, let's think geometrically. The point (0,1) is outside the parabola y²=4x. For a parabola, from an external point, there are two real tangent lines. But in this case, since the point is vertically above the vertex, perhaps one tangent is the vertical line and the other is the sloped line. So yes, two lines. Hence, answer is 2.Therefore, the answer should be C:2.Final Answerboxed{C}</think>To determine how many lines pass through the point (0, 1) and intersect the parabola ( y^2 = 4x ) at only one point, we need to find the tangent lines from the point (0, 1) to the parabola.1. Check if the point (0, 1) is outside the parabola: - The parabola ( y^2 = 4x ) opens to the right. - For a point (x, y) to be outside the parabola, ( y^2 geq 4x ). For (0, 1), ( 1^2 = 1 geq 4 cdot 0 = 0 ), so the point is outside.2. Equation of tangent lines to the parabola: - The general form of a tangent line to ( y^2 = 4x ) at point ( (t^2, 2t) ) is ( ty = x + t^2 ). - Substitute the point (0, 1) into the tangent line equation: ( t cdot 1 = 0 + t^2 ) which simplifies to ( t^2 - t = 0 ). Solving this gives ( t = 0 ) or ( t = 1 ).3. Find the tangent lines: - For ( t = 0 ): The tangent line is ( x = 0 ). This line intersects the parabola at (0, 0). - For ( t = 1 ): The tangent line is ( y = x + 1 ). Substituting into the parabola equation gives a single intersection point (1, 2).4. Verify using the condition for tangency: - For a line ( y = mx + 1 ), substituting into ( y^2 = 4x ) and setting the discriminant to zero confirms ( m = 1 ), giving the line ( y = x + 1 ). - The vertical line ( x = 0 ) is separately verified as a tangent line.Thus, there are two tangent lines from (0, 1) to the parabola ( y^2 = 4x ): ( x = 0 ) and ( y = x + 1 ).The answer is (boxed{C}).,<think>Okay, let me try to figure out this problem. So, the question is: A line passes through the point (0, 1) and intersects the parabola y² = 4x at only one point. How many such lines are there? The options are A: 0, B: 1, C: 2, D: 3.Hmm, first, I need to recall what it means for a line to intersect a parabola at only one point. That should be when the line is tangent to the parabola, right? Because a tangent line touches the parabola at exactly one point. So, the problem is essentially asking how many tangent lines to the parabola y² = 4x pass through the point (0, 1). Wait, but I need to make sure. Is every line that intersects at only one point necessarily a tangent? Or could there be a vertical or horizontal line that also only intersects once without being tangent? Let me check.The parabola y² = 4x opens to the right. Its vertex is at (0,0). The point (0,1) is on the y-axis, one unit above the vertex. So, maybe a vertical line through (0,1) would be x=0. Let's see if that intersects the parabola. If x=0, then plugging into the parabola equation, y² = 4*0 = 0, so y=0. So, x=0 intersects the parabola only at (0,0). But the line x=0 passes through (0,1) as well? Wait, x=0 is the y-axis, which passes through (0,1), but intersects the parabola at (0,0). So, that's two points on the line x=0: (0,1) and (0,0). But wait, the parabola only intersects at (0,0), right? Because when x=0, y=0. So, the line x=0 passes through (0,1) and intersects the parabola at (0,0). But does that count as intersecting at only one point? So, the line x=0 would intersect the parabola at (0,0), but it also passes through (0,1), which is not on the parabola. So, does that count as intersecting the parabola at one point? The problem says "intersects the parabola y²=4x at only one point". So, even though the line passes through (0,1), which is not on the parabola, the intersection points with the parabola are only (0,0). So, is that considered intersecting at only one point? Then x=0 would be such a line. But wait, is x=0 a tangent to the parabola? Let's see. For the parabola y² = 4x, the derivative would give the slope of the tangent at any point. Differentiating both sides, 2y dy/dx = 4, so dy/dx = 2/y. At the point (0,0), the derivative is undefined, which makes sense because the parabola has a vertical tangent at the vertex? Wait, but the vertex is at (0,0), and the parabola opens to the right. So, actually, at the vertex (0,0), the tangent line is vertical, which is x=0. So, x=0 is indeed the tangent line at the vertex. Therefore, x=0 is a tangent line to the parabola, passing through (0,1). So that's one such line. But wait, is x=0 passing through (0,1)? Yes, because x=0 is the y-axis, which includes all points where x=0, so (0,1) is on that line. So, x=0 is one tangent line that passes through (0,1) and touches the parabola at (0,0). But are there other tangent lines to the parabola that pass through (0,1)? The problem is asking how many such lines there are. So, maybe there's the vertical tangent line x=0, and then other tangent lines with different slopes. Let me check.Let me recall that the general equation of a tangent to the parabola y² = 4x can be written as y = mx + 1/m, where m is the slope. Wait, no, let's see. For the standard parabola y² = 4ax, the tangent line at point (at², 2at) is ty = x + at². So, in this case, since 4a = 4, so a=1. Therefore, the tangent line at point (t², 2t) is ty = x + t². So, rearranged, that's y = (1/t)x + t. So, for each t ≠ 0, the tangent line has slope 1/t and y-intercept t. So, if we want a tangent line that passes through (0,1), we can substitute (0,1) into the equation of the tangent line. Wait, but the tangent line is ty = x + t². So, plugging in x=0, y=1, we get t*1 = 0 + t² ⇒ t = t² ⇒ t² - t = 0 ⇒ t(t - 1) = 0 ⇒ t=0 or t=1. But t=0 corresponds to the point (0,0) on the parabola, which is the vertex, and the tangent line there is x=0, as we saw earlier. The other solution is t=1, which gives the tangent line at the point (1², 2*1) = (1,2). The tangent line there is 1*y = x + 1² ⇒ y = x + 1. So, this line has slope 1 and passes through (0,1). So, that's another tangent line. Therefore, we have two tangent lines: x=0 and y = x + 1. Wait, so that's two lines. But the answer options include C:2. Hmm, but wait, let me check again.Wait, when t=0, the tangent line is x=0. When t=1, the tangent line is y = x + 1. But are there any other lines that pass through (0,1) and are tangent to the parabola? Let me see. Maybe another tangent line with a different slope. Let's check.Alternatively, maybe there's another approach. Let's consider the general equation of a line passing through (0,1). Since it passes through (0,1), its equation can be written as y = mx + 1, where m is the slope. However, vertical lines can't be expressed in this form, but we already considered the vertical line x=0, which is a tangent.So, suppose the line is y = mx + 1. To find where this intersects the parabola y² = 4x, substitute y = mx + 1 into the parabola equation:(mx + 1)² = 4xExpanding the left side:m²x² + 2mx + 1 = 4xBring all terms to one side:m²x² + (2m - 4)x + 1 = 0This is a quadratic equation in x. For the line to intersect the parabola at exactly one point, the discriminant of this quadratic must be zero. The discriminant D is:D = [2m - 4]² - 4 * m² * 1Compute D:D = (4m² - 16m + 16) - 4m²Simplify:D = -16m + 16Set discriminant equal to zero for tangency:-16m + 16 = 0 ⇒ -16m = -16 ⇒ m = 1So, m=1. Therefore, the line is y = x + 1. That's the same as we found earlier with t=1. So, that's one tangent line with slope 1, and the vertical line x=0. Wait, but when we substituted the general line y = mx + 1, we didn't consider the vertical line x=0. So, that vertical line is a separate case. So, in total, how many tangent lines are there passing through (0,1)? The vertical line x=0, which is one, and the line y = x + 1, which is another. So, two lines. Therefore, the answer should be C:2. But wait, let me check again. Let me make sure there are not more.Wait, perhaps when t=0, the tangent line is x=0, as we saw. When t=1, we get y = x + 1. Is there another value of t that could give a different tangent line passing through (0,1)? Let's see. From the parametrization earlier, we substituted (0,1) into the tangent line equation ty = x + t² and found t=0 and t=1. So, only two solutions. Therefore, only two tangent lines. So, that's consistent. Therefore, the answer is C:2.Wait, but wait, let's also verify the line y = x + 1. Let's check if it's indeed tangent to the parabola y² = 4x. If we substitute y = x + 1 into y² = 4x, we get (x + 1)^2 = 4x ⇒ x² + 2x + 1 = 4x ⇒ x² - 2x + 1 = 0 ⇒ (x - 1)^2 = 0. So, x=1 is a double root, so the line touches the parabola at x=1, y=2. So, that's correct, one point of contact. So, that line is indeed tangent. Similarly, the vertical line x=0 intersects the parabola at (0,0). Since the parabola is y² = 4x, when x=0, y=0. So, only one point of intersection. And the line x=0 is the vertical tangent at the vertex. So, that's another valid tangent line. Therefore, there are two such lines: x=0 and y = x + 1. So, the answer is 2, which is option C.But wait, let me think again. Is there a possibility of another tangent line? For example, maybe a horizontal line? Let's see. A horizontal line would have equation y = k. Let's see if such a line passes through (0,1). Then k=1, so y=1. Let's see if y=1 intersects the parabola y² = 4x at only one point. Substitute y=1 into the equation: 1 = 4x ⇒ x=1/4. So, the line y=1 intersects the parabola at (1/4, 1). Is that the only intersection? Yes, because if y=1, then x=1/4. So, it's a horizontal line intersecting at one point. But is this line tangent to the parabola? Let's check.For the line y=1, the slope is zero. The derivative of the parabola at the point (1/4, 1) is dy/dx. Differentiating implicitly: 2y dy/dx = 4 ⇒ dy/dx = 2/y. At y=1, dy/dx = 2/1 = 2. So, the slope of the tangent line at (1/4, 1) is 2. But the line y=1 has slope 0, which is not equal to 2. Therefore, the line y=1 is not tangent to the parabola at that point. Therefore, it's a secant line intersecting at one point, but not tangent. Wait, but how can a horizontal line intersect a right-opening parabola at only one point? Wait, the parabola y²=4x opens to the right. A horizontal line y=1 would intersect it at x=(1)^2/4 = 1/4, which is one point. But since the parabola is symmetric about the x-axis, but opening to the right. So, actually, no, the horizontal line y=1 will only intersect the parabola once because for each y, there's only one x. Wait, yes, because the parabola is a function if we solve for x: x = y²/4. So, each y corresponds to exactly one x. Therefore, a horizontal line y=k will intersect the parabola exactly once at (k²/4, k). Therefore, all horizontal lines intersect the parabola exactly once. But they are not tangent lines unless the slope matches. So, the line y=1 is a horizontal line intersecting the parabola at one point but is not a tangent line. Therefore, such a line is not considered a tangent, even though it intersects at one point. Therefore, only the lines that are tangent (i.e., their slope matches the derivative at the point of contact) count. Therefore, the line y=1 is not a tangent, so it's not counted. Therefore, only the two lines we found earlier: x=0 and y = x + 1. So, the answer is 2. Wait, but the problem says "intersects the parabola at only one point". So, does that include lines that are not tangent but still intersect at one point? For example, the horizontal line y=1 intersects at one point but is not tangent. So, if the problem counts such lines, then the answer would be more than two. But according to the problem statement, does it require the line to be tangent, or just to intersect at one point? Let me check the problem again.The problem states: "A line passes through the point (0, 1) and intersects the parabola y²=4x at only one point. How many such lines are there?"So, the problem is asking for lines that pass through (0,1) and intersect the parabola at only one point, regardless of whether they are tangent or not. So, in this case, we need to consider both tangent lines and other lines that intersect the parabola at exactly one point but are not tangent. However, for a parabola, any line that is not parallel to the axis will either intersect at two points, one point (if tangent), or not intersect. Wait, but the parabola y²=4x is a right-opening parabola. Let me consider different cases.For a parabola that opens to the right, like y²=4x, the axis is the x-axis. So, lines can be of various slopes:1. Vertical lines (parallel to the y-axis): These have the equation x = k. For the vertical line x=0, which passes through (0,1), we already saw it intersects the parabola at (0,0). So, only one point. So, this is a vertical line that intersects at one point.2. Non-vertical lines: These have equations y = mx + 1 (since they pass through (0,1)). When we substitute into the parabola equation, we get (mx + 1)^2 = 4x, leading to a quadratic equation. For such a line to intersect the parabola at only one point, the quadratic must have exactly one solution, i.e., discriminant zero. So, that gives us the tangent lines. But we found only one such line (m=1) in addition to the vertical line. But earlier, when I considered the horizontal line y=1, which is a non-vertical line with m=0, we saw that it intersects the parabola at one point (1/4, 1). However, in that case, the discriminant calculation would be different.Wait, let me check. If m=0, the line is y=1. Then substituting into the parabola equation gives 1² = 4x ⇒ x=1/4. So, only one intersection point. So, even though the line is not tangent, it still intersects the parabola at only one point. Therefore, according to the problem statement, such a line should be counted as well. But wait, in our previous discriminant calculation, when we set up the quadratic equation for y = mx + 1, we found that the discriminant is -16m + 16. Setting discriminant to zero gives m=1. So, only the line with m=1 is tangent. However, the horizontal line y=1 (with m=0) would not have been considered in that discriminant case, but when m=0, the line is horizontal, and substituting gives a linear equation, not quadratic. Wait, wait, no. If the line is horizontal, then when you substitute y=1 into the parabola equation, you get 1 = 4x, which is a linear equation, so only one solution. Therefore, in this case, for horizontal lines, since they are not captured by the general y = mx + 1 (since m=0 is allowed), substituting gives a linear equation, so only one solution. Therefore, in this case, the line y=1 intersects the parabola at one point and is not tangent. So, such a line also satisfies the problem's condition. Therefore, are there three such lines: x=0 (vertical), y = x + 1 (tangent with m=1), and y=1 (horizontal with m=0)? Let me check.Wait, but according to the problem statement, how exactly is the line intersecting the parabola? For non-vertical lines, if the line is horizontal, like y=1, then substituting into the parabola equation gives x= (y²)/4, so x=1/4. So, one point. So, that's one intersection. So, that's a valid line. So, if the problem counts such lines, even if they are not tangent, then the answer would be three: vertical x=0, horizontal y=1, and tangent y = x +1. But wait, but according to our earlier discriminant calculation, only m=1 gives a tangent line. So, the horizontal line y=1 is not a tangent but still intersects at one point. Therefore, the problem is not restricted to tangent lines but any lines passing through (0,1) that intersect the parabola at only one point. Therefore, the answer is three? But the options are 0,1,2,3. Let's check.Wait, let's test the horizontal line y=1. It passes through (0,1) and intersects the parabola at (1/4, 1). So, only one point. Is there another horizontal line? No, only y=1 passes through (0,1). Similarly, the vertical line x=0 passes through (0,1) and intersects at (0,0). Then, the tangent line y = x +1 passes through (0,1) and is tangent at (1,2). So, that's three lines? But according to the discriminant calculation, we only found m=1 and the vertical line. Wait, but the horizontal line y=1 is a separate case. Let me think again.Wait, in our earlier discriminant approach, we considered lines of the form y = mx + 1 (excluding vertical lines). When we substituted into the parabola equation, we obtained a quadratic equation in x. For there to be exactly one solution, the discriminant must be zero. However, when m=0, the line is horizontal, and substituting gives a linear equation in x, which has exactly one solution. Therefore, in this case, the discriminant approach doesn't apply because the equation becomes linear. Therefore, horizontal lines (with m=0) are a special case. So, for non-vertical lines, we have two cases:1. Lines with m ≠ 0: These result in a quadratic equation when intersected with the parabola. For these, to have exactly one solution, discriminant must be zero, leading to m=1. So, one line: y = x +1.2. Lines with m =0: The horizontal line y=1, which intersects the parabola at exactly one point (1/4,1). So, this is another line.Then, vertical lines: x=0, which intersects the parabola at exactly one point (0,0). So, this is another line.Therefore, total three lines: x=0, y=1, and y = x +1. Therefore, the answer should be D:3. But the options are A:0, B:1, C:2, D:3. Wait, this contradicts my previous conclusion. So, which is correct?Wait, this is a crucial point. The problem says "intersects the parabola at only one point". So, we need to consider all lines passing through (0,1) that intersect the parabola at exactly one point, regardless of whether they are tangent or not. So, this includes:1. Vertical line x=0: Passes through (0,1) and intersects the parabola at (0,0). Only one point.2. Horizontal line y=1: Passes through (0,1) and intersects the parabola at (1/4,1). Only one point.3. Tangent line y = x +1: Passes through (0,1) and is tangent at (1,2). Only one point.So, three lines. Therefore, the answer should be D:3. But earlier, using the parametrization of the tangent lines, we found two lines: x=0 and y = x +1. However, the horizontal line y=1 is another line that intersects at one point but is not tangent. Therefore, the answer depends on whether the problem counts non-tangent lines that intersect at one point. Wait, but for a parabola, can a non-tangent line intersect at exactly one point? In the case of a horizontal line y=1 and the parabola y²=4x, which is a right-opening parabola. For each y, there is exactly one x. So, a horizontal line y=k will always intersect the parabola at exactly one point (k²/4, k). So, these lines are not tangent but still intersect at only one point. Therefore, they should be counted. So, if the problem is not restricting to tangent lines, then the answer is three. However, if the problem is asking for tangent lines, then the answer is two. So, we need to check the problem statement again.The problem says: "A line passes through the point (0, 1) and intersects the parabola y²=4x at only one point. How many such lines are there?"So, it's asking for lines that pass through (0,1) and intersect the parabola at only one point. It does not specify that they have to be tangent lines. Therefore, both tangent lines and non-tangent lines that intersect at exactly one point should be considered. Therefore, in that case, there are three such lines: vertical x=0, horizontal y=1, and tangent y = x +1. Hence, answer D:3.But wait, this contradicts my previous conclusion. Let me verify again:1. Vertical line x=0: Passes through (0,1), intersects parabola at (0,0). One point.2. Horizontal line y=1: Passes through (0,1), intersects parabola at (1/4,1). One point.3. Tangent line y = x +1: Passes through (0,1), intersects parabola at (1,2). One point (tangent).Therefore, three lines. So, answer D:3. But the options don't have E, so D is the correct answer.But wait, let me check again if the horizontal line y=1 is actually intersecting the parabola at only one point. For the parabola y² =4x, solving for x when y=1 gives x=1/4. So, only one point. So, yes. So, the line y=1 is a valid line. Similarly, the vertical line x=0. And the tangent line y =x +1. So, three lines. Therefore, the answer is D:3.But wait, this contradicts the earlier calculation where substituting the line equation into the parabola gave only two solutions. Wait, no. When substituting y = mx +1 into the parabola equation, we considered m ≠0, leading to quadratic equation. For m=0, we get a linear equation, which has one solution. So, that's another line. Therefore, total three lines. Therefore, answer D:3.But in the options given, D is 3. But previously, I thought the answer was C:2. Hmm, maybe my initial assumption that only tangent lines count is incorrect. The problem says "intersects the parabola at only one point", not necessarily tangent. Therefore, the answer is three. Therefore, the correct answer is D:3.But wait, according to the parabola's geometry, a horizontal line y=k for k ≠0 should intersect the parabola at exactly one point. But in the case of a right-opening parabola, a horizontal line is not parallel to the axis of the parabola. Wait, the axis of the parabola y²=4x is the x-axis. A line parallel to the axis would be horizontal, but since the parabola opens to the right, the axis is the x-axis. Therefore, a line parallel to the axis would be horizontal. Wait, no. Wait, the axis of the parabola is the line about which the parabola is symmetric. For y²=4x, the axis is the x-axis. So, lines parallel to the axis are horizontal lines. But a horizontal line is not parallel to the directrix. Wait, in any case, regardless of that, substituting y=1 into the parabola gives x=1/4. So, only one point. So, according to the problem statement, that line should be counted.Therefore, the answer is three lines: vertical, horizontal, and tangent. So, D:3.But wait, I need to confirm this. Let's see:1. Vertical line x=0: Intersects at (0,0).2. Horizontal line y=1: Intersects at (1/4,1).3. Tangent line y = x +1: Intersects at (1,2).Therefore, three lines. So, answer D:3. But according to the initial parametrization of tangent lines, we found two tangent lines passing through (0,1): x=0 and y=x+1. The horizontal line y=1 is not a tangent line. Therefore, if the problem is only about tangent lines, answer is two. If it's about all lines passing through (0,1) intersecting the parabola at one point (whether tangent or not), the answer is three.Therefore, the problem statement says: "intersects the parabola y²=4x at only one point". It does not specify tangency, so the answer should be three. Therefore, the correct answer is D:3.But now I'm confused because different approaches give different answers. Let me check another way.Another approach: A line passing through (0,1) can be written as y = mx + 1 or x=0.Case 1: x=0. As before, intersects at (0,0). One point.Case 2: y = mx +1. Substitute into parabola:(mx +1)^2 =4x ⇒ m²x² + 2mx +1 =4x ⇒ m²x² + (2m -4)x +1=0.This is quadratic in x. For only one intersection, discriminant D=0 or the equation is linear (which would be when m²=0, i.e., m=0).Subcase 2a: m=0. Then equation becomes 0 + (0 -4)x +1=0 ⇒ -4x +1=0 ⇒ x=1/4. So, one solution. Therefore, line y=1 is included here.Subcase 2b: m≠0. Then quadratic equation. D=(2m -4)^2 -4*m²*1= 4m² -16m +16 -4m²= -16m +16. Set D=0 ⇒ -16m +16=0 ⇒ m=1. So, line y=x +1.Therefore, total solutions:- x=0- y=1 (from m=0)- y=x +1 (from m=1)Therefore, three lines. So, answer D:3.Therefore, the correct answer is D:3.But why did the tangent line parametrization only give two lines? Because when we parametrized the tangent lines as ty = x + t², substituting (0,1) gives t=0 and t=1. So, two tangent lines. But the horizontal line y=1 is not a tangent line, so it's a separate case. Therefore, the problem is not restricted to tangent lines. Hence, three lines in total.Therefore, the answer should be D:3. Therefore, the options given are A:0, B:1, C:2, D:3. So, correct answer is D.But wait, the problem says "intersects the parabola at only one point". However, for the horizontal line y=1, even though it intersects at one point, is there a geometrical interpretation where it's considered as intersecting once, but not tangent? Yes. Therefore, the answer is three.But let me check some references. For a parabola, a line can intersect it at 0, 1, or 2 points. When does a line intersect a parabola at exactly one point? Either when it's tangent, or when the line is parallel to the axis of the parabola? Wait, no. Wait, for a parabola that opens to the right, like y²=4x, the axis is the x-axis. A line parallel to the axis would be horizontal. However, a horizontal line y=k will intersect the parabola at one point (k²/4, k). So, such lines are not tangent but still intersect at one point. Therefore, those are valid. Therefore, three lines: vertical tangent, horizontal non-tangent, and the tangent line with slope 1.Hence, the answer is D:3. Therefore, the correct option is D.But wait, the original problem might have a typo or different configuration. Wait, but according to standard definitions, the answer should be three. However, some might argue that only tangent lines are considered as intersecting at one point. But according to the problem statement, it's clear: intersect at only one point. So, whether tangent or not, as long as the intersection is one point. Therefore, three lines.But now I need to check with an example. Let me take the horizontal line y=1. When I plot the parabola y²=4x, which opens to the right, and the horizontal line y=1, it should intersect at x=(1)^2 /4 = 1/4. So, the point (0.25,1). That's a single point. The line is not tangent there, since the tangent at that point would have slope dy/dx = 2/y = 2/1 = 2. So, the tangent line at (0.25,1) would have equation y -1 = 2(x -0.25), which simplifies to y = 2x -0.5 +1 ⇒ y = 2x +0.5. This line has slope 2 and passes through (0.25,1). It does not pass through (0,1). Therefore, the line y=1 is a secant line intersecting at one point but not passing through any other points of the parabola, hence intersecting only once. Therefore, it's a valid line for the problem's condition.Therefore, the three lines are:1. x=0: vertical line through (0,1), intersects at (0,0).2. y=1: horizontal line through (0,1), intersects at (0.25,1).3. y =x +1: tangent line through (0,1), intersects at (1,2).Thus, three lines. Therefore, the answer is D:3.But I initially thought the answer was C:2 because I considered only tangent lines. But the problem doesn't specify tangency. Therefore, the correct answer is D:3.Wait, but in the original parametrization of the tangent lines, when we substituted (0,1) into the tangent line equation ty = x + t², we found t=0 and t=1. So, two tangent lines. But the horizontal line y=1 is not a tangent line, so it's a third line. Hence, three lines in total. Therefore, the answer is D:3.I think the confusion arises from whether non-tangent lines intersecting at one point are considered. In this problem, they are. Therefore, answer D:3.Final Answerboxed{D}</think>To determine how many lines pass through the point (0, 1) and intersect the parabola ( y^2 = 4x ) at only one point, we need to consider both tangent lines and non-tangent lines that intersect the parabola at exactly one point.1. Vertical Line ( x = 0 ): - This line passes through (0, 1) and intersects the parabola at (0, 0). Since the parabola ( y^2 = 4x ) has a vertical tangent at the vertex (0, 0), this line is a tangent line.2. Horizontal Line ( y = 1 ): - This line passes through (0, 1) and intersects the parabola at ( (1/4, 1) ). Although this line is not tangent to the parabola, it intersects the parabola at exactly one point.3. Tangent Line ( y = x + 1 ): - This line is tangent to the parabola at the point (1, 2). Substituting ( y = x + 1 ) into the parabola equation ( y^2 = 4x ) results in a quadratic equation with a discriminant of zero, confirming it is a tangent line.To verify, we consider the general equation of a line passing through (0, 1) as ( y = mx + 1 ). Substituting into the parabola equation gives:[(mx + 1)^2 = 4x implies m^2x^2 + (2m - 4)x + 1 = 0]For the line to intersect the parabola at exactly one point, the discriminant must be zero:[(2m - 4)^2 - 4m^2 = 0 implies -16m + 16 = 0 implies m = 1]This gives the tangent line ( y = x + 1 ).Additionally, considering the horizontal line ( y = 1 ) (which is not covered by the quadratic discriminant method as it results in a linear equation) and the vertical line ( x = 0 ), we find three lines in total.Thus, the number of such lines is (boxed{D}).]

🔑:From the given conditions, when the line is x=0 or y=1, which means the line is perpendicular to the x-axis or y-axis, it obviously intersects the parabola y^2=4x at only one point.When the slope of the line is k, the equation of the line can be written as y-1=k(x-0). Substituting this into the equation of the parabola y^2=4x, we get k^2x^2+(2k-4)x+1=0.Therefore, Delta=(2k-4)^2-4k^2=0, solving this gives k=1. Hence, there are a total of boxed{3} lines that meet the condition.Therefore, the correct answer is boxed{text{D}}.

🧠:[<think>Okay, so I have this problem here about a sum that's greater than 19/20 and less than 20/21. The sum is given as a series of fractions: 1/(1×√2 + 2√1) + 1/(2√3 + 3√2) + ... + 1/(n√(n+1) + (n+1)√n). And the question is asking for the difference between the maximum and minimum values of the positive integer n that satisfy these inequalities. Alright, let me try to unpack this step by step.First, I need to understand the structure of each term in the series. Each term is of the form 1/(k√(k+1) + (k+1)√k). Let me write that down for clarity: the k-th term is 1 divided by [k√(k+1) + (k+1)√k]. Hmm. Maybe I can simplify each term somehow? Rationalize the denominator? That often helps with expressions involving square roots.Let me try rationalizing. If I have a denominator like a + b, where a and b are terms with square roots, multiplying numerator and denominator by (a - b) would give me (a - b)/(a² - b²). Let's see if that works here.Take the first term: 1/(1×√2 + 2×√1). Let's set a = 1×√2 and b = 2×√1. Then, multiplying numerator and denominator by (a - b):[1×(√2 - 2√1)] / [(1×√2)^2 - (2×√1)^2].Calculating the denominator: (1²×2) - (2²×1) = 2 - 4 = -2. So the first term becomes (√2 - 2)/-2 = (2 - √2)/2. Wait, let me check that:Wait, numerator is (√2 - 2) and denominator is -2. So (√2 - 2)/(-2) = (2 - √2)/2. Yeah, that's correct. So the first term simplifies to (2 - √2)/2.Let me check another term to see if there's a pattern. Take the second term: 1/(2√3 + 3√2). Let's apply the same process. Multiply numerator and denominator by (2√3 - 3√2):Numerator becomes (2√3 - 3√2), denominator becomes (2√3)^2 - (3√2)^2.Calculating denominator: (4×3) - (9×2) = 12 - 18 = -6. So the second term is (2√3 - 3√2)/(-6) = (3√2 - 2√3)/6.Hmm. So each term can be rewritten as [ (k+1)√k - k√(k+1) ] divided by [ (k√(k+1))² - ((k+1)√k )² ].Wait, but from the first two examples, the denominator after rationalization was -[k²(k+1) - (k+1)^2 k] = -[k(k+1)(k - (k+1))] = -[k(k+1)(-1)] = k(k+1). So maybe in general, when I rationalize, the denominator simplifies to -[k(k+1)].Wait, let me verify with the first term:Denominator after rationalization was -2, which is -1×2, i.e., -1×(1×2). Similarly, the second term denominator was -6, which is -2×3. So in general, denominator after rationalization is -k(k+1). So each term becomes:[ (k+1)√k - k√(k+1) ] / [ -k(k+1) ].Which can be rewritten as [ (k√(k+1) - (k+1)√k ) ] / [k(k+1) ].Wait, but in the first term, we had (2 - √2)/2. Let me see if that fits into this.Wait, (k=1):[1×√2 - 2×√1]/[1×2] = (√2 - 2)/2, but with the denominator being negative. Wait, maybe I need to check the sign.Wait, in the first example, after rationalization, the denominator was -2, so the term became (√2 - 2)/(-2) = (2 - √2)/2. Similarly, in the second term, denominator was -6, so (2√3 - 3√2)/(-6) = (3√2 - 2√3)/6.So the general term after rationalization is [(k+1)√k - k√(k+1)] divided by [ -k(k+1) ] multiplied by the original numerator 1. So that gives:[ (k+1)√k - k√(k+1) ] / [ -k(k+1) ] = [ k√(k+1) - (k+1)√k ] / [k(k+1) ].But when I calculate for k=1, this would be [1×√2 - 2×√1]/[1×2] = (√2 - 2)/2, but actually we had (2 - √2)/2. So perhaps there's a sign error here. Let me re-examine.Original term is 1/[k√(k+1) + (k+1)√k]. When I multiply numerator and denominator by [k√(k+1) - (k+1)√k], which is the conjugate, the denominator becomes [k√(k+1)]² - [(k+1)√k]².Calculating denominator:k²(k+1) - (k+1)^2 k = k(k+1)[k - (k+1)] = k(k+1)(-1) = -k(k+1).So denominator is -k(k+1). The numerator is [k√(k+1) - (k+1)√k]. Therefore, the entire term becomes:[k√(k+1) - (k+1)√k] / [-k(k+1)] = [ (k+1)√k - k√(k+1) ] / [k(k+1)].So the term is positive? Wait, but in the first term, we had (2 - √2)/2 which is positive. Let's check for k=1:[ (1+1)√1 - 1√(1+1) ] / [1×2] = [2×1 - 1×√2]/2 = (2 - √2)/2, which is correct. Similarly, for k=2:[3√2 - 2√3]/6. So yes, the numerator is positive because 3√2 ≈ 4.24 and 2√3 ≈ 3.464, so 4.24 - 3.464 ≈ 0.776, which is positive. So each term is positive, which makes sense because the original terms were positive.So, general term simplifies to [ (k+1)√k - k√(k+1) ] / [k(k+1)].Wait, that can be further simplified. Let's see:[ (k+1)√k - k√(k+1) ] / [k(k+1) ] = [ (k+1)√k ] / [k(k+1) ] - [ k√(k+1) ] / [k(k+1) ].Simplify each term:First term: [ (k+1)√k ] / [k(k+1) ] = √k / k = 1/√k.Second term: [ k√(k+1) ] / [k(k+1) ] = √(k+1)/(k+1) = 1/√(k+1).Therefore, the entire term simplifies to 1/√k - 1/√(k+1).Oh! That's a telescoping series! So each term is the difference of reciprocals of square roots. So when we add up the series from k=1 to k=n, the sum telescopes:Sum = (1/√1 - 1/√2) + (1/√2 - 1/√3) + ... + (1/√n - 1/√(n+1)).All the intermediate terms cancel out, leaving Sum = 1 - 1/√(n+1).Wow, that's a neat simplification. So the sum from k=1 to n of 1/[k√(k+1) + (k+1)√k] is equal to 1 - 1/√(n+1).So the problem states that this sum is greater than 19/20 and less than 20/21. Therefore:19/20 < 1 - 1/√(n+1) < 20/21.Our goal is to find the positive integer n that satisfies both inequalities, then find the difference between the maximum and minimum such n.First, let's solve the inequalities.Starting with the left inequality: 19/20 < 1 - 1/√(n+1).Subtract 19/20 from both sides: 0 < 1/20 - 1/√(n+1).Wait, that's not helpful. Let me rearrange the inequality:1 - 1/√(n+1) > 19/20Subtract 1 from both sides:-1/√(n+1) > 19/20 - 1 = -1/20Multiply both sides by -1 (remembering to reverse the inequality):1/√(n+1) < 1/20Then, take reciprocals (since both sides are positive, inequality reverses again):√(n+1) > 20Square both sides:n + 1 > 400So n > 399. Therefore, n must be at least 400.Now, the right inequality: 1 - 1/√(n+1) < 20/21Subtract 1 from both sides:-1/√(n+1) < 20/21 - 1 = -1/21Multiply both sides by -1 (reverse inequality):1/√(n+1) > 1/21Take reciprocals (reverse inequality again):√(n+1) < 21Square both sides:n + 1 < 441Thus, n < 440. So n must be less than 440, i.e., n ≤ 439.Therefore, combining both inequalities:399 < n < 440.Since n is a positive integer, n must be in 400, 401, ..., 439. So the minimum n is 400 and maximum n is 439. The difference between them is 439 - 400 = 39.Wait, hold on. Wait, n has to be greater than 399, so the smallest integer n is 400. And n has to be less than 440, so the largest integer n is 439. Therefore, the difference is 439 - 400 = 39. So the answer should be 39.But let me double-check the calculations to be sure.Starting with the left inequality:1 - 1/√(n+1) > 19/201/√(n+1) < 1 - 19/20 = 1/20So √(n+1) > 20n+1 > 400n > 399, so n ≥ 400.Right inequality:1 - 1/√(n+1) < 20/211/√(n+1) > 1 - 20/21 = 1/21√(n+1) < 21n+1 < 441n < 440, so n ≤ 439.So n ranges from 400 to 439 inclusive. The number of terms is 439 - 400 + 1 = 40. But the question asks for the difference between the maximum and minimum n, which is 439 - 400 = 39.Hence, the answer is 39.Wait, but let me confirm if the sum is indeed equal to 1 - 1/√(n+1). Let me test with a small n. Let's take n=1.Sum should be first term: 1/(1×√2 + 2×√1) = 1/(√2 + 2). According to the telescoping formula, it should be 1 - 1/√(1+1) = 1 - 1/√2. Let's compute both:Original term: 1/(√2 + 2). Multiply numerator and denominator by (2 - √2):(2 - √2)/[(√2 + 2)(2 - √2)] = (2 - √2)/(4 - 2) = (2 - √2)/2 = 1 - (√2)/2 ≈ 1 - 0.707 ≈ 0.293.Telescoping sum: 1 - 1/√2 ≈ 1 - 0.707 ≈ 0.293. So they match. Good.Another test: n=2.Sum is first term + second term. First term: 1 - 1/√2 ≈ 0.293. Second term: 1/(2√3 + 3√2). Let's compute that:Multiply numerator and denominator by (2√3 - 3√2):(2√3 - 3√2)/[(2√3)^2 - (3√2)^2] = (2√3 - 3√2)/(12 - 18) = (2√3 - 3√2)/(-6) = (3√2 - 2√3)/6 ≈ (4.242 - 3.464)/6 ≈ 0.778/6 ≈ 0.130.So total sum is 0.293 + 0.130 ≈ 0.423.Telescoping formula: 1 - 1/√3 ≈ 1 - 0.577 ≈ 0.423. That matches. So the formula is correct.Therefore, the sum S(n) = 1 - 1/√(n+1). So the inequalities:19/20 < 1 - 1/√(n+1) < 20/21Solving for n:First inequality:1 - 1/√(n+1) > 19/20=> 1/√(n+1) < 1 - 19/20 = 1/20=> √(n+1) > 20=> n+1 > 400=> n > 399Second inequality:1 - 1/√(n+1) < 20/21=> 1/√(n+1) > 1 - 20/21 = 1/21=> √(n+1) < 21=> n+1 < 441=> n < 440Thus, n is an integer satisfying 400 ≤ n ≤ 439. Therefore, the maximum n is 439, the minimum n is 400. The difference is 439 - 400 = 39.So the answer should be 39. Therefore, the difference between maximum and minimum n is 39.Final Answerboxed{39}</think>Given the problem, we need to find the difference between the maximum and minimum values of the positive integer ( n ) such that the sum frac{1}{1 times sqrt{2} + 2 sqrt{1}} + frac{1}{2 sqrt{3} + 3 sqrt{2}} + cdots + frac{1}{n sqrt{n+1} + (n+1) sqrt{n}}is greater than (frac{19}{20}) and less than (frac{20}{21}).First, we simplify each term in the series. By rationalizing the denominator, each term (frac{1}{k sqrt{k+1} + (k+1) sqrt{k}}) can be rewritten as (frac{1}{sqrt{k}} - frac{1}{sqrt{k+1}}). This results in a telescoping series:left(1 - frac{1}{sqrt{2}}right) + left(frac{1}{sqrt{2}} - frac{1}{sqrt{3}}right) + cdots + left(frac{1}{sqrt{n}} - frac{1}{sqrt{n+1}}right)This series telescopes to:1 - frac{1}{sqrt{n+1}}We need this sum to satisfy the inequalities:frac{19}{20} < 1 - frac{1}{sqrt{n+1}} < frac{20}{21}Solving the left inequality:1 - frac{1}{sqrt{n+1}} > frac{19}{20} implies frac{1}{sqrt{n+1}} < frac{1}{20} implies sqrt{n+1} > 20 implies n+1 > 400 implies n > 399Solving the right inequality:1 - frac{1}{sqrt{n+1}} < frac{20}{21} implies frac{1}{sqrt{n+1}} > frac{1}{21} implies sqrt{n+1} < 21 implies n+1 < 441 implies n < 440Thus, ( n ) must satisfy ( 400 leq n leq 439 ). The difference between the maximum and minimum values of ( n ) is:439 - 400 = 39The final answer is (boxed{39}).,<think>Okay, let's see. The problem is about a sum that's greater than 19/20 and less than 20/21. We need to find the difference between the maximum and minimum values of the positive integer n that satisfy this. Hmm, okay. Let me start by understanding the given series.The series is:1/(1×√2 + 2√1) + 1/(2√3 + 3√2) + ... + 1/(n√(n+1) + (n+1)√n)So each term in the series is of the form 1/(k√(k+1) + (k+1)√k). The sum starts from k=1 and goes up to k=n. The problem states that this sum is greater than 19/20 and less than 20/21. We need to find the possible values of n and then take the difference between the maximum and minimum n.First, maybe I can try to simplify the general term. Let's look at one term: 1/(k√(k+1) + (k+1)√k). That denominator looks complicated. Maybe rationalizing it would help? Let's try.To rationalize the denominator, we can multiply the numerator and denominator by the conjugate. The conjugate of the denominator (a + b) is (a - b), so multiplying by (a - b)/(a - b) would rationalize it.Let me denote the denominator as D = k√(k+1) + (k+1)√k. Then the conjugate would be k√(k+1) - (k+1)√k. Wait, but actually, let's confirm:Wait, the denominator is A + B where A = k√(k+1) and B = (k+1)√k. So the conjugate would be A - B. Therefore, multiplying numerator and denominator by (A - B)/(A - B) would give us:[1 * (A - B)] / [(A + B)(A - B)] = (A - B)/(A² - B²)Let me compute A² and B².A = k√(k+1), so A² = k²(k + 1)B = (k + 1)√k, so B² = (k + 1)² kTherefore, A² - B² = k²(k + 1) - (k + 1)^2 k = (k + 1)k [k - (k + 1)] = (k + 1)k [ -1 ] = -k(k + 1)So the denominator becomes -k(k + 1). Therefore, the expression after rationalization is:(A - B)/(-k(k + 1)) = (k√(k+1) - (k + 1)√k)/(-k(k + 1))Simplify numerator:k√(k + 1) - (k + 1)√k = k√(k + 1) - k√k - √kWait, maybe factor something out? Let's see:= k[√(k + 1) - √k] - √kHmm, not sure if that's helpful. Alternatively, factor out √k from both terms:Wait, but the terms are k√(k+1) and -(k+1)√k. Let's factor out √k:= √k [k√( (k+1)/k ) - (k + 1) ]√( (k+1)/k ) = √(1 + 1/k )So:= √k [k √(1 + 1/k) - (k + 1)]Hmm, not sure. Maybe another approach. Let's go back to the rationalized expression:After rationalization, each term becomes:[ k√(k + 1) - (k + 1)√k ] / [ -k(k + 1) ]Let me factor out √k from the numerator:√k [ k √( (k + 1)/k ) - (k + 1) ] / [ -k(k + 1) ]Wait, √( (k + 1)/k ) = √(1 + 1/k ). Let's write that:= √k [ k √(1 + 1/k ) - (k + 1) ] / [ -k(k + 1) ]Hmm, maybe we can simplify inside the brackets:k √(1 + 1/k ) - (k + 1) = k [ √(1 + 1/k ) - 1 ] - 1But this might not be helpful. Alternatively, perhaps we can approximate √(1 + 1/k ) using a Taylor series, but maybe that's overcomplicating.Alternatively, let's note that [k√(k+1) - (k + 1)√k] can be written as √k√k√(k + 1) - √(k + 1)√(k + 1)√kWait, that might not help. Let me compute the numerator:k√(k + 1) - (k + 1)√kFactor out √k:√k [ k √( (k + 1)/k ) - (k + 1) / √k ]Wait, that seems messy. Let's try plugging in specific values to see if there's a pattern.Let's take k = 1:Numerator: 1*√2 - 2*√1 = √2 - 2 ≈ 1.414 - 2 ≈ -0.586Denominator after rationalization: -1*2 = -2So the term becomes (-0.586)/(-2) ≈ 0.293. Wait, but let's compute it exactly:[1√2 - 2√1]/(-1*2) = (√2 - 2)/(-2) = (2 - √2)/2 ≈ (2 - 1.414)/2 ≈ 0.586/2 ≈ 0.293So the first term after rationalization is (2 - √2)/2 ≈ 0.293.But let's check: the original term was 1/(1√2 + 2√1) = 1/(√2 + 2). Let's compute that:√2 ≈ 1.414, so √2 + 2 ≈ 3.414. 1/3.414 ≈ 0.293. So same result. So rationalizing gives us (2 - √2)/2, which is the same as 1/(√2 + 2). So maybe after rationalization, the term can be expressed as (something).But perhaps there's a telescoping sum here. If each term can be written as a difference of two terms, then the sum might collapse. Let's see.Looking back at the expression after rationalization:[ k√(k + 1) - (k + 1)√k ] / [ -k(k + 1) ]Let's write this as:[ (k + 1)√k - k√(k + 1) ] / [k(k + 1) ]Which is the same as:( (k + 1)√k - k√(k + 1) ) / [k(k + 1) ]Let me split the fraction:= (k + 1)√k / [k(k + 1)] - k√(k + 1)/[k(k + 1)]Simplify each term:First term: (k + 1)√k / [k(k + 1)] = √k / k = 1/√kSecond term: k√(k + 1)/[k(k + 1)] = √(k + 1)/(k + 1) = 1/√(k + 1)Therefore, the entire expression simplifies to:1/√k - 1/√(k + 1)Wait, that's a telescoping term! So each term in the original sum, after rationalization, is equal to 1/√k - 1/√(k + 1). Therefore, the sum becomes telescoping.So the original sum:Sum_{k=1}^n [1/(k√(k+1) + (k+1)√k)] = Sum_{k=1}^n [1/√k - 1/√(k + 1) ]Which telescopes to:[1/√1 - 1/√2] + [1/√2 - 1/√3] + ... + [1/√n - 1/√(n + 1)]All the intermediate terms cancel out, leaving:1 - 1/√(n + 1)Therefore, the sum is equal to 1 - 1/√(n + 1).So the problem states that this sum is greater than 19/20 and less than 20/21. Therefore:19/20 < 1 - 1/√(n + 1) < 20/21Let's solve these inequalities for n.First, let's subtract 1 from all parts:19/20 - 1 < -1/√(n + 1) < 20/21 - 1Which simplifies to:-1/20 < -1/√(n + 1) < -1/21Multiply all parts by -1, which reverses the inequalities:1/20 > 1/√(n + 1) > 1/21Which can be written as:1/21 < 1/√(n + 1) < 1/20Take reciprocals, remembering to reverse the inequalities again:21 > √(n + 1) > 20Square all parts:21² > n + 1 > 20²Compute 20² = 400, 21² = 441. So:441 > n + 1 > 400Subtract 1 from all parts:440 > n > 399Therefore, n must satisfy 399 < n < 440. But n is a positive integer, so n can be from 400 up to 439 inclusive.Wait, but hold on. Let me check again. The inequalities after squaring:√(n + 1) > 20 ⇒ n + 1 > 400 ⇒ n > 399√(n + 1) < 21 ⇒ n + 1 < 441 ⇒ n < 440So since n is an integer, n must satisfy 400 ≤ n ≤ 439.Therefore, the minimum value of n is 400, and the maximum value is 439.The difference between maximum and minimum is 439 - 400 = 39.Wait, but let's confirm:If n is greater than 399, so n >= 400.If n is less than 440, so n <= 439.Therefore, n ranges from 400 to 439 inclusive. The number of terms is 439 - 400 + 1 = 40. But the problem asks for the difference between maximum and minimum values of n. So maximum n is 439, minimum is 400. Difference is 439 - 400 = 39.But let's double-check if the inequalities are strict. The original inequalities are:19/20 < 1 - 1/√(n + 1) < 20/21So solving 1 - 1/√(n + 1) > 19/20 ⇒ 1/√(n + 1) < 1 - 19/20 = 1/20 ⇒ √(n + 1) > 20 ⇒ n + 1 > 400 ⇒ n > 399. Since n is integer, n ≥ 400.Similarly, 1 - 1/√(n + 1) < 20/21 ⇒ 1/√(n + 1) > 1 - 20/21 = 1/21 ⇒ √(n + 1) < 21 ⇒ n + 1 < 441 ⇒ n < 440 ⇒ n ≤ 439.Hence, n can be 400 to 439 inclusive. Therefore, the difference between maximum and minimum n is 439 - 400 = 39.So the answer should be 39. Let me check if I made any mistakes in the telescoping sum part.Original sum was each term 1/(k√(k+1)+(k+1)√k). After rationalization, each term became [1/√k - 1/√(k + 1)], which telescopes to 1 - 1/√(n + 1). That seems right.Testing with n=1: sum is 1/(1√2 + 2√1) = 1/(√2 + 2). Rationalize: (2 - √2)/ ( (√2 + 2)(2 - √2) ) = (2 - √2)/(4 - 2) = (2 - √2)/2 = 1 - (√2)/2 ≈ 1 - 0.707 ≈ 0.293. On the other hand, 1 - 1/√(1 + 1) = 1 - 1/√2 ≈ same value. So that's correct.Therefore, the answer is 39.Final Answerboxed{39}</think>Given the series:begin{array}{l}frac{1}{1 times sqrt{2}+2 sqrt{1}}+frac{1}{2 sqrt{3}+3 sqrt{2}}+cdots+ frac{1}{n sqrt{n+1}+(n+1) sqrt{n}}end{array}we need to find the difference between the maximum and minimum values of the positive integer ( n ) such that the sum is greater than ( frac{19}{20} ) and less than ( frac{20}{21} ).First, we simplify each term by rationalizing the denominator. Each term ( frac{1}{k sqrt{k+1} + (k+1) sqrt{k}} ) can be rationalized to ( frac{1}{sqrt{k}} - frac{1}{sqrt{k+1}} ). This results in a telescoping series:sum_{k=1}^n left( frac{1}{sqrt{k}} - frac{1}{sqrt{k+1}} right)The series telescopes to:1 - frac{1}{sqrt{n+1}}We need this sum to satisfy the inequalities:frac{19}{20} < 1 - frac{1}{sqrt{n+1}} < frac{20}{21}Solving these inequalities, we start by isolating ( frac{1}{sqrt{n+1}} ):frac{19}{20} < 1 - frac{1}{sqrt{n+1}} < frac{20}{21}Subtracting 1 from all parts:-frac{1}{20} < -frac{1}{sqrt{n+1}} < -frac{1}{21}Multiplying by -1 and reversing the inequalities:frac{1}{21} < frac{1}{sqrt{n+1}} < frac{1}{20}Taking reciprocals and squaring:21^2 > n + 1 > 20^2This simplifies to:441 > n + 1 > 400Subtracting 1 from all parts:440 > n > 399Thus, ( n ) must be an integer between 400 and 439 inclusive. The difference between the maximum and minimum values of ( n ) is:439 - 400 = 39The final answer is:boxed{39}]

🔑:Consider the general term. The k-th term isbegin{array}{l}frac{1}{k sqrt{k+1}+(k+1) sqrt{k}} =frac{1}{sqrt{k} sqrt{k+1}(sqrt{k}+sqrt{k+1})} =frac{sqrt{k+1}-sqrt{k}}{sqrt{k} sqrt{k+1}} =frac{1}{sqrt{k}}-frac{1}{sqrt{k+1}}(k=1,2, cdots, n) .end{array}Thus, the original expression isbegin{array}{l}=left(frac{1}{sqrt{1}}-frac{1}{sqrt{2}}right)+left(frac{1}{sqrt{2}}-frac{1}{sqrt{3}}right)+cdots+left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right) =1-frac{1}{sqrt{n+1}} .end{array}From the given information,begin{array}{l}frac{19}{20}frac{1}{sqrt{n+1}}>frac{1}{21} Rightarrow 20<sqrt{n+1}<21 Rightarrow 20^{2}-1<n<21^{2}-1 .end{array}Since n is a positive integer,n_{max }=21^{2}-2, n_{text {min }}=20^{2} text {. }Therefore, 21^{2}-2-20^{2}=39 is the answer.

🧠:[<think>Okay, let's try to figure out this problem. So, we have a circle with radius 3 and a square with side length 2. The circle crosses the center of the square. We need to find the positive difference between the areas of the nonoverlapping portions of the two figures. Hmm, okay.First, let me visualize this. The square has side length 2, so each side is 2 units long. The center of the square is where the circle's center is located, right? Wait, the problem says the circle crosses the center of the square. Wait, maybe the center of the square is a point that lies on the circle? Or is the center of the square coinciding with the center of the circle? Hmm, the wording says "a circle of radius 3 crosses the center of a square of side length 2". So maybe the circle passes through the center of the square, but the circle's center isn't necessarily the same as the square's center. Wait, but if the circle passes through the center of the square, but the square has a side length of 2, so its center is at some point. Wait, maybe the circle is centered at a different point such that it passes through the square's center. But the problem doesn't specify where the circle is located. Hmm, this is confusing.Wait, maybe the problem states that the circle's center is at the square's center. Because otherwise, how else would the circle cross the center of the square? If the circle has radius 3 and the square has side length 2, then if the circle is centered at the square's center, then the entire square would be inside the circle, right? Because the distance from the center of the square to any of its corners is half the diagonal, which is (sqrt(2^2 + 2^2))/2 = sqrt(8)/2 = 2*sqrt(2)/2 = sqrt(2) ≈ 1.414. So the square's center is only about 1.414 units away from its corners, but the circle has radius 3, which is much larger. So the entire square would be inside the circle. Then the overlapping area would be the entire square, and the non-overlapping portions would be the circle's area minus the square's area, and the square's area minus the circle's area. But since the square is entirely inside the circle, the non-overlapping portion of the square would be zero, and the non-overlapping portion of the circle would be the circle's area minus the square's area. Then the positive difference between these two would just be that value. But wait, let's check again.Wait, the problem says "the positive difference between the areas of the nonoverlapping portions of the figures." So for the square, the non-overlapping portion is the part of the square not inside the circle, and for the circle, the non-overlapping portion is the part of the circle not inside the square. Then we have to compute |Area_nonoverlap_circle - Area_nonoverlap_square|. But if the square is entirely inside the circle, then Area_nonoverlap_square is zero, and Area_nonoverlap_circle is Area_circle - Area_overlap. But since the square is entirely inside the circle, Area_overlap is Area_square, so Area_nonoverlap_circle = π*3² - 2² = 9π - 4. Then the difference between 9π - 4 and 0 would be 9π - 4, so the positive difference is 9π - 4. But let me verify if the square is indeed entirely inside the circle.The square has side length 2, so from center to any side is 1 unit. The distance from the center to any corner is sqrt(1² + 1²) = sqrt(2) ≈ 1.414. The circle has radius 3, which is much larger than sqrt(2). Therefore, the entire square is inside the circle. Therefore, the overlapping area is the entire area of the square. So the non-overlapping portion of the square is zero, and the non-overlapping portion of the circle is the circle's area minus the square's area, which is 9π - 4. Therefore, the positive difference is |9π - 4 - 0| = 9π - 4. Is that right?But wait, the problem says "the positive difference between the areas of the nonoverlapping portions of the figures." So nonoverlapping portions would be (circle area - overlap) and (square area - overlap). But if the square is entirely within the circle, then the square's nonoverlapping portion is zero, and the circle's nonoverlapping portion is circle area - square area. Therefore, the positive difference is (9π - 4) - 0 = 9π - 4. Therefore, the answer is 9π - 4.Wait, but let me check if that's the case.Alternatively, maybe the circle is centered at the square's center, but the square is not entirely inside the circle? Wait, but the radius is 3, which is much larger than the square's half-diagonal. So the square must be entirely inside the circle. Therefore, overlapping area is the entire square. So non-overlapping portions: circle has 9π - 4, square has 0. Difference is 9π - 4.But the problem says "the positive difference". So even if it's 4 - 9π, we take absolute value. But since 9π is about 28.27, and 4 is 4, so 28.27 - 4 is positive, so 9π - 4 is already positive. Therefore, the answer is 9π - 4.But wait, maybe I made a wrong assumption here. The problem states: "a circle of radius 3 crosses the center of a square of side length 2". So maybe the circle is not centered at the square's center. Maybe the circle is passing through the square's center but is centered somewhere else. Then the square might be partially inside and partially outside the circle. Wait, but the problem says "crosses the center", so perhaps the center of the square is on the circumference of the circle. So the circle's center is at a distance of 3 from the square's center. Wait, but then the square's center is on the circle.But if the circle's center is 3 units away from the square's center, then depending on the orientation, the square might be partially inside the circle.Wait, perhaps I misinterpreted the problem. Let me re-read it: "A circle of radius 3 crosses the center of a square of side length 2. Find the positive difference between the areas of the nonoverlapping portions of the figures."Hmm. So the circle crosses the center of the square. That could mean that the center of the square is a point on the circle. So the distance from the circle's center to the square's center is equal to the radius of the circle, which is 3. Therefore, the square is located such that its center is on the circumference of the circle. So the circle is centered 3 units away from the square's center.In that case, the square is partially inside and partially outside the circle. So we need to compute the overlapping area between the square and the circle, then subtract that from both the circle and the square to get their non-overlapping portions, and then take the positive difference between those two areas.That makes the problem more complex. So perhaps I initially misunderstood the problem. If the circle's center is 3 units away from the square's center, then the square is only partially inside the circle. Then we need to compute the area of overlap.But how would the square be positioned? The square has side length 2, so its center is at (0,0) for simplicity. Then the circle is centered at (3,0), radius 3. So the square extends from (-1, -1) to (1, 1). The circle is centered at (3,0) with radius 3. The distance from the circle's center to the square's center is 3 units. Then the square is just touching the circle? Wait, no. The distance from (3,0) to the square's rightmost point at (1,0) is 2 units. So the rightmost point of the square is 2 units away from the circle's center, and the circle's radius is 3, so the square extends from x = -1 to x = 1, y = -1 to y = 1. The circle is centered at (3,0), so the leftmost point of the circle is at (0,0). Therefore, the square's right half from x=0 to x=1 is inside the circle? Wait, the circle's leftmost point is at (0,0), so any point to the left of that is outside. The square is from x=-1 to x=1, so the part from x=0 to x=1 is within the circle's left half? Wait, no. Wait, the circle is centered at (3,0) with radius 3, so it goes from x=0 to x=6. So the square, which is from x=-1 to x=1, is entirely to the left of the circle's center. The circle's leftmost point is at (0,0), so the square's right edge is at x=1, which is 2 units away from the circle's center at (3,0). The distance from (3,0) to (1,0) is 2, which is less than the radius 3, so the point (1,0) is inside the circle. Similarly, points (1,1) and (1,-1) are at distance sqrt((3-1)^2 + (0-1)^2) = sqrt(4 + 1) = sqrt(5) ≈ 2.236, which is less than 3, so those points are also inside the circle. The square's top right corner is at (1,1), which is inside the circle. Then, how much of the square is inside the circle?Wait, if the square is centered at (0,0) and the circle is centered at (3,0) with radius 3, then the entire square is inside the circle? Because the furthest point of the square from the circle's center is the leftmost point (-1,0), which is 4 units away from (3,0). But 4 is greater than the radius 3, so that point is outside. Therefore, part of the square is inside the circle, and part is outside.Therefore, the overlapping area is the part of the square that is inside the circle.So to compute the non-overlapping portions:- For the square: Area of square minus overlapping area.- For the circle: Area of circle minus overlapping area.Then take the positive difference between these two.Therefore, the problem requires computing the overlapping area between the square and the circle, then computing non-overlapping areas for each, then their difference.This seems more complicated. Let me try to set up coordinates.Let me place the square with its center at the origin (0,0). The square has side length 2, so its corners are at (1,1), (1,-1), (-1,1), (-1,-1). The circle is such that it crosses the center of the square, which is at (0,0). Therefore, the center of the circle must be 3 units away from (0,0), because the radius is 3, so the circle passes through (0,0). Let's assume the circle is centered at (3,0) for simplicity. So the equation of the circle is (x - 3)^2 + y^2 = 9.We need to find the area of overlap between the square and the circle.To compute the overlapping area, we can integrate over the square where it's inside the circle. But since the square is symmetric with respect to the x-axis, we can compute the area in the upper half and double it.The square extends from x = -1 to x = 1, and y = -1 to y = 1. The circle is centered at (3,0) with radius 3. The leftmost point of the circle is at x=0, which is the center of the square. So the overlapping region is the part of the square from x=0 to x=1, and y=-1 to y=1, plus maybe a small part to the left of x=0?Wait, let me check. For points in the square from x=-1 to x=0, the distance from (3,0) would be sqrt((3 - x)^2 + y^2). The minimum distance at x=0, y=0 is 3 units, which is exactly the radius. So the point (0,0) is on the circle. Points left of x=0 in the square would be at x negative, so the distance from (3,0) would be more than 3, hence outside the circle. Therefore, only the part of the square where x >= 0 is inside the circle.Therefore, the overlapping area is the part of the square from x=0 to x=1, y=-1 to y=1. But wait, even at x=1, the distance from (3,0) is sqrt((3-1)^2 + 0^2) = 2, which is less than 3, so the entire right half of the square (from x=0 to x=1) is inside the circle. However, the left half (x from -1 to 0) is outside the circle.Therefore, the overlapping area is half of the square's area? Wait, the square's total area is 4 (since side length 2, area 2x2=4). Wait, no, wait. Wait, the square has side length 2, so each side is 2 units, so from -1 to 1 in both x and y. So area is (2)^2 = 4. But if overlapping region is x from 0 to 1, which is half the square along the x-axis, then overlapping area is 2. But let me check.Wait, the square from x=0 to x=1, y=-1 to y=1 is a rectangle of length 1 and height 2, so area 2. But is the entire rectangle inside the circle?At x=0, y ranges from -1 to 1. The distance from (3,0) to (0, y) is sqrt(3^2 + y^2) = sqrt(9 + y^2). Since y is up to 1, so sqrt(9 + 1) = sqrt(10) ≈ 3.16, which is greater than 3. Therefore, points (0,1) and (0,-1) are outside the circle. So the overlapping region is not the entire rectangle from x=0 to x=1, but a region where the distance from (3,0) is less than or equal to 3.Therefore, the overlapping area is the part of the square where x >= 0 and (x - 3)^2 + y^2 <= 9.To compute this area, let's switch to integrating.We can set up the integral in Cartesian coordinates. For x from 0 to 1, and y from -1 to 1, but within the circle.But the circle equation is (x - 3)^2 + y^2 = 9.Solving for y in terms of x, we get y = sqrt(9 - (x - 3)^2). But since the square restricts y between -1 and 1, the overlapping region for each x is y between -1 and 1, but only if that range is within the circle's bounds.But let's check for x from 0 to 1:For x in [0,1], the circle's y ranges are from -sqrt(9 - (x - 3)^2) to sqrt(9 - (x - 3)^2). But since the square restricts y to [-1,1], the overlapping y-values are the intersection of these intervals.So, the overlapping area for each x is:y_top = min(1, sqrt(9 - (x - 3)^2))y_bottom = max(-1, -sqrt(9 - (x - 3)^2))But since sqrt(9 - (x - 3)^2) is the upper half of the circle, and negative sqrt is the lower half.But given that x is in [0,1], let's compute sqrt(9 - (x - 3)^2):Let me compute (x - 3) for x in [0,1] is [-3, -2], so (x - 3)^2 is [9, 4]. Therefore, 9 - (x - 3)^2 is [0, 5]. Therefore, sqrt(9 - (x - 3)^2) is [0, sqrt(5) ≈ 2.236]. But since the square restricts y to [-1,1], the overlapping region for y is from -1 to 1. Therefore, for all x in [0,1], the circle's y-range includes the square's y-range. Because sqrt(9 - (x - 3)^2) >= sqrt(5) ≈ 2.236, which is greater than 1. Therefore, for all x in [0,1], the circle's y goes from -2.236 to 2.236, which includes the square's y from -1 to 1. Therefore, the overlapping area in the square is the entire right half of the square, which is area 2.Wait, that can't be, because when x=0, the circle at x=0 has y^2 = 9 - (0 - 3)^2 = 9 - 9 = 0, so y=0. So at x=0, the only point on the circle is (0,0). So the overlapping region at x=0 is just the point (0,0). Wait, but earlier when I considered x=0, the circle's equation gives y=0. Therefore, at x=0, only the point (0,0) is on both the square and the circle. Then, for x slightly more than 0, the circle allows some y-values.Wait, perhaps my earlier reasoning is flawed. Let's re-express the circle's equation.The circle is (x - 3)^2 + y^2 = 9. So for a given x, the y-values are between -sqrt(9 - (x - 3)^2) and sqrt(9 - (x - 3)^2).So let's plug x=0 into the circle's equation:(0 - 3)^2 + y^2 = 9 => 9 + y^2 = 9 => y^2 = 0 => y=0. So at x=0, the only point on the circle is (0,0). Therefore, at x=0, the overlapping region is just that single point. For x > 0, the circle allows more y-values. For example, x=1:(1 - 3)^2 + y^2 = 4 + y^2 = 9 => y^2 = 5 => y=±sqrt(5) ≈ ±2.236. But the square only goes up to y=1, so at x=1, the overlapping y-values are from -1 to 1.Wait, so for x from 0 to 1, the circle's y-values go from -sqrt(9 - (x - 3)^2) to sqrt(9 - (x - 3)^2). However, since the square's y ranges from -1 to 1, the overlapping region's y-values are the intersection of these intervals.Therefore, for x between 0 and 1, the overlapping y-values are from -1 to 1 only if sqrt(9 - (x - 3)^2) >= 1.Let's check when sqrt(9 - (x - 3)^2) >= 1.Squaring both sides: 9 - (x - 3)^2 >= 1 => (x - 3)^2 <= 8 => |x - 3| <= sqrt(8) ≈ 2.828. Since x is in [0,1], |x - 3| = 3 - x, which is between 2 and 3. Since sqrt(8) ≈ 2.828, so 3 - x <= sqrt(8) => x >= 3 - sqrt(8) ≈ 3 - 2.828 ≈ 0.172. Therefore, for x >= 0.172, sqrt(9 - (x - 3)^2) >= 1. Therefore, in this interval x >= 0.172, the overlapping y is from -1 to 1. For x < 0.172, sqrt(9 - (x - 3)^2) < 1, so the overlapping y is from -sqrt(9 - (x - 3)^2) to sqrt(9 - (x - 3)^2).Therefore, the overlapping area can be split into two parts:1. x from 0 to 3 - sqrt(8) ≈ 0.172: the overlapping region is a vertical slice where y ranges from -sqrt(9 - (x - 3)^2) to sqrt(9 - (x - 3)^2).2. x from 3 - sqrt(8) to 1: the overlapping region is y from -1 to 1.Therefore, the total overlapping area is the integral from x=0 to x=3 - sqrt(8) of 2*sqrt(9 - (x - 3)^2) dx plus the integral from x=3 - sqrt(8) to x=1 of 2*1 dx.This seems complicated, but let's compute it step by step.First, let's compute 3 - sqrt(8). sqrt(8) is 2*sqrt(2) ≈ 2.828, so 3 - 2.828 ≈ 0.172.Let’s denote a = 3 - sqrt(8). Then, the overlapping area is:A = ∫ from 0 to a [2*sqrt(9 - (x - 3)^2)] dx + ∫ from a to 1 [2*1] dx.Let’s compute the second integral first:∫ from a to 1 2 dx = 2*(1 - a).Now, the first integral:∫ from 0 to a 2*sqrt(9 - (x - 3)^2) dx.Let’s make a substitution: let u = x - 3. Then, du = dx. When x=0, u=-3; when x=a, u=a - 3 = -sqrt(8).Wait, but integrating from u=-3 to u=-sqrt(8):∫ from u=-3 to u=-sqrt(8) 2*sqrt(9 - u^2) du.This is the integral of 2*sqrt(9 - u^2) du from -3 to -sqrt(8). But sqrt(9 - u^2) is the equation for a semicircle of radius 3. The integral of sqrt(9 - u^2) du from a to b is (1/2)( (u sqrt(9 - u^2) + 9 arcsin(u/3)) ) evaluated from a to b.Therefore, the integral from -3 to -sqrt(8) of 2*sqrt(9 - u^2) du is 2*( (1/2)(u sqrt(9 - u^2) + 9 arcsin(u/3)) ) evaluated from -3 to -sqrt(8).Simplifying:= [u sqrt(9 - u^2) + 9 arcsin(u/3)] from -3 to -sqrt(8)First, evaluate at u = -sqrt(8):First term: (-sqrt(8)) * sqrt(9 - 8) = (-sqrt(8)) * sqrt(1) = -sqrt(8)Second term: 9 arcsin(-sqrt(8)/3)Note that sqrt(8) = 2*sqrt(2), so sqrt(8)/3 ≈ 2.828/3 ≈ 0.9428. But arcsin of that is valid because sqrt(8)/3 ≈ 0.9428 < 1.But arcsin(-sqrt(8)/3) = - arcsin(sqrt(8)/3)So the total at u = -sqrt(8):= -sqrt(8) + 9*(-arcsin(sqrt(8)/3)) = -sqrt(8) - 9 arcsin(sqrt(8)/3)Now, evaluate at u = -3:First term: (-3)*sqrt(9 - 9) = 0Second term: 9 arcsin(-3/3) = 9 arcsin(-1) = 9*(-π/2)Therefore, at u = -3:= 0 + 9*(-π/2) = -9π/2Thus, the integral from -3 to -sqrt(8):[ -sqrt(8) - 9 arcsin(sqrt(8)/3) ] - [ -9π/2 ] = -sqrt(8) - 9 arcsin(sqrt(8)/3) + 9π/2Therefore, the first integral is equal to this value.Putting it all together, the overlapping area A is:A = [ -sqrt(8) - 9 arcsin(sqrt(8)/3) + 9π/2 ] + 2*(1 - a)But a = 3 - sqrt(8), so 1 - a = 1 - 3 + sqrt(8) = -2 + sqrt(8)Therefore, the second integral is 2*(-2 + sqrt(8)) = -4 + 2*sqrt(8)Thus, total overlapping area:A = [ -sqrt(8) - 9 arcsin(sqrt(8)/3) + 9π/2 ] + [ -4 + 2*sqrt(8) ]Simplify:= (-sqrt(8) + 2*sqrt(8)) - 4 - 9 arcsin(sqrt(8)/3) + 9π/2= sqrt(8) - 4 - 9 arcsin(sqrt(8)/3) + 9π/2But sqrt(8) is 2*sqrt(2), so:= 2*sqrt(2) - 4 - 9 arcsin(2*sqrt(2)/3) + (9π)/2So this is the overlapping area.Now, the non-overlapping area of the square is Area_square - A = 4 - AThe non-overlapping area of the circle is Area_circle - A = 9π - ATherefore, the positive difference between these two is |(4 - A) - (9π - A)| = |4 - A -9π + A| = |4 -9π| = 9π -4Wait, but that's the same result as before when we assumed the entire square was inside the circle. But that contradicts our previous analysis where part of the square is outside the circle.Hmm, this suggests that despite the overlapping area being complicated, the difference ends up being the same as if the entire square was inside the circle. That seems odd. Let me check.Wait, the positive difference is |(4 - A) - (9π - A)| = |4 -9π|. The A cancels out. So regardless of the overlapping area A, the difference between the non-overlapping areas is |4 -9π|, which is 9π -4. Therefore, the answer is 9π -4, which is approximately 28.27 -4 = 24.27.But this seems counterintuitive. How can the difference be the same regardless of overlapping area? Because when you take the difference between (square's non-overlap) and (circle's non-overlap), you have (4 - A) - (9π - A) = 4 - A -9π + A = 4 -9π. The overlapping area cancels out. Therefore, the positive difference is |4 -9π| =9π -4.Therefore, regardless of how much the square and circle overlap, the positive difference between their non-overlapping areas is always 9π -4. That seems mathematically consistent, but is it logically sound?Wait, let's think differently. If the square is entirely inside the circle, then non-overlapping areas are 0 (for the square) and 9π -4 (for the circle). The difference is 9π -4. If the square is partially inside, then non-overlapping part of the square is 4 - A, and non-overlapping part of the circle is 9π -A. The difference is (9π -A) - (4 -A) =9π -4. So regardless of A, the difference remains 9π -4. Therefore, the answer is 9π -4.Therefore, even if the square is partially overlapping, the difference between the non-overlapping areas is 9π -4. So the answer is 9π -4.But why is that? Because when you subtract the same overlapping area from both the square and the circle, the difference between their non-overlapping areas becomes (Area_circle - A) - (Area_square - A) = Area_circle - Area_square. Therefore, the overlapping area cancels out, and the difference is simply the difference between the total areas of the circle and the square.Therefore, regardless of how they overlap, the positive difference between the non-overlapping portions is just the difference between their total areas. That makes sense, because the overlapping area is subtracted from both, so when you take the difference, it cancels.Therefore, the answer is |9π -4| =9π -4.Wow, that was a long process, but I think that's the key insight. The positive difference between the non-overlapping portions is equal to the positive difference between the total areas of the two figures. Because when you subtract the same overlapping area from both, their difference remains the same as the difference of the total areas.Therefore, the answer is 9π -4.Final AnswerThe positive difference is boxed{9pi - 4}.</think>To solve the problem of finding the positive difference between the areas of the nonoverlapping portions of a circle of radius 3 and a square of side length 2, where the circle crosses the center of the square, we can follow these steps:1. Determine the areas of the circle and the square: - The area of the circle is ( pi times 3^2 = 9pi ). - The area of the square is ( 2 times 2 = 4 ).2. Understand the overlap: - The circle is centered such that it crosses the center of the square. However, the key insight is that the positive difference between the nonoverlapping portions of the figures depends only on the difference between their total areas, regardless of the overlap. This is because the overlapping area is subtracted from both the circle and the square, and thus cancels out when taking the difference.3. Calculate the positive difference: - The nonoverlapping portion of the circle is ( 9pi - text{overlapping area} ). - The nonoverlapping portion of the square is ( 4 - text{overlapping area} ). - The positive difference between these nonoverlapping portions is ( |(9pi - text{overlapping area}) - (4 - text{overlapping area})| = |9pi - 4| ).4. Final result: - Since ( 9pi ) is greater than 4, the positive difference is ( 9pi - 4 ).Thus, the positive difference between the areas of the nonoverlapping portions of the figures is (boxed{9pi - 4}).,<think>Alright, so I have this problem here: There's a circle with radius 3 and a square with side length 2. The circle crosses the center of the square. I need to find the positive difference between the areas of the non-overlapping parts of the circle and the square. Hmm, okay. Let me try to visualize this.First, let me recall the formulas for the areas. The area of the circle is πr², which in this case would be π*(3)² = 9π. The area of the square is side length squared, so 2² = 4. So the total areas are 9π for the circle and 4 for the square. But since they overlap somewhere, I need to subtract the overlapping area from both to find the non-overlapping parts. Then, take the positive difference between those two non-overlapping areas.Wait, the problem says "the positive difference between the areas of the nonoverlapping portions of the figures." So, I need to compute (Area of circle minus overlapping area) minus (Area of square minus overlapping area) and take the absolute value? Or maybe the problem wants the sum of the non-overlapping parts of both figures and then subtract each other? Wait, no. Let me read again: "positive difference between the areas of the nonoverlapping portions of the figures." So each figure has a non-overlapping portion, and we need the positive difference between those two areas. So, compute |(Circle area - overlap) - (Square area - overlap)| = |Circle area - Square area|. Wait, that can't be right. Because if you subtract the overlap from both, then when you subtract those, the overlaps cancel out. So the difference between the non-overlapping areas is just the difference between the original areas. But that seems too straightforward. The problem mentions that the circle crosses the center of the square, so maybe the overlap is not zero. Therefore, maybe my initial thought is wrong.Wait, let me test with numbers. Suppose the circle and square don't overlap at all. Then the non-overlapping areas would be 9π and 4, difference |9π - 4|. If they overlap completely, then the non-overlapping areas would be 9π - 4 and 0, difference |9π - 4 - 0| = same. Wait, but in reality, the circle has radius 3 and the square has side 2, and the circle crosses the center of the square. So the square is inside the circle? Because the square has side length 2, so the distance from its center to any corner is sqrt(1² + 1²) = sqrt(2) ≈ 1.414, which is less than 3. So the entire square is inside the circle. Wait, is that true? Let me check.The square has side length 2, so each half-side is 1. The center of the square is at the intersection of its diagonals. If the circle is centered at the square's center, then the distance from the center to any corner of the square is sqrt(1² + 1²) = sqrt(2) ≈ 1.414, which is indeed less than the radius 3. Therefore, the entire square is inside the circle. But the problem says "a circle of radius 3 crosses the center of a square of side length 2." Wait, does that mean the circle's center is at the square's center? Or just that the circle passes through the square's center? If the circle passes through the square's center but isn't necessarily centered there, then maybe only part of the square is inside the circle.Wait, the problem is ambiguous. Wait, let me check the exact wording: "A circle of radius 3 crosses the center of a square of side length 2." So the circle crosses (passes through) the center of the square. That could mean that the center of the square is on the circumference of the circle? Because if the circle's center was at the square's center, then the circle would contain the entire square. But if the center of the square is a point on the circle's circumference, then the circle's center is at a distance of 3 from the square's center. Then, in that case, the square is partially inside and partially outside the circle.Wait, so this is a crucial point. The problem states that the circle "crosses the center" of the square. So "crosses" here probably means that the center is on the circle's perimeter, i.e., the center of the square is a point on the circle. Therefore, the distance from the circle's center to the square's center is equal to the radius, which is 3. Wait, but the square has side length 2, so if the circle is centered 3 units away from the square's center, how much of the square is inside the circle?Wait, let me parse this again. The problem says: "A circle of radius 3 crosses the center of a square of side length 2." So "crosses" in geometric terms usually would mean that the center is on the circle, so the circle passes through the center. So the center of the square is a point on the circle. Therefore, the distance between the centers of the circle and the square is equal to the radius of the circle, which is 3. But the square is of side length 2, so its center is the midpoint, and the square extends 1 unit in all directions from the center.So, the circle has radius 3, and its center is 3 units away from the square's center. So the square is entirely within the circle only if the distance from the circle's center to the square's farthest point is less than or equal to 3. The square's farthest point from the circle's center would be on the side opposite to the circle's center. Wait, let me visualize this. Let's say the square is centered at point S, and the circle is centered at point C, which is 3 units away from S. The square has side length 2, so from S, it extends 1 unit in all directions. The maximum distance from C to any point in the square would be the distance from C to the farthest corner or side.Wait, to find the maximum distance from point C to the square, since the square is centered at S, and C is 3 units away from S, then the maximum distance from C to a point in the square would be the distance from C to S plus the maximum distance from S to the square's points. The maximum distance from S to any square point is sqrt(1² + 1²) = sqrt(2) ≈ 1.414. So the maximum distance from C to the square is 3 + sqrt(2) ≈ 4.414, which is greater than the radius 3. Therefore, parts of the square are outside the circle. Conversely, the minimum distance from C to the square is 3 - sqrt(2) ≈ 1.586, so the closest points of the square to C are about 1.586 units away, which is still greater than zero, so the circle does not fully enclose the square. Wait, but the circle has radius 3, so points on the square can be as close as 3 - sqrt(2) or as far as 3 + sqrt(2) from C. But since the circle's radius is 3, any part of the square beyond 3 units from C is outside the circle, and within 3 units is inside.Therefore, the square is partially inside and partially outside the circle. So the overlapping area is the part of the square that is inside the circle, and the non-overlapping portions are the part of the square outside the circle and the part of the circle outside the square. Then, the problem asks for the positive difference between the areas of the non-overlapping portions of the figures. That is, compute (Area of square - overlapping area) and (Area of circle - overlapping area), then take the absolute difference between these two.So, let me break it down:1. Calculate the overlapping area (the part of the square that is inside the circle).2. Subtract the overlapping area from both the square and the circle to get their non-overlapping portions.3. Find the absolute difference between these two non-overlapping areas.Therefore, the key challenge here is to compute the overlapping area, which is the area of the square that lies inside the circle.To compute this overlapping area, we need to figure out how the circle and square intersect. Given that the center of the square is on the circle (since the circle crosses the center of the square), the square is positioned such that its center is on the circumference of the circle. The square has side length 2, so it extends 1 unit in all directions from its center. The circle has radius 3 and is centered 3 units away from the square's center.Wait, so if the square's center is on the circle, then the distance between the centers of the circle and the square is 3. The square is of side 2, centered at S, and the circle is centered at C, 3 units away from S. So we have two centers separated by 3 units, the square has half-side length 1, and the circle has radius 3. So to find the overlapping area, we need to find the area of the square that is inside the circle.This seems like a geometry problem involving a circle and a square offset from each other. The intersection area can be calculated by integrating or by finding the overlapping regions, but it might require some geometric analysis.Let me try to set up a coordinate system. Let's place the square such that its center S is at the origin (0,0). The circle is centered at point C, which is 3 units away from S. Let's place point C along the positive x-axis for simplicity, so C is at (3,0). The square is centered at (0,0) with sides of length 2, so its vertices are at (±1, ±1). The circle has radius 3 and is centered at (3,0). We need to find the area of the square that lies inside this circle.So, the square extends from x = -1 to x = 1, and y = -1 to y = 1. The circle is centered at (3,0) with radius 3. The equation of the circle is (x - 3)^2 + y^2 = 9. The square is the region |x| ≤ 1 and |y| ≤ 1.We need to find the area within |x| ≤ 1, |y| ≤ 1 that also satisfies (x - 3)^2 + y^2 ≤ 9.Let me visualize this. The circle is centered at (3,0) with radius 3, so it touches the origin (0,0). Because the distance from (3,0) to (0,0) is 3, which is exactly the radius. So the origin is on the circle. Therefore, the center of the square, which is at the origin, is on the circle. The square itself is from x = -1 to 1, y = -1 to 1. The circle extends from x = 0 to 6 (since center at 3, radius 3). Therefore, the square is on the leftmost part of the circle. The overlapping region would be the left half of the square, maybe? But not exactly.Wait, let's check if the entire square is inside the circle. The farthest point of the square from the circle's center (3,0) would be the leftmost point of the square at (-1,0). The distance from (3,0) to (-1,0) is 4, which is greater than the radius 3, so that point is outside the circle. The closest point would be the rightmost point of the square at (1,0), which is 2 units away from (3,0), so inside the circle. So part of the square is inside the circle, and part is outside. Specifically, the right part of the square is inside the circle, and the left part is outside.Therefore, the overlapping area is the portion of the square where (x - 3)^2 + y^2 ≤ 9. Let's find the region within the square that satisfies this inequality.First, let's write the inequality:(x - 3)^2 + y^2 ≤ 9Expanding this:x² - 6x + 9 + y² ≤ 9Simplify:x² - 6x + y² ≤ 0But since x² + y² is always non-negative, the only way this inequality can hold is if x² + y² ≤ 6x. So, the region inside the square where x² + y² ≤ 6x.But since the square is from x = -1 to 1, and y = -1 to 1, and 6x would range from -6 to 6. However, x in the square is between -1 and 1, so 6x is between -6 and 6. But x² + y² is between 0 and (1 + 1) = 2. Therefore, the inequality x² + y² ≤ 6x can be rewritten as 6x ≥ x² + y². Since x is between -1 and 1, let's analyze this.For x in [-1,1], we have 6x. So for x negative (from -1 to 0), 6x is negative, but x² + y² is non-negative, so the inequality 6x ≥ x² + y² would never hold because left side is negative and right side is non-negative. Therefore, in the left half of the square (x from -1 to 0), there is no overlap with the circle.For x from 0 to 1, 6x is positive. So the inequality 6x ≥ x² + y² can hold here. Therefore, the overlapping area is the region in the square where x is from 0 to 1, and y satisfies y² ≤ 6x - x².So, the overlapping area is the integral over x from 0 to 1 of the length in y-direction, which is 2*sqrt(6x - x²) (since y can be from -sqrt(6x - x²) to sqrt(6x - x²)), integrated with respect to x.Therefore, overlapping area A = ∫ from 0 to 1 of 2*sqrt(6x - x²) dx.Let me compute this integral. First, let's simplify the expression under the square root:6x - x² = -x² + 6x = -(x² - 6x) = -(x² -6x + 9 - 9) = -(x - 3)^2 + 9.So, 6x - x² = 9 - (x - 3)^2. So sqrt(6x -x²) = sqrt(9 - (x -3)^2). Wait, that seems like the equation of a circle with radius 3 centered at x =3, y=0. Which makes sense, since that's our original circle equation. So integrating sqrt(9 - (x -3)^2) from x=0 to 1. But integrating this function from 0 to 1 would give the area under the curve, which is the area of the circle segment from x=0 to x=1. Since our overlapping region is exactly this segment within the square.Alternatively, maybe there is a better way to compute this integral.Let me perform a substitution. Let u = x - 3. Then du = dx. But when x =0, u = -3; when x=1, u= -2. Then the integral becomes ∫ from u=-3 to u=-2 of sqrt(9 - u²) du. But integrating sqrt(9 - u²) du is the same as the area of a circular segment from angle θ1 to θ2, where u = 3 sinθ or something. Wait, but integrating sqrt(a² - u²) du is (u/2 sqrt(a² - u²) + (a²/2) sin^{-1}(u/a)) ) + C. So, using that formula, the integral of sqrt(9 - u²) du from -3 to -2 would be:[ (u/2) sqrt(9 - u²) + (9/2) sin^{-1}(u/3) ) ] evaluated from u = -3 to u = -2.First, let's compute at u = -2:(-2/2) sqrt(9 - (-2)^2) + (9/2) sin^{-1}(-2/3)= (-1) sqrt(9 -4) + (9/2)(- sin^{-1}(2/3))= (-1) sqrt(5) - (9/2) sin^{-1}(2/3)At u = -3:(-3/2) sqrt(9 - (-3)^2) + (9/2) sin^{-1}(-3/3)= (-3/2) sqrt(0) + (9/2) sin^{-1}(-1)= 0 + (9/2)(-π/2)= - (9π)/4Therefore, the integral from u=-3 to u=-2 is [ (-sqrt(5) - (9/2) sin^{-1}(2/3)) ] - [ -9π/4 ] = -sqrt(5) - (9/2) sin^{-1}(2/3) + 9π/4But remember, the original integral was ∫ from x=0 to 1 of 2*sqrt(6x - x²) dx, which we transformed into ∫ from u=-3 to u=-2 of sqrt(9 - u²) du, so the overlapping area A = 2 * [ integral from x=0 to 1 sqrt(6x -x²) dx ] = 2 * [ integral from u=-3 to u=-2 sqrt(9 - u²) du ] = 2 * [ -sqrt(5) - (9/2) sin^{-1}(2/3) + 9π/4 ]Wait, no. Wait, when we did the substitution, u = x - 3, du = dx, so the integral becomes ∫ sqrt(9 - u²) du from u = -3 to u = -2. Then, the overlapping area A = 2 times this integral? Wait, original integral was A = ∫ from 0 to 1 of 2*sqrt(6x -x²) dx. But when we did u = x -3, then x = u +3, and dx = du. So substituting:sqrt(6x - x²) = sqrt(6(u +3) - (u +3)^2 )= sqrt(6u + 18 - (u² +6u +9))= sqrt(6u +18 - u² -6u -9)= sqrt(9 - u²)Therefore, the integral becomes ∫ from x=0 to1 sqrt(6x -x²) dx = ∫ from u=-3 to u=-2 sqrt(9 - u²) duTherefore, the overlapping area A = 2 * ∫ from 0 to1 sqrt(6x -x²) dx = 2 * ∫ from u=-3 to u=-2 sqrt(9 - u²) duSo, computing the integral as above:Integral from -3 to -2 of sqrt(9 - u²) du = [ (-sqrt(5) - (9/2) sin^{-1}(2/3)) + 9π/4 ]Therefore, A = 2 * [ (-sqrt(5) - (9/2) sin^{-1}(2/3)) + 9π/4 ] = 2*(-sqrt(5)) - 2*(9/2) sin^{-1}(2/3) + 2*(9π/4 )Simplify:= -2 sqrt(5) - 9 sin^{-1}(2/3) + (9π/2 )Therefore, overlapping area A = (9π/2) - 2 sqrt(5) - 9 sin^{-1}(2/3)But this seems complicated. Maybe there's a better way?Alternatively, perhaps using polar coordinates. Let me try that.But since the square is axis-aligned and the circle is offset, polar coordinates might complicate things. Let me think.Alternatively, we can model the intersection as a circular segment. The overlapping area is the area of the part of the square that is inside the circle.Given that the square is centered at (0,0) with sides from -1 to 1 in x and y, and the circle is centered at (3,0) with radius 3. The overlapping region is the set of points (x,y) where |x| ≤1, |y| ≤1, and (x-3)^2 + y^2 ≤9.As we saw before, for x from 0 to1, the circle overlaps the square, and for x from -1 to0, there is no overlap. So, the overlapping region is from x=0 to1, and for each x in that interval, y ranges from -sqrt(6x -x²) to sqrt(6x -x²), but also limited by the square's boundaries at y= -1 and y=1.Wait, but we need to check if sqrt(6x -x²) is less than or equal to1. Let's see. For x in [0,1], sqrt(6x -x²) is sqrt(x(6 -x)). When x=0, sqrt(0) =0; when x=1, sqrt(6*1 -1) = sqrt(5) ≈2.236, which is greater than1. Therefore, in the interval x ∈ [0,1], the upper and lower bounds for y given by the circle are from -sqrt(6x -x²) to sqrt(6x -x²), but since the square only extends from y=-1 to y=1, the actual overlapping region in y is from y=-1 to y=1, but only where sqrt(6x -x²) ≥1. Wait, no. Wait, for a given x in [0,1], the circle allows y from -sqrt(6x -x²) to sqrt(6x -x²). However, since the square restricts y to be between -1 and1, the overlapping y's would be the intersection of these intervals.Therefore, if sqrt(6x -x²) ≤1, then the overlapping y is from -sqrt(6x -x²) to sqrt(6x -x²). If sqrt(6x -x²) >1, then the overlapping y is from -1 to1.Wait, so we need to find the x in [0,1] where sqrt(6x -x²) =1. Solving for x:sqrt(6x -x²) =1Squaring both sides:6x -x² =1x² -6x +1 =0Using quadratic formula:x = [6 ± sqrt(36 -4)] /2 = [6 ± sqrt(32)] /2 = [6 ±4*sqrt(2)] /2 = 3 ±2*sqrt(2)So the solutions are x =3 +2√2 ≈3 +2.828≈5.828 and x=3 -2√2≈3 -2.828≈0.172. Since x is in [0,1], only x≈0.172 is relevant.Therefore, at x≈0.172, sqrt(6x -x²)=1. So for x from0 to≈0.172, sqrt(6x -x²) ≤1, so the overlapping y is from -sqrt(6x -x²) to sqrt(6x -x²). For x from≈0.172 to1, sqrt(6x -x²) >1, so overlapping y is from -1 to1.Therefore, the overlapping area can be split into two parts:1. From x=0 tox=c (where c≈0.172), the vertical slices have height 2*sqrt(6x -x²).2. From x=c tox=1, the vertical slices have height 2*1=2.Therefore, overlapping area A = ∫ from0 toc 2*sqrt(6x -x²) dx + ∫ fromc to1 2*1 dx.So first, compute c=3 -2√2≈0.172. Let me confirm:c=3 -2√2≈3 -2.828≈0.172, yes.Therefore, the overlapping area A = 2∫0^{3 -2√2} sqrt(6x -x²) dx + 2∫_{3 -2√2}^1 1 dxFirst, compute the first integral: ∫0^{3 -2√2} sqrt(6x -x²) dx. Let's use substitution again.Let me set u = x -3 again, but perhaps another substitution is better. Let me complete the square for 6x -x².6x -x² = -x² +6x = -(x² -6x) = -(x² -6x +9 -9) = -(x -3)^2 +9.So, sqrt(6x -x²) = sqrt(9 - (x -3)^2)Therefore, the integral ∫ sqrt(6x -x²) dx = ∫ sqrt(9 - (x -3)^2) dxThis is the equation of a circle with radius 3 centered at (3,0). Therefore, the integral from x=0 tox=c is the area of the circular segment from x=0 tox=c.Alternatively, using substitution. Let’s set θ such that (x -3) =3 sinθ. Then dx =3 cosθ dθ.But when x=0, 0 -3=3 sinθ => sinθ = -1 => θ=-π/2.When x=c=3 -2√2, (3 -2√2 -3)= -2√2=3 sinθ => sinθ= -2√2/3.But 2√2≈2.828, so sinθ= -2.828/3≈-0.9428. Therefore θ= arcsin(-2√2/3). Let's denote α=arcsin(2√2/3), so θ=-α.Therefore, the integral becomes:∫ sqrt(9 - (x -3)^2) dx from0 toc = ∫θ=-π/2 toθ=-α sqrt(9 -9 sin²θ) *3 cosθ dθ= ∫ from -π/2 to -α sqrt(9 cos²θ) *3 cosθ dθ= ∫ from -π/2 to -α 3 |cosθ| *3 cosθ dθSince θ is between -π/2 and -α, where α is positive and less than π/2, so θ is in the range where cosθ is positive (since between -π/2 and 0, cosine is positive). Therefore, |cosθ|=cosθ.Therefore, the integral becomes:∫ from -π/2 to -α 3 cosθ *3 cosθ dθ =9 ∫ from -π/2 to -α cos²θ dθUsing the identity cos²θ = (1 + cos2θ)/2:=9/2 ∫ from -π/2 to -α (1 + cos2θ) dθ=9/2 [ θ + (sin2θ)/2 ] evaluated from -π/2 to -α=9/2 [ (-α + (sin(-2α))/2 ) - ( -π/2 + (sin(-π))/2 ) ]Simplify:=9/2 [ -α - (sin2α)/2 + π/2 -0 ] (since sin(-π)=0)=9/2 [ (π/2 - α) - (sin2α)/2 ]Now, we need to compute α=arcsin(2√2/3). Let's compute sin2α:sin2α=2 sinα cosα=2*(2√2/3)*sqrt(1 - (8/9))=2*(2√2/3)*(1/3)=2*(2√2)/9=4√2/9Therefore:Integral =9/2 [ (π/2 - α) - (4√2/9)/2 ] =9/2 [ π/2 - α - 2√2/9 ]=9/2*(π/2 - α) - 9/2*(2√2/9)= (9/2)(π/2 - α) - √2So, the first integral ∫0^{c} sqrt(6x -x²) dx=9/2*(π/2 - α) - √2But α=arcsin(2√2/3). Therefore:= (9/2)(π/2 - arcsin(2√2/3)) - √2Now, the first part of the overlapping area is 2 times this integral:2*[ (9/2)(π/2 - arcsin(2√2/3)) - √2 ] =9*(π/2 - arcsin(2√2/3)) - 2√2The second part of the overlapping area is 2*∫_{c}^1 1 dx=2*(1 -c)=2*(1 -(3 -2√2))=2*( -2 +2√2)= -4 +4√2Therefore, total overlapping area A= [9*(π/2 - arcsin(2√2/3)) -2√2] + [ -4 +4√2 ]=9*(π/2 - arcsin(2√2/3)) -2√2 -4 +4√2Simplify:=9*(π/2 - arcsin(2√2/3)) +2√2 -4So, A= (9π/2 -9 arcsin(2√2/3)) +2√2 -4Therefore, the overlapping area is (9π/2 -9 arcsin(2√2/3) +2√2 -4 )This seems quite complicated. Maybe there's a simpler way to calculate this area?Alternatively, using symmetry or geometric properties. Let me consider the geometry again.The circle is centered at (3,0) with radius 3, so the point (0,0) is on the circle. The square is from (-1, -1) to (1,1). The overlapping region is the part of the square where (x-3)^2 + y^2 ≤9.This area can also be calculated as the area of the circle sector minus the area of the triangle formed by the circle's center, the intersection points, and so on. Wait, but the overlapping region with the square complicates this approach.Alternatively, perhaps using polar coordinates with the circle's center at (3,0). Let me try polar coordinates.In polar coordinates, the circle's equation is r=6 cosθ (since centered at (3,0), which in polar is (3,0), so the equation is r=6 cosθ). The square, however, is more complicated in polar coordinates. The square is from x=-1 to1 and y=-1 to1. So in polar coordinates, x= r cosθ, y= r sinθ. So the square is the region where |r cosθ| ≤1 and |r sinθ| ≤1. Therefore, r ≤ min(1/|cosθ|, 1/|sinθ|). But integrating this in polar coordinates might be difficult, especially because the circle is r=6 cosθ. So overlapping region is where r ≤6 cosθ and r ≤ min(1/|cosθ|, 1/|sinθ|).But this seems too complex. Maybe another approach.Wait, perhaps using coordinate transformation. Let me shift the coordinate system so that the circle is centered at the origin. Let me translate the system 3 units to the left. So, the circle's center becomes (0,0), and the square's center becomes (-3,0). The square, originally from x=-1 to1 and y=-1 to1, after translation becomes x from -4 to-2 and y from-1 to1. The circle has equation x² + y² =9.Therefore, the overlapping area is the region within the square (translated to x from -4 to-2, y from-1 to1) that is inside the circle x² + y² ≤9.But this might not simplify much. The translated square is from x=-4 to-2 and y=-1 to1. The circle of radius 3 is centered at (0,0). The overlapping region would be where x is between -4 and-2, y between -1 and1, and x² + y² ≤9.But since x is from -4 to-2, and the circle only extends to x=-3 (since radius 3). So, x from -4 to-3 is outside the circle, and x from-3 to-2 is inside the circle. Wait, no. x=-3 is exactly on the circle (distance from center is 3). x=-3. So for x from-3 to-2, the x-coordinate is between -3 and-2, y between -1 and1. So x² + y² ranges from (-3)^2 + (-1)^2 =10 to (-2)^2 +1^2=5. Since the circle's radius squared is9, so x² + y² ≤9. Therefore, the points in the translated square (x from-4 to-2, y from-1 to1) that are inside the circle must satisfy x² + y² ≤9. Since x is from-4 to-2, which is left of the circle's center.Wait, but x from-3 to-2 is within the circle? Let me check x=-3, y=0: x² + y²=9, which is on the circle. x=-2, y=0: x² + y²=4 <9, so inside. So, the overlapping region is x from-3 to-2, y from-1 to1, and also x from-3 to-2, y must satisfy x² + y² ≤9. Wait, but for x from-3 to-2, x² is from9 to4. So x² + y² ≤9 => y² ≤9 -x². Therefore, y ranges from -sqrt(9 -x²) to sqrt(9 -x²). But since the square restricts y to -1 to1, the overlapping y is the intersection of these intervals. So for x from-3 to-2, the y-interval is from-1 to1, because sqrt(9 -x²) ≥ sqrt(9 -4)=sqrt(5)≈2.236>1. Therefore, the overlapping region is x from-3 to-2, y from-1 to1, which is a rectangle. But wait, x from-3 to-2 is part of the translated square (original square after translation is x from-4 to-2). But in the translated system, the overlapping region is x from-3 to-2, y from-1 to1. So this is a rectangle of width1 (from x=-3 to-2) and height2 (from y=-1 to1). Therefore, area=2*1=2.But wait, this can’t be right because when we transformed back to the original coordinate system, this area is the overlapping region. Wait, but in the translated system, the overlapping area is a rectangle of area2. But when we shift back, this rectangle corresponds to the original square from x=0 to1 (since we shifted left by3, x=-3 becomes x=0). Wait, this is getting confusing.Alternatively, maybe after translation, the overlapping region is x from-3 to-2, y from-1 to1. Translating back, x becomes x +3. So in original coordinates, x from0 to1, y from-1 to1. Which matches our previous analysis. So, the overlapping area in translated coordinates is2 (the rectangle), but in original coordinates, is this correct?But wait, in translated coordinates, the circle is centered at origin, and the square is shifted left by3. So overlapping region is the rectangle x from-3 to-2, y from-1 to1. But in the translated system, this rectangle is inside both the circle and the square. But in the translated system, the circle has radius3, so x from-3 to-2 is on the edge, but does the rectangle lie entirely within the circle?At x=-3, y can be from-1 to1: x² + y²=9 + y², which ranges from9 to10, which is outside the circle (radius3, so radius squared9). Therefore, only part of the rectangle is inside the circle.Wait, this contradicts our earlier conclusion. There must be a mistake here.Wait, in the translated system, the square is from x=-4 to-2, y=-1 to1. The circle is x² + y² ≤9. So overlapping region is the intersection of x between-4 to-2, y between-1 to1, and x² + y² ≤9.But for x between-4 to-3, x² is between16 and9, so x² + y² is already greater than9, so no overlap. For x between-3 to-2, x² is between9 and4. Therefore, for x in[-3, -2], y must satisfy y² ≤9 -x². Since x is negative here, x=-3 to-2, 9 -x² is 9 - (between4 and9) = between0 and5. Therefore, y² ≤9 -x². For example, at x=-3, y² ≤0 => y=0. At x=-2, y² ≤9 -4=5 => y between-√5 and√5. But the square restricts y to-1 to1. Therefore, for x in[-3, -2], the overlapping region in y is from -sqrt(9 -x²) to sqrt(9 -x²), but limited by the square's y from-1 to1. Therefore, for each x in[-3, -2], the overlapping y is from max(-sqrt(9 -x²), -1) to min(sqrt(9 -x²),1).But since sqrt(9 -x²) at x=-3 is0, and at x=-2 is sqrt(5)≈2.236. Therefore, for x in[-3, c], where c is some x where sqrt(9 -x²)=1, the overlapping y is from -sqrt(9 -x²) to sqrt(9 -x²). For x in[c, -2], the overlapping y is from-1 to1.Wait, solving sqrt(9 -x²)=1:sqrt(9 -x²)=1 =>9 -x²=1 =>x²=8 =>x=±2√2≈±2.828. But in our case, x is in[-3, -2]. So x=-2√2≈-2.828 is in[-3, -2]. So c=-2√2≈-2.828.Therefore, the overlapping region is:1. For x from-3 to-2√2 (≈-2.828), y from- sqrt(9 -x²) to sqrt(9 -x²).2. For x from-2√2 to-2, y from-1 to1.Therefore, overlapping area A=∫ from-3 to-2√2 2*sqrt(9 -x²) dx + ∫ from-2√2 to-2 2*1 dxBut this is similar to the integral we computed earlier but in the translated coordinate system. Translating back to original coordinates, this would correspond to the same overlapping area.But this seems redundant. Let me compute this integral.First, integral from-3 to-2√2 of 2*sqrt(9 -x²) dx. Let’s use substitution. Let x=3 sinθ, then dx=3 cosθ dθ.But when x=-3, θ= -π/2. When x=-2√2≈-2.828, x=3 sinθ => sinθ= x/3≈-2.828/3≈-0.9428=>θ≈-1.230 radians.So the integral becomes:2 * ∫ fromθ=-π/2 toθ≈-1.230 sqrt(9 -9 sin²θ)*3 cosθ dθ=2 * ∫ from-π/2 to-1.230 3 cosθ *3 cosθ dθ=2 *9 ∫ from-π/2 to-1.230 cos²θ dθ=18 ∫ cos²θ dθUsing the identity cos²θ=(1 +cos2θ)/2:=18*(1/2) ∫ (1 +cos2θ) dθ=9[θ + (sin2θ)/2] from-π/2 to-1.230At θ=-1.230:9[ -1.230 + (sin(-2.46))/2 ]sin(-2.46)≈sin(-2.46 + π)≈sin(0.6816)≈0.632. But actually, since sin(-2.46)= -sin(2.46). Let me compute 2.46 radians:2.46 radians is approximately 141 degrees. sin(141 degrees)=sin(180-39)=sin(39)=≈0.6293. So sin(-2.46)= -0.6293.Therefore:9[ -1.230 + (-0.6293)/2 ]≈9[ -1.230 -0.3146 ]≈9*(-1.5446)≈-13.9014At θ=-π/2:9[ -π/2 + (sin(-π))/2 ]=9[ -π/2 +0 ]≈-9*(1.5708)≈-14.1372Therefore, the integral≈-13.9014 -(-14.1372)=0.2358But this can't be right. Wait, this is the evaluation of the integral from-π/2 to-1.230. So it's [F(-1.230) - F(-π/2)]≈0.2358. So the first integral is≈0.2358.Second integral is ∫ from-2√2 to-2 2*1 dx=2*( -2 -(-2√2))=2*( -2 +2√2)= -4 +4√2≈-4 +5.656≈1.656Therefore, total overlapping area A≈0.2358 +1.656≈1.8918But this is in the translated system. But overlapping area should be same regardless of coordinate system. Wait, but we translated the system, so overlapping area remains the same. Therefore, in original coordinates, the overlapping area is≈1.8918.But in original coordinates, the square has area4, the circle has area9π≈28.274. So non-overlapping areas would be4 -1.8918≈2.1082 for the square and9π -1.8918≈28.274 -1.8918≈26.382 for the circle. The positive difference would be≈26.382 -2.1082≈24.2738. But this answer seems approximate. The problem probably expects an exact answer.Wait, the problem says "positive difference between the areas of the nonoverlapping portions of the figures." So the exact value is |(4 - A) - (9π - A)|= |4 -9π|. But this contradicts our previous reasoning. Wait, but if you think about it:The non-overlapping portion of the square is4 -A.The non-overlapping portion of the circle is9π -A.Then the difference between them is |(4 -A) - (9π -A)|= |4 -9π|.But that can't be, because the difference would be independent of A. This suggests that regardless of the overlapping area, the difference is |4 -9π|≈ |4 -28.274|≈24.274. But this contradicts our previous calculation where the difference was approximately24.2738, which is close to24.274. So maybe the difference is indeed |4 -9π|, which is exactly9π -4, since9π >4.But that seems odd. Why would the problem mention the overlapping if it cancels out? Maybe my initial assumption is wrong. Let me think again.Wait, if the square is entirely inside the circle, then the non-overlapping parts would be the entire square (non-overlapping with the circle) and the circle minus the square (non-overlapping with the square). But in that case, the difference would be |4 - (9π -4)|=|8 -9π|. But in our case, the square is partially overlapping. However, if the overlapping area cancels out when taking the difference, which is |(4 -A) - (9π -A)|= |4 -9π|, then the answer is simply9π -4. But that seems too straightforward, and the problem mentions that the circle crosses the center of the square, implying that the overlapping area is non-trivial.But according to the earlier approximate calculation, the difference is approximately24.274, which is exactly9π -4≈28.274 -4=24.274. Therefore, this suggests that the difference is indeed9π -4, regardless of the overlapping area. But how?Wait, the non-overlapping portion of the square is the part of the square not inside the circle, which is4 -A. The non-overlapping portion of the circle is the part of the circle not inside the square, which is9π -A. Then, the difference between these two areas is (9π -A) - (4 -A)=9π -4. So regardless of A, the overlapping area cancels out. Therefore, the positive difference is9π -4.But this seems to be a trick question. The key insight is that when you take the difference between the non-overlapping areas, the overlapping parts subtract out, leaving just the difference between the original areas. But that would be true only if the overlapping area is counted once in both figures. Wait, no. Wait, the non-overlapping area of the square is4 -A, and non-overlapping area of the circle is9π -A. The difference between these is (9π -A) - (4 -A)=9π -4. So yes, the difference is simply the difference between the original areas.But this feels counterintuitive. For example, imagine if two figures overlap partially; the non-overlapping parts would be each original area minus the overlap. The difference between these non-overlapping areas would be the difference between the original areas. So if Figure A has areaA, Figure B has areaB, overlapping areaC. Then non-overlapping areas areA -C andB -C. The difference is |(A -C) - (B -C)|=|A -B|. So yes, the difference is just the difference between the original areas, regardless of the overlapping area.Therefore, in this problem, even though the square and circle partially overlap, the positive difference between the non-overlapping portions is simply |9π -4|, which is9π -4≈28.274 -4=24.274.But this contradicts the earlier detailed calculation where the overlapping area affected the result. However, according to the mathematical reasoning, the difference should indeed be9π -4. Therefore, perhaps the problem is designed to show that the difference between the non-overlapping areas is the same as the difference between the original areas.But why does the problem mention that the circle crosses the center of the square? If the entire square were inside the circle, then the non-overlapping part of the square would be0, and the non-overlapping part of the circle would be9π -4, so difference9π -4. If the square is partially overlapping, then non-overlapping parts are4 -A and9π -A, difference9π -4. If the square is entirely outside, then non-overlapping areas are4 and9π, difference9π -4. So regardless of the overlap, the difference is the same. Therefore, the answer is9π -4.But why would the problem specify the position of the circle crossing the center of the square? Because if the square were entirely inside the circle, then the non-overlapping portion of the square is0, and the difference would be9π -4 -0=9π -4. If the square is partially overlapping, the difference is still9π -4. So regardless of how they overlap, the difference remains the same. Therefore, maybe the answer is simply9π -4.But I need to confirm this with the earlier approximate calculation. If the overlapping area A≈1.8918, then non-overlapping square area≈4 -1.8918≈2.1082, non-overlapping circle area≈9π -1.8918≈26.382. Difference≈26.382 -2.1082≈24.274, which is exactly9π -4≈28.274 -4=24.274. Therefore, even with the overlapping area considered, the difference is indeed9π -4. Therefore, the overlapping area cancels out when taking the difference between the non-overlapping portions.Therefore, the positive difference is9π -4. So the answer isboxed{9pi -4}.But wait, the problem states "positive difference", so we need to ensure it's positive. Since9π is about28.274, so28.274 -4=24.274, which is positive. So the positive difference is9π -4.But this seems too straightforward for a problem that mentions the circle crossing the center of the square. Maybe I missed something?Wait, if the square is entirely inside the circle, then the non-overlapping portion of the square is0, and the non-overlapping portion of the circle is9π -4. So the difference is9π -4 -0=9π -4. If the square is partially overlapping, then non-overlapping portions are4 -A and9π -A, difference(9π -A) - (4 -A)=9π -4. So regardless of A, the difference is9π -4. Therefore, the answer is9π -4.But the problem says "the positive difference between the areas of the nonoverlapping portions of the figures". So it's the absolute value of (non-overlapping circle area - non-overlapping square area). Since9π -4 is positive, the answer is9π -4.But why does the problem mention the specific configuration where the circle crosses the center of the square? Because if the square were entirely outside the circle, then the non-overlapping areas would be4 and9π, difference|9π -4|. If the square is entirely inside the circle, non-overlapping areas are0 and9π -4, difference9π -4. If it's partially overlapping, same difference. Therefore, regardless of the position, as long as there is some overlap (i.e., the figures are not completely separate), the difference is the same. But wait, if the square and circle don't overlap at all, then the difference is|9π -4|. But if they do overlap, difference is9π -4. But since the problem states that the circle crosses the center of the square, there must be some overlap. Therefore, the answer is9π -4.But to confirm, let's imagine another scenario: suppose the circle and square don't overlap at all. Then non-overlapping areas are9π and4, difference|9π -4|. But in this problem, since the circle crosses the center of the square, they must overlap, so the difference is9π -4. So depending on the overlap, the difference is either9π -4 or|9π -4|. But since the problem says "positive difference", it's the absolute value, so it should be9π -4 because9π >4.Therefore, the answer is9π -4.But this seems to ignore the specific geometric configuration described. However, mathematically, regardless of the overlapping area, the difference remains the same. Therefore, the answer isboxed{9pi -4}.Final AnswerThe positive difference between the areas of the nonoverlapping portions is boxed{9pi - 4}.</think>To solve the problem of finding the positive difference between the areas of the nonoverlapping portions of a circle of radius 3 and a square of side length 2, where the circle crosses the center of the square, we can follow these steps:1. Calculate the areas of the circle and the square: - The area of the circle is (9pi) (since the radius is 3, the area is (pi times 3^2 = 9pi)). - The area of the square is (4) (since the side length is 2, the area is (2^2 = 4)).2. Determine the overlapping area: - The circle crosses the center of the square, meaning the center of the square is on the circumference of the circle. The distance from the center of the circle to the center of the square is 3 (the radius of the circle). - The square is partially inside and partially outside the circle. However, the key insight is that the overlapping area (A) cancels out when calculating the difference between the nonoverlapping portions.3. Calculate the nonoverlapping portions: - The nonoverlapping portion of the square is (4 - A). - The nonoverlapping portion of the circle is (9pi - A).4. Find the positive difference between these nonoverlapping areas: - The difference is (|(9pi - A) - (4 - A)| = |9pi - 4|).Since (9pi) is greater than 4, the positive difference is simply (9pi - 4).Thus, the positive difference between the areas of the nonoverlapping portions of the figures is (boxed{9pi - 4}).]

🔑:Solution: Call the area of the square s, the area of the circle c, and the area of the overlapping portion x. The area of the circle not overlapped by the square is c-x and the are of the square not overlapped by the circle is s-x, so the difference between these two is (c-x)-(s-x)=c-s=9 pi^{2}-4.

🧠:[<think>Okay, let me try to work through this problem step by step. So, we have an odd function f defined on all real numbers. That means f(-x) = -f(x) for every x. Got that. Then, the derivative of f is f', and for x not equal to zero, the inequality f'(x) + f(x)/x > 0 holds. We need to compare the values of a, b, c defined as:a = (1/2)f(1/2)b = -2f(-2)c = (ln 1/2)f(ln 1/2)And figure out the order between them, choosing the correct option from A to D.First, let me note that since f is odd, we can use that property to simplify some of these expressions. For example, in the definition of b: -2f(-2) can be rewritten using the oddness: f(-2) = -f(2), so substituting that in, b becomes -2*(-f(2)) = 2f(2). So b is 2f(2). That might be easier to work with.Similarly, for c, ln(1/2) is equal to -ln 2, right? Because ln(1/2) = ln(2^{-1}) = -ln 2. So ln(1/2) is negative. Therefore, c is (-ln2)*f(-ln2). Again, using the oddness of f, f(-ln2) = -f(ln2). So substituting that in, c becomes (-ln2)*(-f(ln2)) = (ln2)f(ln2). So c simplifies to (ln2)f(ln2). Wait, but ln(1/2) is negative, so the original expression is (negative number)*f(negative number). Then, using f is odd, the negative inside f becomes a negative outside, so two negatives multiply to a positive. So c is (ln2)f(ln2). So c is (ln2)f(ln2). That's helpful.Similarly, a is (1/2)f(1/2). So all three expressions a, b, c are of the form k*f(k), where k is a positive number. Because:a: k = 1/2b: k = 2 (since b = 2f(2))c: k = ln2 (since c = (ln2)f(ln2))So, ln2 is approximately 0.693, so ln2 is between 0 and 1, specifically about 0.693. So the k's are 1/2 (0.5), ln2 (~0.693), and 2. So ordering them from smallest to largest k: 1/2 < ln2 < 2.But we need to compare a, b, c which are k*f(k) for these k's. So the problem reduces to comparing 0.5*f(0.5), 0.693*f(0.693), and 2*f(2). Then, depending on how f(k) behaves, we can order them.But the key here is the differential inequality given: for x ≠ 0, f'(x) + f(x)/x > 0. Let me write that down:f'(x) + (f(x)/x) > 0 for x ≠ 0.Hmm. Let's see. Maybe we can rewrite this inequality in a more manageable form. Let's multiply both sides by x (but we need to be careful about the sign of x because multiplying by a negative number reverses the inequality). Wait, but x can be positive or negative. Let me think. Alternatively, maybe consider the expression x*f'(x) + f(x) > 0, since multiplying both sides by x. Wait, but the original inequality is f'(x) + f(x)/x > 0. So if I multiply both sides by x, we get x*f'(x) + f(x) > 0 when x > 0, and x*f'(x) + f(x) < 0 when x < 0 because multiplying by a negative number flips the inequality.But since the original inequality is given for x ≠ 0, perhaps it's better to consider x > 0 and x < 0 separately. Since the function is odd, maybe we can focus on x > 0 and use symmetry for x < 0.But let's see. Let me consider x > 0 first. For x > 0, the inequality becomes x*f'(x) + f(x) > 0. Let's see if we can recognize this as the derivative of some product. Let's check:d/dx [x*f(x)] = f(x) + x*f'(x). Exactly! So the left-hand side of the inequality (x*f'(x) + f(x)) is the derivative of x*f(x). Therefore, the inequality tells us that for x > 0, the derivative of x*f(x) is positive, meaning that x*f(x) is increasing on (0, ∞). Similarly, for x < 0, the inequality is f'(x) + f(x)/x > 0, but x is negative. So let's see:If x < 0, multiplying both sides by x (negative) reverses the inequality:x*f'(x) + f(x) < 0.But x < 0. Let's let t = -x > 0. Then, f'(x) = f'(-t) = derivative at -t. Since f is odd, f(-t) = -f(t). Therefore, f'(-t) = derivative of -f(t) at t, which is -f'(t). So f'(-t) = -f'(t). Therefore, for x = -t < 0,x*f'(x) + f(x) = (-t)*f'(-t) + f(-t) = (-t)*(-f'(t)) + (-f(t)) = t*f'(t) - f(t).So the inequality x*f'(x) + f(x) < 0 for x < 0 becomes t*f'(t) - f(t) < 0, where t > 0.But from the x > 0 case, we know that the derivative of x*f(x) is t*f'(t) + f(t) = d/dt [t*f(t)] > 0. So for t > 0, t*f'(t) + f(t) > 0. But here, for x < 0, we have t*f'(t) - f(t) < 0. Let me note that.So, combining both:For x > 0, d/dx [x*f(x)] > 0, so x*f(x) is increasing on (0, ∞).For x < 0, if we let t = -x > 0, then the expression becomes t*f'(t) - f(t) < 0, which is equivalent to d/dx [x*f(x)] (but for x < 0) related to this?Wait, maybe another approach. Let's check the derivative of x*f(x) for x > 0: increasing. So x*f(x) is increasing on (0, ∞). So as x increases, x*f(x) increases. Therefore, if we have two positive numbers, say, 0 < a < b, then a*f(a) < b*f(b). So since 1/2 < ln2 < 2 (since ln2 ≈ 0.693), then (1/2)*f(1/2) < (ln2)*f(ln2) < 2*f(2). Therefore, a < c < b. Which would correspond to option C.But wait, hold on. Let's make sure. The key here is that x*f(x) is increasing for x > 0. Therefore, since 1/2 < ln2 < 2, then (1/2)f(1/2) < (ln2)f(ln2) < 2f(2). Hence, a < c < b. Therefore, the correct order is a < c < b, which is option C. So answer C.But let me verify this carefully.First, because f is odd, we can focus on x > 0, since for x < 0, the behavior is determined by the oddness. But the critical part is the differential inequality. For x > 0, we established that d/dx [x*f(x)] = x*f'(x) + f(x) > 0. Therefore, the function x*f(x) is strictly increasing on (0, ∞). Therefore, as x increases, x*f(x) increases. Therefore, for any two positive numbers, say, x1 < x2, then x1*f(x1) < x2*f(x2).Given that 1/2 ≈ 0.5, ln2 ≈ 0.693, and 2. So 0.5 < 0.693 < 2. Therefore:(1/2)*f(1/2) < (ln2)*f(ln2) < 2*f(2)Therefore, a < c < b. So the order is a < c < b, which is option C. Therefore, the correct answer is C.Wait, but let me check with x < 0. The problem mentions that the function is defined on all real numbers, but the inequality holds for x ≠ 0. We considered x > 0. For x < 0, as we saw, if we set x = -t where t > 0, the inequality becomes t*f'(t) - f(t) < 0. But how does that affect the behavior?Alternatively, maybe we can consider the function x*f(x) for x < 0. Let me see. Let x < 0, then x*f(x) = x*(-f(-x)) = -x*f(-x). Since x is negative, -x is positive, so x*f(x) = (-x)*f(-x). But since x < 0, let me set t = -x > 0. Then x*f(x) = t*f(-t) = t*(-f(t)) = -t*f(t). So x*f(x) is equal to -t*f(t) where t > 0.So for x < 0, x*f(x) is the negative of t*f(t). Since t*f(t) is increasing in t (as established earlier), then x*f(x) is decreasing in x? Wait, maybe. Let me think. If t increases, then t*f(t) increases, so -t*f(t) decreases. Therefore, as x becomes more negative (i.e., as t = -x increases), x*f(x) becomes more negative. Therefore, x*f(x) is decreasing for x < 0.But how does this affect our problem? Our variables a, b, c all correspond to positive x values because:a = (1/2)f(1/2): x = 1/2 > 0b = 2f(2): x = 2 > 0c = (ln2)f(ln2): x = ln2 > 0So even though f is odd, we don't need to consider negative x values here because a, b, c are all expressed in terms of positive x multiplied by f at positive x. Therefore, the behavior of x*f(x) on the positive real line is sufficient to determine the order.Since x*f(x) is increasing on (0, ∞), then the larger x is, the larger x*f(x) is. So since 1/2 < ln2 < 2, then (1/2)f(1/2) < (ln2)f(ln2) < 2f(2), which translates to a < c < b. Therefore, the answer is option C.But just to ensure there's no trick here, let me check if there's any other consideration. For example, could f(x) be negative for positive x? Since f is odd, if x is positive, f(x) could be positive or negative depending on the function. However, given the differential inequality, perhaps we can deduce the sign of f(x).Let's consider x > 0. The derivative of x*f(x) is positive, so x*f(x) is increasing. Let's suppose x approaches 0 from the right. Then x*f(x) approaches 0*f(0). But since f is odd, f(0) = 0. So as x approaches 0+, x*f(x) approaches 0. Since the function is increasing, then for x > 0, x*f(x) must be greater than 0. Because if at x approaching 0, it's approaching 0, and it's increasing, so for all x > 0, x*f(x) > 0. Therefore, f(x) > 0 for x > 0. Because x is positive, and x*f(x) is positive, so f(x) must be positive. So f(x) is positive for x > 0, and negative for x < 0, which is consistent with being an odd function.Therefore, since f(x) > 0 for x > 0, then a, b, c are all positive because they are positive multiples of positive numbers. So when we say a < c < b, it's in terms of their magnitudes.Therefore, the conclusion remains that a < c < b, which is option C.Final Answerboxed{C}</think>Given an odd function ( y = f(x) ) defined on (mathbb{R}) with the derivative ( y = f'(x) ), we know that for ( x neq 0 ), the inequality ( f'(x) + frac{f(x)}{x} > 0 ) holds. We need to compare the values ( a = frac{1}{2}fleft( frac{1}{2} right) ), ( b = -2f(-2) ), and ( c = left( ln frac{1}{2} right)fleft( ln frac{1}{2} right) ).1. Using the odd function property: - ( b = -2f(-2) = 2f(2) ) - ( ln frac{1}{2} = -ln 2 ), so ( c = (-ln 2)f(-ln 2) = (ln 2)f(ln 2) )2. Simplifying the expressions: - ( a = frac{1}{2}fleft( frac{1}{2} right) ) - ( b = 2f(2) ) - ( c = (ln 2)f(ln 2) )3. Analyzing the differential inequality: - For ( x > 0 ), the inequality ( f'(x) + frac{f(x)}{x} > 0 ) can be rewritten as ( frac{d}{dx}(x f(x)) > 0 ), indicating that ( x f(x) ) is increasing on ( (0, infty) ).4. Ordering the values: - Since ( x f(x) ) is increasing for ( x > 0 ), we compare the values of ( x ) in ( a ), ( b ), and ( c ): - ( frac{1}{2} approx 0.5 ) - ( ln 2 approx 0.693 ) - ( 2 ) - Therefore, ( frac{1}{2} < ln 2 < 2 ), leading to ( frac{1}{2}fleft( frac{1}{2} right) < (ln 2)f(ln 2) < 2f(2) ).Thus, the correct order is ( a < c < b ), corresponding to option (boxed{C}).,<think>Okay, let's try to tackle this problem step by step. So, we have an odd function f defined on all real numbers, and its derivative f' satisfies the inequality f'(x) + f(x)/x > 0 for all x ≠ 0. Then we have three expressions a, b, c involving f evaluated at certain points, and we need to compare their values.First, let's recall what an odd function is. An odd function satisfies f(-x) = -f(x) for all x. That might be useful later when dealing with the points like -2 in expression b.The key inequality given is f'(x) + f(x)/x > 0 for x ≠ 0. Let me think about how to handle this. Maybe we can manipulate this inequality into something more familiar. If I multiply both sides by x, but wait, x could be positive or negative. Hmm. Maybe instead, consider the expression f'(x) + f(x)/x. Let's see if this can be written as the derivative of some product or quotient.Wait, the left-hand side looks like the derivative of f(x) multiplied by some integrating factor. Let's check. Suppose we have an expression like f'(x) + (1/x)f(x). That reminds me of the derivative of f(x) times some function. Let me recall: if we have a linear differential equation y' + P(x)y = Q(x), the integrating factor is e^{∫P(x)dx}. But here, maybe we can write f'(x) + (1/x)f(x) as the derivative of (something). Let's compute d/dx [f(x)/x]. Using the quotient rule: derivative of f(x)/x is [f'(x)x - f(x)] / x². Hmm, that's (f'(x)x - f(x)) / x². But our inequality is f'(x) + f(x)/x > 0, which can be rewritten as (f'(x)x + f(x)) / x > 0. So, (d/dx [f(x)/x] * x²) = f'(x)x - f(x). But in our inequality, it's f'(x)x + f(x) > 0. Wait, that's different. Hmm.Alternatively, maybe multiply both sides by x to get x f'(x) + f(x) > 0 for x ≠ 0. So, x f'(x) + f(x) > 0. Let's think about this expression. If we have x f'(x) + f(x), that's actually the derivative of x f(x). Let's check: d/dx [x f(x)] = f(x) + x f'(x). Yes! So, the inequality is equivalent to d/dx [x f(x)] > 0 for x ≠ 0. That's a crucial observation. So, the derivative of x f(x) is positive everywhere except at x = 0. Therefore, the function x f(x) is strictly increasing on the entire real line except at x = 0. Since x f(x) is defined for all real numbers except x = 0, but the original function f is defined on all real numbers, so x f(x) is defined for all x except x=0? Wait, no. If f is defined on all real numbers, then x f(x) is also defined on all real numbers. The derivative of x f(x) exists except at x=0? Wait, the original problem states that the derivative f' exists, so x f(x) is differentiable everywhere, and its derivative is f(x) + x f'(x), which is given to be positive for all x ≠ 0. So, the function x f(x) is strictly increasing on (-∞, 0) and (0, ∞). But since x=0 is a point where x f(x) is zero, we need to check the behavior around x=0.Wait, but since the function is odd, f(-x) = -f(x), so x f(x) is an even function? Let me check: If we replace x with -x, then (-x) f(-x) = (-x)(-f(x)) = x f(x). So yes, x f(x) is an even function. Therefore, it's symmetric about the y-axis. And since it's strictly increasing on (0, ∞), it must also be strictly increasing on (-∞, 0) as an even function. Wait, no. If a function is even and differentiable, its derivative is odd. So, the derivative of x f(x) is even function's derivative, which is odd. Wait, but we already found that the derivative of x f(x) is positive for all x ≠ 0. Since derivative is positive for x > 0 and since the function is even, the derivative for x < 0 would be negative? Wait, that contradicts. Wait, no. Wait, the derivative of x f(x) is f(x) + x f'(x), which is given to be greater than 0 for all x ≠ 0. So, regardless of x being positive or negative, the derivative is positive. But for an even function, the derivative should be an odd function. Wait, this seems contradictory. Let me think again.Suppose g(x) = x f(x). Since f is odd, then g(-x) = (-x) f(-x) = (-x)(-f(x)) = x f(x) = g(x). So, g(x) is even. Therefore, its derivative g'(x) should be an odd function. But according to the given condition, g'(x) = f(x) + x f'(x) > 0 for all x ≠ 0. But an odd function that's positive for all x ≠ 0 would require that g'(x) is positive for x > 0 and negative for x < 0. But here, it's given that g'(x) is positive for all x ≠ 0. Therefore, unless g'(x) is both positive and odd, which would require that for x < 0, g'(x) = -g'(-x). But if g'(x) is positive for x > 0, then for x < 0, g'(x) = -g'(-x) would be negative. But this contradicts the given condition that g'(x) > 0 for all x ≠ 0. Therefore, there must be a mistake in my reasoning.Wait, let's check again. The function g(x) = x f(x) is even, so its derivative g'(x) is odd. The problem states that g'(x) > 0 for all x ≠ 0. However, since g'(x) is odd, if g'(x) > 0 for x > 0, then g'(x) = -g'(-x) would imply that g'(x) < 0 for x < 0. But the problem says that for all x ≠ 0, g'(x) > 0. That is a contradiction unless my previous steps are wrong.Wait, the problem states that f is odd, defined on ℝ, and its derivative is f'. The given inequality is f'(x) + f(x)/x > 0 for x ≠ 0. So, in terms of g(x) = x f(x), we have g'(x) > 0 for x ≠ 0. However, g(x) is even, so its derivative is odd. Therefore, if g'(x) > 0 for x > 0, then for x < 0, g'(x) = -g'(-x) < 0. But the problem states that g'(x) > 0 for all x ≠ 0. Therefore, this is only possible if g'(x) is both positive and negative for x ≠ 0, which is impossible unless g'(x) = 0, which contradicts the inequality. Therefore, there must be an error in my reasoning.Wait, maybe I made a mistake in the calculation. Let's re-express the derivative:Given that f is odd, so f(-x) = -f(x). Then, g(x) = x f(x). Then, compute g(-x) = (-x) f(-x) = (-x)(-f(x)) = x f(x) = g(x), so g is even. Therefore, g is even, so its derivative g' is odd. Then, if g'(x) > 0 for all x ≠ 0, but since g' is odd, for x > 0, g'(x) > 0, and for x < 0, g'(x) = -g'(-x) < 0. This contradicts the given condition. Therefore, perhaps there is a mistake in the problem statement, or maybe my interpretation is wrong.Wait, hold on, the problem states that the derivative of f exists and the inequality holds for x ≠ 0. But perhaps the problem is not saying that the inequality holds for all x ≠ 0, but rather, when x ≠ 0. Wait, the problem says: "When x ≠ 0, f'(x) + f(x)/x > 0". So, perhaps the problem is saying that for x ≠ 0, the inequality holds. But as we saw, if g'(x) = f(x) + x f'(x) > 0 for x ≠ 0, and since g is even, then g'(x) is odd, which would require that if g'(x) > 0 for x > 0, then g'(x) < 0 for x < 0. Therefore, this would imply that the inequality f'(x) + f(x)/x > 0 holds only for x > 0, and for x < 0, since g'(x) = f(x) + x f'(x) < 0. But the problem states that for x ≠ 0, f'(x) + f(x)/x > 0. This is impossible unless f'(x) + f(x)/x is both positive and negative depending on the sign of x. Wait, but the problem states that for all x ≠ 0, f'(x) + f(x)/x > 0. That seems contradictory.Wait, perhaps the problem is written incorrectly? Or maybe there's a miscalculation here. Let me check again:If f is odd, then f(-x) = -f(x). Let's compute f'(-x). Since f is differentiable, the derivative of f(-x) is -f'(-x). But also, since f(-x) = -f(x), differentiating both sides gives -f'(-x) = -f'(x), so f'(-x) = f'(x). Therefore, f' is an even function. So, f' is even.Given that, let's look at the inequality f'(x) + f(x)/x > 0 for x ≠ 0. Let's plug in -x instead of x:For x ≠ 0, f'(-x) + f(-x)/(-x) > 0.But f'(-x) = f'(x) (since f' is even) and f(-x)/(-x) = (-f(x))/(-x) = f(x)/x. Therefore, the inequality becomes f'(x) + f(x)/x > 0, which is the same as the original inequality. So, the inequality holds for both x and -x. Therefore, the original inequality f'(x) + f(x)/x > 0 holds for all x ≠ 0, regardless of the sign of x.But earlier, we had that the derivative of g(x) = x f(x) is g'(x) = f(x) + x f'(x) = x [f'(x) + f(x)/x] > 0 for all x ≠ 0. Wait, but x can be positive or negative. So, if x is positive, then x [f'(x) + f(x)/x] > 0 implies that f'(x) + f(x)/x > 0. If x is negative, then x [f'(x) + f(x)/x] > 0 would imply that f'(x) + f(x)/x < 0, because multiplying by a negative number reverses the inequality. But the problem states that for all x ≠ 0, f'(x) + f(x)/x > 0. Therefore, this would imply that for x negative, x [f'(x) + f(x)/x] > 0 is equivalent to f'(x) + f(x)/x < 0. But since the problem states that f'(x) + f(x)/x > 0 for all x ≠ 0, then for x negative, we have x [f'(x) + f(x)/x] > 0 ⇒ negative times positive is negative, which contradicts the given that g'(x) > 0 for all x ≠ 0.Wait, this is getting confusing. Let's clarify:If g(x) = x f(x), then g'(x) = f(x) + x f'(x). The problem states that f'(x) + f(x)/x > 0 for all x ≠ 0. Therefore, multiplying both sides by x (which can be positive or negative), we get:If x > 0: multiplying by x doesn't change the inequality, so x [f'(x) + f(x)/x] = x f'(x) + f(x) = g'(x) > 0.If x < 0: multiplying by x (negative) reverses the inequality, so x [f'(x) + f(x)/x] < 0. But x [f'(x) + f(x)/x] = g'(x). Therefore, for x < 0, g'(x) < 0.But the original problem states that f'(x) + f(x)/x > 0 for all x ≠ 0, which leads to g'(x) > 0 when x > 0 and g'(x) < 0 when x < 0. However, we also know that g(x) is even, so g'(x) is odd, which aligns with this: for x > 0, g'(x) > 0; for x < 0, g'(x) < 0.But wait, the problem states that f'(x) + f(x)/x > 0 for all x ≠ 0. So when x is negative, even though the expression f'(x) + f(x)/x is positive, multiplying by x (negative) gives a negative result, so g'(x) < 0 for x < 0.But the user's problem statement is correct as given. So, putting this together, the function g(x) = x f(x) has derivative g'(x) which is positive when x > 0 and negative when x < 0. Therefore, the function g(x) is increasing on (0, ∞) and decreasing on (-∞, 0). Since it's even, symmetric about the y-axis, so it has a minimum at x = 0? Wait, but at x = 0, g(x) = 0 * f(0) = 0. Since f is odd, f(0) = 0.Wait, let me analyze the behavior of g(x). For x > 0, since g'(x) > 0, g(x) is increasing on (0, ∞). For x < 0, since g'(x) < 0, g(x) is decreasing on (-∞, 0). Therefore, the function g(x) has a minimum at x = 0 (since it's increasing to the right of 0 and decreasing to the left of 0). But since g(0) = 0, and since g is even, this suggests that g(x) ≥ 0 for all x. Wait, is that necessarily true?Suppose x > 0, then since g(x) is increasing on (0, ∞) and g(0) = 0, then for x > 0, g(x) > 0. Similarly, for x < 0, since g(x) is decreasing on (-∞, 0) and g(0) = 0, then for x < 0, g(x) > 0 as well? Wait, but how?Wait, if g(x) is decreasing on (-∞, 0), then as x approaches 0 from the left, g(x) approaches g(0) = 0 from above. Therefore, for x < 0, g(x) > 0. Therefore, overall, g(x) = x f(x) is positive for all x ≠ 0, and zero at x = 0. Therefore, f(x) = g(x)/x. For x > 0, f(x) = g(x)/x, which is positive because both g(x) and x are positive. For x < 0, f(x) = g(x)/x, which is negative because g(x) is positive and x is negative. Therefore, f(x) is positive on (0, ∞) and negative on (-∞, 0), which aligns with f being an odd function.So, to recap, g(x) = x f(x) is positive for x ≠ 0, even function, increasing on (0, ∞), decreasing on (-∞, 0). Therefore, for x1 < x2 in (0, ∞), g(x1) < g(x2). Similarly, for x1 < x2 in (-∞, 0), since g is decreasing, g(x1) > g(x2).But since we need to compare a, b, c, which are:a = (1/2) f(1/2) = (1/2) * [g(1/2) / (1/2)] = g(1/2).Similarly, b = -2 f(-2) = -2 * [g(-2)/(-2)] = -2 * [g(2)/(-2)] because g(-2) = g(2) (since g is even). Therefore, g(2)/(-2) = [g(2)/2] * (-1). Wait, let's compute step by step.First, for a:a = (1/2) f(1/2). Since f(1/2) = g(1/2)/(1/2) = 2 g(1/2). Therefore, a = (1/2) * 2 g(1/2) = g(1/2).Similarly, for b:b = -2 f(-2). Since f(-2) = -f(2) (because f is odd). Therefore, b = -2*(-f(2)) = 2 f(2). But f(2) = g(2)/2. So, b = 2*(g(2)/2) = g(2).For c:c = (ln(1/2)) f(ln(1/2)). Let's compute ln(1/2) = -ln 2 ≈ -0.693. So, c = (-ln 2) f(-ln 2). Since f is odd, f(-ln 2) = -f(ln 2). Therefore, c = (-ln 2)*(-f(ln 2)) = ln 2 * f(ln 2). And f(ln 2) = g(ln 2)/ln 2. So, c = ln 2 * (g(ln 2)/ln 2) = g(ln 2).Therefore, a = g(1/2), b = g(2), c = g(ln 2). Therefore, comparing a, b, c is equivalent to comparing g(1/2), g(ln 2), and g(2).But we know that g(x) is increasing on (0, ∞). Therefore, since 1/2 ≈ 0.5, ln 2 ≈ 0.693, and 2. So, 0.5 < 0.693 < 2. Therefore, since g is increasing on (0, ∞), g(0.5) < g(0.693) < g(2). Therefore, a < c < b.So, the correct order is a < c < b, which is option C.Wait, let me verify again. The key steps were realizing that a, b, c can be written in terms of g(x), which is x f(x). Then, since g is increasing on (0, ∞), the larger the x, the larger the g(x). The points in question are 1/2, ln 2 ≈ 0.693, and 2. So, 1/2 ≈ 0.5 < ln 2 ≈ 0.693 < 2, hence g(0.5) < g(0.693) < g(2), so a < c < b. Therefore, the answer is option C.I need to make sure there are no mistakes here. Let's check each step again.1. g(x) = x f(x) is even: Yes, because f is odd, so g(-x) = -x f(-x) = -x*(-f(x)) = x f(x) = g(x).2. The derivative g'(x) = f(x) + x f'(x), which is given to be positive for x > 0 and negative for x < 0. Therefore, g is increasing on (0, ∞) and decreasing on (-∞, 0).3. Therefore, for x > 0, the larger x is, the larger g(x) is. Since a = g(1/2), c = g(ln 2), and b = g(2), and 1/2 < ln 2 < 2, we have g(1/2) < g(ln 2) < g(2), so a < c < b, which is option C.Everything checks out.Final Answerboxed{C}</think>Given the problem, we need to determine the correct order of (a), (b), and (c) where (a = frac{1}{2}fleft(frac{1}{2}right)), (b = -2f(-2)), and (c = left(ln frac{1}{2}right)fleft(ln frac{1}{2}right)). The function (f) is odd and defined on (mathbb{R}), and its derivative satisfies (f'(x) + frac{f(x)}{x} > 0) for (x neq 0).1. Key Observations: - The function (g(x) = x f(x)) is even because (f) is odd. - The derivative (g'(x) = f(x) + x f'(x)) is given to be positive for all (x neq 0). - Since (g(x)) is even, (g'(x)) is odd. However, the inequality (f'(x) + frac{f(x)}{x} > 0) implies that (g'(x) > 0) for (x > 0) and (g'(x) < 0) for (x < 0). This means (g(x)) is increasing on ((0, infty)) and decreasing on ((-∞, 0)).2. Expressing (a), (b), and (c) in terms of (g(x)): - (a = frac{1}{2}fleft(frac{1}{2}right) = gleft(frac{1}{2}right)) - (b = -2f(-2) = 2f(2) = g(2)) - (c = left(ln frac{1}{2}right)fleft(ln frac{1}{2}right) = ln 2 cdot f(ln 2) = g(ln 2))3. Comparing (g(x)) Values: - Since (g(x)) is increasing on ((0, infty)), the values of (g) at different points depend on their magnitude. - The points to compare are (frac{1}{2} approx 0.5), (ln 2 approx 0.693), and (2). Since (0.5 < 0.693 < 2), it follows that (gleft(frac{1}{2}right) < g(ln 2) < g(2)).Thus, the correct order is (a < c < b), which corresponds to option (boxed{C}).]

🔑:AnalysisThis problem examines the ability to use the derivative to study the monotonicity of a function constructed for comparison, testing reasoning skills. Let g(x)=xf(x), then g'(x)=f(x)+xf'(x). Since when x neq 0, f'(x)+frac{f(x)}{x} > 0, it can be concluded that when x > 0, xf'(x)+f(x) > 0. That is, when x > 0, g'(x) > 0, therefore, when x > 0, the function g(x) is monotonically increasing.SolutionLet g(x)=xf(x),then g'(x)=f(x)+xf'(x).Since when x neq 0, f'(x)+frac{f(x)}{x} > 0,it follows that when x > 0, xf'(x)+f(x) > 0.That is, when x > 0, g'(x) > 0,therefore, when x > 0, the function g(x) is monotonically increasing.Since f(x) is an odd function,it follows that b=-2f(-2)=2f(2),and c=ln frac{1}{2}f(ln frac{1}{2})=ln 2f(ln 2),Since 2 > ln 2 > frac{1}{2},it follows that g(2) > g(ln 2) > g(frac{1}{2}),that is, b > c > a.Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, let's see. I have this problem here about the sign function and absolute values. The question gives me four options, A to D, and I need to figure out which one correctly represents |x| in terms of the sgn function. Hmm, let me start by recalling what the sgn function is. So, the sign function, sgn(x), is defined as 1 when x is positive, 0 when x is zero, and -1 when x is negative. That's the standard definition, right? So, sgn(x) basically tells us the sign of x. Now, the problem is asking which of the given equations correctly expresses the absolute value of x using this sgn function. Let me look at each option one by one.Option A says |x| = x |sgn(x)|. Let me parse that. The right-hand side is x multiplied by the absolute value of sgn(x). Since sgn(x) can be -1, 0, or 1, the absolute value of sgn(x) would be |sgn(x)|. Let's see:If x > 0, sgn(x) is 1, so |sgn(x)| is 1. Then x * 1 = x, which is equal to |x|. If x = 0, sgn(x) is 0, so |sgn(x)| is 0, then x * 0 = 0, which is |0|. If x < 0, sgn(x) is -1, so |sgn(x)| is 1, and x * 1 = x. Wait, but here x is negative, so x * 1 would be negative, which can't be |x|. For example, if x = -2, then |x| = 2, but x * |sgn(x)| would be -2 * 1 = -2, which is not equal to 2. So Option A doesn't hold when x is negative. Therefore, A is incorrect.Moving on to Option B: |x| = x sgn|x|. Let's analyze this. First, sgn|x|. Since |x| is always non-negative, sgn|x| is either 1 when |x| > 0 or 0 when |x| = 0. But |x| = 0 only when x = 0. So, if x ≠ 0, sgn|x| is 1, and if x = 0, it's 0. Therefore, the right-hand side becomes x * 1 when x ≠ 0, and 0 when x = 0. Let's check:If x > 0: x * 1 = x = |x|. If x = 0: 0 = |0|. If x < 0: x * 1 = x, which is negative, but |x| is positive. So again, for negative x, this would give x instead of |x|. So same problem as A. Hence, B is also incorrect.Option C: |x| = |x| sgn x. Let's see. Here, the right-hand side is |x| multiplied by sgn x. Let's break it down:If x > 0: sgn x = 1, so |x| * 1 = |x|. That works. If x = 0: sgn x = 0, so |x| * 0 = 0. Which is |0|. If x < 0: sgn x = -1, so |x| * (-1) = -|x|. But |x| is positive, so this would be -|x|, which is equal to x (since x is negative). However, |x| is supposed to be positive. For example, if x = -3, then |x| = 3, but |x| * sgn x = 3 * (-1) = -3, which is not equal to |x|. So Option C is incorrect because when x is negative, it gives -|x| instead of |x|.Now Option D: |x| = x sgn x. Let's check this. For each case:If x > 0: sgn x = 1, so x * 1 = x = |x|. That works. If x = 0: 0 * 0 = 0 = |0|. If x < 0: sgn x = -1, so x * (-1) = -x. But since x is negative, -x is positive, which is the absolute value of x. For example, x = -5: x * sgn x = (-5) * (-1) = 5 = | -5 |. So in all cases, this works. Therefore, Option D is correct. Let me just verify once again to make sure I didn't make a mistake. For positive x: x * 1 = x. For zero: 0. For negative x: x * (-1) = -x = |x|. Yes, that seems right. The other options either fail when x is negative or in some other case. So the correct answer should be D.Final Answerboxed{D}</think>To determine which equation correctly represents ( |x| ) using the sign function ( text{sgn}(x) ), we analyze each option:- Option A: ( |x| = x |text{sgn}(x)| ) - For ( x > 0 ), ( text{sgn}(x) = 1 ), so ( |text{sgn}(x)| = 1 ). This gives ( x cdot 1 = x ), which is correct. - For ( x = 0 ), ( text{sgn}(x) = 0 ), so ( |text{sgn}(x)| = 0 ). This gives ( 0 cdot 0 = 0 ), which is correct. - For ( x < 0 ), ( text{sgn}(x) = -1 ), so ( |text{sgn}(x)| = 1 ). This gives ( x cdot 1 = x ), which is incorrect since ( x ) is negative.- Option B: ( |x| = x text{sgn}|x| ) - For ( x neq 0 ), ( text{sgn}|x| = 1 ). This gives ( x cdot 1 = x ), which is incorrect for ( x < 0 ). - For ( x = 0 ), ( text{sgn}|x| = 0 ). This gives ( 0 cdot 0 = 0 ), which is correct.- Option C: ( |x| = |x| text{sgn}(x) ) - For ( x > 0 ), ( text{sgn}(x) = 1 ). This gives ( |x| cdot 1 = |x| ), which is correct. - For ( x = 0 ), ( text{sgn}(x) = 0 ). This gives ( 0 cdot 0 = 0 ), which is correct. - For ( x < 0 ), ( text{sgn}(x) = -1 ). This gives ( |x| cdot (-1) = -|x| ), which is incorrect.- Option D: ( |x| = x text{sgn}(x) ) - For ( x > 0 ), ( text{sgn}(x) = 1 ). This gives ( x cdot 1 = x ), which is correct. - For ( x = 0 ), ( text{sgn}(x) = 0 ). This gives ( 0 cdot 0 = 0 ), which is correct. - For ( x < 0 ), ( text{sgn}(x) = -1 ). This gives ( x cdot (-1) = -x ), which is correct since ( -x = |x| ).Thus, the correct answer is (boxed{D}).,<think>Okay, let me try to figure out this problem step by step. So, we have a function called sgn(x), which is the sign function. It's defined as 1 if x is positive, 0 if x is zero, and -1 if x is negative. The question is asking which of the options A, B, C, or D correctly expresses the absolute value of x, which is |x|, in terms of x and the sgn function. Let me recall that the absolute value of x is x when x is non-negative and -x when x is negative. So, maybe the sgn function can help capture that sign change.Let me look at each option one by one.Option A: |x| = x |sgn x|. Hmm. Let's see. The right-hand side is x multiplied by the absolute value of sgn x. Since sgn x is either 1, 0, or -1, the absolute value of sgn x would be |1| = 1, |0| = 0, or |-1| = 1. So |sgn x| is 1 when x is positive or negative, and 0 when x is zero. Therefore, x |sgn x| would be x*1 when x ≠ 0, which is just x, and 0 when x is zero. But |x| is x when x ≥ 0 and -x when x < 0. Wait, if we use x |sgn x|, then for negative x, sgn x is -1, |sgn x| is 1, so x *1 is x, which is negative. But |x| should be positive. So this doesn't work for negative x. For example, if x = -2, then |x| = 2, but x |sgn x| = -2 *1 = -2, which is not equal to 2. So Option A is incorrect.Option B: |x| = x sgn |x|. Let's see. First, |x| is always non-negative, so sgn |x| would be 1 if |x| > 0 and 0 if |x| = 0. But |x| is 0 only when x is 0. So sgn |x| is 1 for all x ≠ 0 and 0 when x = 0. Therefore, x multiplied by sgn |x| would be x*1 = x when x ≠ 0, and x*0 = 0 when x = 0. But |x| is equal to x when x ≥ 0, so if x is positive, this works, but if x is negative, then x sgn |x| would be x*1 = x, which is negative, but |x| is positive. So for x negative, this would give x, which is wrong. For example, x = -3, |x| = 3, but x sgn |x| = -3*1 = -3 ≠ 3. So Option B is also incorrect.Option C: |x| = |x| sgn x. Let's check. If x is positive, sgn x = 1, so RHS is |x| *1 = |x|, which is correct. If x is zero, sgn x = 0, so RHS is |x|*0 = 0, which is correct. If x is negative, sgn x = -1, so RHS is |x|*(-1). But |x| is positive, so this would be -|x|. However, |x| is supposed to be positive, but here when x is negative, it's giving -|x| which is negative. For example, x = -4, |x| = 4, but |x| sgn x = 4*(-1) = -4 ≠ 4. So Option C is incorrect.Option D: |x| = x sgn x. Let's analyze this. If x is positive, sgn x = 1, so RHS is x*1 = x, which is |x|. If x is zero, sgn x = 0, so RHS is 0*0 = 0, which is |0|. If x is negative, sgn x = -1, so RHS is x*(-1) = -x. But since x is negative, -x is positive, which is |x|. For example, x = -5, RHS is (-5)*(-1) = 5 = |x|. So in all cases, this works. Therefore, Option D seems to be correct.Wait, let me double-check each case again.Case 1: x > 0. sgn x = 1. Then x sgn x = x*1 = x = |x|. Correct.Case 2: x = 0. sgn x = 0. Then x sgn x = 0*0 = 0 = |0|. Correct.Case 3: x < 0. sgn x = -1. Then x sgn x = x*(-1) = -x. Since x is negative, -x is positive, so -x = |x|. Correct.So all three cases check out. Therefore, Option D is the correct identity.Just to make sure none of the other options could also be correct, let me think again.Option A fails for negative x, as shown.Option B also fails for negative x.Option C fails for negative x because it gives a negative value.Option D works in all cases. So the answer must be D.Final Answerboxed{D}</think>To determine which option correctly expresses the absolute value of ( x ) in terms of the sign function ( text{sgn}(x) ), we analyze each option:1. Option A: ( |x| = x |text{sgn}(x)| ) - For ( x > 0 ), ( text{sgn}(x) = 1 ), so ( x |text{sgn}(x)| = x cdot 1 = x ), which is correct. - For ( x = 0 ), ( text{sgn}(x) = 0 ), so ( x |text{sgn}(x)| = 0 cdot 0 = 0 ), which is correct. - For ( x < 0 ), ( text{sgn}(x) = -1 ), so ( x |text{sgn}(x)| = x cdot 1 = x ), which is incorrect because ( |x| ) should be positive. - Conclusion: Incorrect for negative ( x ).2. Option B: ( |x| = x text{sgn}(|x|) ) - ( text{sgn}(|x|) ) is 1 for ( x neq 0 ) and 0 for ( x = 0 ). - For ( x > 0 ), ( x text{sgn}(|x|) = x cdot 1 = x ), which is correct. - For ( x = 0 ), ( x text{sgn}(|x|) = 0 cdot 0 = 0 ), which is correct. - For ( x < 0 ), ( x text{sgn}(|x|) = x cdot 1 = x ), which is incorrect because ( |x| ) should be positive. - Conclusion: Incorrect for negative ( x ).3. Option C: ( |x| = |x| text{sgn}(x) ) - For ( x > 0 ), ( text{sgn}(x) = 1 ), so ( |x| cdot 1 = |x| ), which is correct. - For ( x = 0 ), ( text{sgn}(x) = 0 ), so ( |x| cdot 0 = 0 ), which is correct. - For ( x < 0 ), ( text{sgn}(x) = -1 ), so ( |x| cdot (-1) = -|x| ), which is incorrect because ( |x| ) should be positive. - Conclusion: Incorrect for negative ( x ).4. Option D: ( |x| = x text{sgn}(x) ) - For ( x > 0 ), ( text{sgn}(x) = 1 ), so ( x cdot 1 = x = |x| ), which is correct. - For ( x = 0 ), ( text{sgn}(x) = 0 ), so ( 0 cdot 0 = 0 = |0| ), which is correct. - For ( x < 0 ), ( text{sgn}(x) = -1 ), so ( x cdot (-1) = -x = |x| ), which is correct. - Conclusion: Correct for all cases.Thus, the correct answer is (boxed{D}).]

🔑:For option A, the right side =x|sgnx|= begin{cases} x, & xneq 0 0, & x=0end{cases}, while the left side =|x|= begin{cases} x, & xgeqslant 0 -x, & x 0 0, & x=0 -x, & x < 0end{cases}, while the left side =|x|= begin{cases} x, & xgeqslant 0 -x, & x < 0end{cases}, which is obviously correct;Therefore, the correct answer is boxed{D}.To solve this problem, remove the absolute value symbols and compare each option individually. This question tests the comparison of function expressions, and the key to solving it is to correctly remove the absolute value symbols. Pay attention to the accumulation of problem-solving methods, and this question is of moderate difficulty.

🧠:[<think>Okay, let me try to solve part (I) first. The problem is to calculate sin(5π/3) + cos(11π/2) + tan(-11π/6). Hmm, trigonometric functions with different angles. I need to remember the unit circle and the values of sine, cosine, and tangent at these specific angles. Let me take each term one by one.Starting with sin(5π/3). 5π/3 is in radians, right? Let me recall that 2π radians is a full circle, so 5π/3 is just 60 degrees less than 2π, since π/3 is 60 degrees. So, 5π/3 is equivalent to 360° - 60° = 300°. The sine of 300°... sine is positive in the first and second quadrants, but 300° is in the fourth quadrant where sine is negative. The reference angle here is 60°, so sin(300°) = -sin(60°) = -√3/2. So, sin(5π/3) should be -√3/2. Let me double-check that. Yes, 5π/3 is in the fourth quadrant, x is positive, y is negative, so sine is negative, and the reference angle is π/3. So sin(5π/3) = -√3/2. Got that.Next term is cos(11π/2). Hmm, 11π/2 radians. Let me see, since 2π is a full circle, 11π/2 divided by 2π is 11/4, which is 2 and 3/4. So that's 2 full circles plus 3π/2. So cos(11π/2) is the same as cos(3π/2). Cos(3π/2) is 0, right? Because 3π/2 is 270 degrees, where cosine is zero. Let me confirm. Yes, cos(3π/2) is indeed 0 because the coordinates at 3π/2 are (0, -1), so cosine (x-coordinate) is 0. So cos(11π/2) = 0. That term is zero.Third term is tan(-11π/6). Tangent is sine over cosine, and it's an odd function, so tan(-x) = -tan(x). Therefore, tan(-11π/6) = -tan(11π/6). Now, 11π/6 is in radians. Let's see, 11π/6 is equal to 2π - π/6, so that's 330 degrees. The reference angle is π/6. Tangent is sine over cosine. At 11π/6, which is in the fourth quadrant, sine is negative and cosine is positive, so tangent (sine/cosine) is negative. But since we have tan(11π/6) = tan(330°) = -tan(π/6) = -1/√3. Therefore, tan(-11π/6) = -tan(11π/6) = -(-1/√3) = 1/√3. Let me verify that. Yes, because tan is periodic with period π, so tan(-11π/6) is the same as tan(π/6) but adjusted for the negative angle. Wait, another way: -11π/6 is the same as rotating clockwise 11π/6, which is the same as rotating counterclockwise π/6, because 11π/6 clockwise is equivalent to 2π - 11π/6 = π/6 counterclockwise. So tan(-11π/6) is tan(π/6) which is 1/√3. Wait, but since tangent has a period of π, but -11π/6 is equivalent to π/6 in the first quadrant? Let me confirm. Let's add 2π to -11π/6 to get the positive equivalent angle. -11π/6 + 2π = -11π/6 + 12π/6 = π/6. So yes, tan(-11π/6) is the same as tan(π/6) = 1/√3. So that term is 1/√3.So putting all together: sin(5π/3) + cos(11π/2) + tan(-11π/6) = (-√3/2) + 0 + (1/√3). Now, I need to add these two terms. Let me get a common denominator. The terms are -√3/2 and 1/√3. Let me rationalize the denominator of 1/√3: that's √3/3. So, -√3/2 + √3/3. Factor out √3: √3*(-1/2 + 1/3) = √3*(-3/6 + 2/6) = √3*(-1/6) = -√3/6.Wait, so the total sum is -√3/6? Let me check each step again to be sure.First term: sin(5π/3). 5π/3 is 300°, reference angle 60°, sine negative, so -√3/2. Correct.Second term: cos(11π/2). 11π/2 divided by 2π is 11/4, which is 2 full circles (8π/4) plus 3π/2. So cos(11π/2) = cos(3π/2) = 0. Correct.Third term: tan(-11π/6). Converted to positive angle by adding 2π: -11π/6 + 2π = π/6. So tan(π/6) is 1/√3. Correct.Adding them: -√3/2 + 0 + 1/√3. Convert 1/√3 to √3/3. Then, -√3/2 is equal to -3√3/6 and √3/3 is 2√3/6. Wait, no. Wait, √3/3 is equal to (√3)/3. If we write both over 6:-√3/2 = (-3√3)/6√3/3 = (2√3)/6Adding them: (-3√3 + 2√3)/6 = (-√3)/6. So, yes, total is -√3/6. That seems right.So part (I) answer is -√3/6. Let me just confirm once more.Alternatively, maybe I made a mistake in the sign of the tangent term. Let's see. tan(-11π/6). The angle -11π/6. If we measure it clockwise, that's 11π/6 radians clockwise, which is 330 degrees clockwise, which is equivalent to 30 degrees in the fourth quadrant. But tangent in the fourth quadrant is negative, because sine is negative and cosine is positive. But since we have tan(-11π/6) = tan(π/6) but in the fourth quadrant. Wait, but when we added 2π to -11π/6, we got π/6, which is in the first quadrant. But tangent has period π, so tan(-11π/6) = tan(π/6) because adding π multiple times: -11π/6 + 2π = π/6, which is correct. But in terms of the original angle, -11π/6, which is the same as rotating 11π/6 clockwise, which brings you to the same terminal angle as π/6 in the first quadrant. However, tangent is periodic with period π, so tan(-11π/6) = tan(-11π/6 + 2π) = tan(π/6). So tan(-11π/6) is tan(π/6) = 1/√3. So positive. So that term is positive 1/√3. Correct. Then the sum is -√3/2 + 1/√3. Converting to common denominator: which is 6. So -3√3/6 + 2√3/6 = (-√3)/6. Yes, that's correct.Okay, so part (I) is -√3/6. Now moving to part (II): Simplify the expression [sinθ / (1 - 1/tanθ)] + [cosθ / (1 - tanθ)].Alright, let's see. Let me first parse the expression. The first term is sinθ divided by (1 - 1/tanθ), and the second term is cosθ divided by (1 - tanθ). We need to simplify this sum.First, let's recall that tanθ = sinθ / cosθ, so 1/tanθ = cotθ = cosθ / sinθ. Let me rewrite the denominators in terms of sine and cosine to make it easier.Starting with the first denominator: 1 - 1/tanθ. Since 1/tanθ is cotθ, which is cosθ/sinθ. So 1 - cosθ/sinθ. Let me combine the terms over a common denominator. 1 can be written as sinθ/sinθ, so:1 - cosθ/sinθ = (sinθ - cosθ)/sinθ.Therefore, the first term becomes sinθ divided by [(sinθ - cosθ)/sinθ], which is equal to sinθ * [sinθ / (sinθ - cosθ)] = sin²θ / (sinθ - cosθ).Now the second term: cosθ / (1 - tanθ). Again, tanθ is sinθ/cosθ, so 1 - tanθ = 1 - sinθ/cosθ. Let's combine over common denominator:1 - sinθ/cosθ = (cosθ - sinθ)/cosθ.Therefore, the second term becomes cosθ divided by [(cosθ - sinθ)/cosθ] = cosθ * [cosθ / (cosθ - sinθ)] = cos²θ / (cosθ - sinθ).But notice that the denominator in the second term is (cosθ - sinθ), which is the negative of (sinθ - cosθ). So we can write:cos²θ / (cosθ - sinθ) = -cos²θ / (sinθ - cosθ).Therefore, the entire expression becomes:[sin²θ / (sinθ - cosθ)] + [-cos²θ / (sinθ - cosθ)] = [sin²θ - cos²θ] / (sinθ - cosθ).Now, the numerator is sin²θ - cos²θ, which is a difference of squares. Recall that a² - b² = (a - b)(a + b). Therefore:sin²θ - cos²θ = (sinθ - cosθ)(sinθ + cosθ).Therefore, the entire expression becomes:[(sinθ - cosθ)(sinθ + cosθ)] / (sinθ - cosθ).We can cancel out (sinθ - cosθ) from numerator and denominator, provided that sinθ ≠ cosθ. So the simplified expression is sinθ + cosθ.Wait, let me check each step again.First term: sinθ / (1 - 1/tanθ). Let me compute 1 - 1/tanθ.1/tanθ = cotθ = cosθ/sinθ, so 1 - cosθ/sinθ = (sinθ - cosθ)/sinθ. Then sinθ divided by that is sinθ * [sinθ / (sinθ - cosθ)] = sin²θ / (sinθ - cosθ). Correct.Second term: cosθ / (1 - tanθ). 1 - tanθ = 1 - sinθ/cosθ = (cosθ - sinθ)/cosθ. Then cosθ divided by that is cosθ * [cosθ / (cosθ - sinθ)] = cos²θ / (cosθ - sinθ). Which is the same as -cos²θ / (sinθ - cosθ). Correct.Adding the two terms: sin²θ/(sinθ - cosθ) - cos²θ/(sinθ - cosθ) = (sin²θ - cos²θ)/(sinθ - cosθ). Factor numerator as (sinθ - cosθ)(sinθ + cosθ), cancel with denominator, left with sinθ + cosθ. Perfect.Therefore, the simplified expression is sinθ + cosθ. Let me verify with an example. Let’s take θ = 45°, so tanθ = 1. Then original expression:First term: sin45 / (1 - 1/1) = sin45 / 0. Uh-oh, division by zero. Hmm, so for θ = 45°, the original expression is undefined because denominator becomes zero. Similarly, in the simplified expression sinθ + cosθ, for θ = 45°, sin45 + cos45 = √2/2 + √2/2 = √2. But original expression is undefined here. So the simplification holds where the original expression is defined, i.e., when 1 - 1/tanθ ≠ 0 and 1 - tanθ ≠ 0. So 1 - 1/tanθ ≠ 0 => 1/tanθ ≠ 1 => tanθ ≠ 1. Similarly, 1 - tanθ ≠ 0 => tanθ ≠ 1. So as long as tanθ ≠ 1, which is when θ ≠ 45° + kπ, the expression is defined and simplifies to sinθ + cosθ. So the simplification is valid for θ where tanθ ≠ 1. So the answer is sinθ + cosθ.Yes, that seems correct. So part (II) simplifies to sinθ + cosθ.Final Answer(I) boxed{-dfrac{sqrt{3}}{6}} (II) boxed{sin theta + cos theta}</think>(I) To calculate (sin frac{5pi}{3} + cos frac{11pi}{2} + tan(-frac{11pi}{6})):1. (sin frac{5pi}{3}): - (frac{5pi}{3}) is in the fourth quadrant with reference angle (frac{pi}{3}). - (sin frac{5pi}{3} = -sin frac{pi}{3} = -frac{sqrt{3}}{2}).2. (cos frac{11pi}{2}): - (frac{11pi}{2}) is equivalent to (frac{3pi}{2}) after subtracting (2pi times 2). - (cos frac{3pi}{2} = 0).3. (tan(-frac{11pi}{6})): - (-frac{11pi}{6}) is equivalent to (frac{pi}{6}) after adding (2pi). - (tan frac{pi}{6} = frac{1}{sqrt{3}}). - Since (tan) is odd, (tan(-frac{11pi}{6}) = frac{1}{sqrt{3}}).Summing these results:[-frac{sqrt{3}}{2} + 0 + frac{1}{sqrt{3}} = -frac{sqrt{3}}{2} + frac{sqrt{3}}{3} = sqrt{3} left(-frac{1}{2} + frac{1}{3}right) = sqrt{3} left(-frac{1}{6}right) = -frac{sqrt{3}}{6}](II) To simplify (frac{sin theta}{1 - frac{1}{tan theta}} + frac{cos theta}{1 - tan theta}):1. Simplify the first term: - (1 - frac{1}{tan theta} = 1 - cot theta = frac{sin theta - cos theta}{sin theta}) - (frac{sin theta}{frac{sin theta - cos theta}{sin theta}} = frac{sin^2 theta}{sin theta - cos theta})2. Simplify the second term: - (1 - tan theta = frac{cos theta - sin theta}{cos theta}) - (frac{cos theta}{frac{cos theta - sin theta}{cos theta}} = frac{cos^2 theta}{cos theta - sin theta} = -frac{cos^2 theta}{sin theta - cos theta})3. Combine the terms: - (frac{sin^2 theta}{sin theta - cos theta} - frac{cos^2 theta}{sin theta - cos theta} = frac{sin^2 theta - cos^2 theta}{sin theta - cos theta}) - Factor the numerator: (sin^2 theta - cos^2 theta = (sin theta - cos theta)(sin theta + cos theta)) - Cancel the common factor: (frac{(sin theta - cos theta)(sin theta + cos theta)}{sin theta - cos theta} = sin theta + cos theta)Final Answers:(I) (boxed{-dfrac{sqrt{3}}{6}})(II) (boxed{sin theta + cos theta}),<think>Okay, let me try to tackle these two problems step by step. I remember some trigonometric identities and unit circle values, so hopefully, that will help. Let's start with part (I).Problem I: Calculate sin(5π/3) + cos(11π/2) + tan(-11π/6)Alright, so I need to compute three trigonometric functions at specific angles and then add them up. Let me break them down one by one.1. Calculating sin(5π/3):First, 5π/3 radians. I know that the unit circle goes up to 2π, which is a full circle. Let me convert 5π/3 to degrees to maybe make it easier. Since π radians is 180 degrees, so 5π/3 is (5/3)*180 = 300 degrees. 300 degrees is in the fourth quadrant, where sine is negative. The reference angle here is 360 - 300 = 60 degrees. So sin(300°) = -sin(60°). Sin(60°) is √3/2, so sin(5π/3) should be -√3/2.Wait, let me confirm that. The coordinates on the unit circle for 5π/3 are (1/2, -√3/2), right? Because cosine is positive in the fourth quadrant, and sine is negative. So yes, sin(5π/3) is indeed -√3/2. Got that.2. Calculating cos(11π/2):Hmm, 11π/2 radians. That seems like a large angle. Let me see how many full rotations that is. Since 2π is a full circle, dividing 11π/2 by 2π gives (11π/2)/(2π) = 11/4 = 2.75. So that's 2 full rotations plus 3/4 of a rotation. So 11π/2 is equivalent to 3π/2 after subtracting the full rotations (since 11π/2 - 2*2π = 11π/2 - 4π = 11π/2 - 8π/2 = 3π/2). Alternatively, in degrees, π/2 is 90 degrees, so 11π/2 is 11*90 = 990 degrees. Subtracting 2*360 = 720, 990 - 720 = 270 degrees, which is 3π/2 radians. At 3π/2 radians, the cosine value is 0, right? Because the coordinates on the unit circle are (0, -1), so cosine is 0. So cos(11π/2) = 0. That seems straightforward.Wait, but let me double-check. If I take 11π/2, subtracting 4π (which is 2 full rotations) gives 11π/2 - 8π/2 = 3π/2. So yes, equivalent to 3π/2. So cos(3π/2) is 0. Correct.3. Calculating tan(-11π/6):Negative angle, so direction is clockwise. Let me first find the positive equivalent by adding 2π. -11π/6 + 2π = -11π/6 + 12π/6 = π/6. So tan(-11π/6) is tan(π/6). But since tangent is an odd function, tan(-x) = -tan(x), so tan(-11π/6) = -tan(11π/6). Wait, hold on. Maybe I need to be careful here.Wait, maybe I should convert -11π/6 to a positive angle. Since angles are periodic with period π for tangent, but actually, the period is π, but for tangent, it's π, right? Wait, no, tangent has a period of π. So adding π to the angle will give the same value. But let's see.Alternatively, adding 2π to -11π/6 gives -11π/6 + 12π/6 = π/6. So tan(-11π/6) is tan(π/6). But let me check the sign. Since -11π/6 is the same as rotating clockwise 11π/6. Alternatively, 11π/6 is in the fourth quadrant (since π/6 is 30 degrees, so 11π/6 is 330 degrees). So -11π/6 would be equivalent to rotating clockwise 330 degrees, which is the same as rotating counterclockwise 30 degrees (since 360 - 330 = 30). Wait, but rotating clockwise 330 degrees is the same as rotating counterclockwise 30 degrees. Hmm, no. Wait, rotating clockwise by θ is equivalent to rotating counterclockwise by 2π - θ. Wait, no. If θ is the clockwise rotation, then the equivalent counterclockwise rotation is 2π - θ. So -11π/6 is a clockwise rotation of 11π/6, which is equivalent to a counterclockwise rotation of 2π - 11π/6 = (12π/6 - 11π/6) = π/6. Therefore, tan(-11π/6) is equal to tan(π/6). But since tangent is odd, tan(-x) = -tan(x), so tan(-11π/6) = -tan(11π/6). But 11π/6 is equivalent to π/6 in the fourth quadrant, where tangent is negative. So tan(11π/6) = -tan(π/6) = -1/√3. Therefore, tan(-11π/6) = -tan(11π/6) = -(-1/√3) = 1/√3.Wait, that seems conflicting. Let me verify:Method 1: Convert -11π/6 to positive by adding 2π: -11π/6 + 2π = π/6. So tan(-11π/6) = tan(π/6) = 1/√3.Method 2: Using tan(-x) = -tan(x). So tan(-11π/6) = -tan(11π/6). Now, 11π/6 is in the fourth quadrant. Tangent there is sin/cos. Sin(11π/6) is -1/2, cos(11π/6) is √3/2, so tan(11π/6) = (-1/2)/(√3/2) = -1/√3. Therefore, -tan(11π/6) = -(-1/√3) = 1/√3. So both methods give the same result. So tan(-11π/6) is 1/√3.So that seems correct.So putting it all together:sin(5π/3) = -√3/2cos(11π/2) = 0tan(-11π/6) = 1/√3Therefore, the sum is -√3/2 + 0 + 1/√3.Now, need to combine these terms. Let me rationalize 1/√3 to have a common denominator:1/√3 = √3/3Similarly, -√3/2 can be written as - (√3)/2So now, we have:-√3/2 + √3/3To add these, find a common denominator, which is 6.-√3/2 = - (3√3)/6√3/3 = (2√3)/6Adding them: (-3√3 + 2√3)/6 = (-√3)/6So the total sum is -√3/6.Wait, let me verify the arithmetic:-√3/2 + √3/3 = √3(-1/2 + 1/3) = √3(-3/6 + 2/6) = √3(-1/6) = -√3/6.Yes, that's correct.So the final result for part (I) is -√3/6.Problem II: Simplify [sinθ / (1 - 1/tanθ)] + [cosθ / (1 - tanθ)]Alright, so need to simplify this expression. Let me first look at each term separately.First term: sinθ / (1 - 1/tanθ)Second term: cosθ / (1 - tanθ)Let me handle them one by one.Starting with the first term: sinθ / (1 - 1/tanθ)First, note that 1/tanθ is cotθ, so 1 - cotθ. But maybe better to express in terms of sine and cosine.Let me write 1/tanθ as cotθ. So 1 - cotθ.Alternatively, tanθ is sinθ/cosθ, so 1/tanθ is cosθ/sinθ. Therefore, the denominator becomes 1 - cosθ/sinθ = (sinθ - cosθ)/sinθ.Therefore, the first term becomes:sinθ / [(sinθ - cosθ)/sinθ] = sinθ * [sinθ / (sinθ - cosθ)] = sin²θ / (sinθ - cosθ)Alright, so the first term simplifies to sin²θ / (sinθ - cosθ)Now, moving to the second term: cosθ / (1 - tanθ)Similarly, tanθ is sinθ/cosθ, so 1 - tanθ = 1 - sinθ/cosθ = (cosθ - sinθ)/cosθ.Therefore, the second term is cosθ / [(cosθ - sinθ)/cosθ] = cosθ * [cosθ / (cosθ - sinθ)] = cos²θ / (cosθ - sinθ)But note that cosθ - sinθ = -(sinθ - cosθ). So we can write this as:cos²θ / (- (sinθ - cosθ)) = -cos²θ / (sinθ - cosθ)Therefore, the second term simplifies to -cos²θ / (sinθ - cosθ)Now, combining both terms:First term: sin²θ / (sinθ - cosθ)Second term: -cos²θ / (sinθ - cosθ)Therefore, adding them together:[sin²θ - cos²θ] / (sinθ - cosθ)Wait, because it's sin²θ/(sinθ - cosθ) + (-cos²θ)/(sinθ - cosθ) = (sin²θ - cos²θ)/(sinθ - cosθ)Now, the numerator is sin²θ - cos²θ, which is a difference of squares and can be factored as (sinθ - cosθ)(sinθ + cosθ)Therefore, the expression becomes:[(sinθ - cosθ)(sinθ + cosθ)] / (sinθ - cosθ)Assuming that sinθ ≠ cosθ, we can cancel out (sinθ - cosθ) from numerator and denominator, resulting in:sinθ + cosθTherefore, the simplified expression is sinθ + cosθ.But let me verify each step again to be sure.Starting with the first term:sinθ / (1 - 1/tanθ) = sinθ / (1 - cotθ) = sinθ / [1 - (cosθ/sinθ)] = sinθ / [(sinθ - cosθ)/sinθ] = sinθ * (sinθ / (sinθ - cosθ)) = sin²θ / (sinθ - cosθ). That seems right.Second term:cosθ / (1 - tanθ) = cosθ / [1 - (sinθ/cosθ)] = cosθ / [(cosθ - sinθ)/cosθ] = cosθ * (cosθ / (cosθ - sinθ)) = cos²θ / (cosθ - sinθ) = -cos²θ / (sinθ - cosθ). Correct.Adding the two terms: sin²θ/(sinθ - cosθ) + (-cos²θ)/(sinθ - cosθ) = (sin²θ - cos²θ)/(sinθ - cosθ)Numerator: sin²θ - cos²θ factors into (sinθ - cosθ)(sinθ + cosθ). Then, divide by (sinθ - cosθ), giving sinθ + cosθ. Yes, that works, provided that sinθ - cosθ ≠ 0, which would be when sinθ ≠ cosθ, i.e., when θ ≠ π/4 + kπ for integer k. But since the original expression is undefined when denominators are zero, the simplification holds where the original expression is defined.Therefore, the simplified form is sinθ + cosθ.So, after all that, problem (II) simplifies to sinθ + cosθ.Final Answer(I) boxed{-dfrac{sqrt{3}}{6}}; (II) boxed{sin theta + cos theta}</think>(I) To calculate (sin frac{5pi}{3} + cos frac{11pi}{2} + tan(-frac{11pi}{6})):1. (sin frac{5pi}{3}): - (5pi/3) is in the fourth quadrant, reference angle (pi/3). - (sin frac{5pi}{3} = -sin frac{pi}{3} = -frac{sqrt{3}}{2}).2. (cos frac{11pi}{2}): - (frac{11pi}{2}) is equivalent to (frac{3pi}{2}) after subtracting full rotations. - (cos frac{3pi}{2} = 0).3. (tan(-frac{11pi}{6})): - Convert to positive angle: (-frac{11pi}{6} + 2pi = frac{pi}{6}). - (tan(-frac{11pi}{6}) = tan frac{pi}{6} = frac{1}{sqrt{3}}).Summing these results:[-frac{sqrt{3}}{2} + 0 + frac{1}{sqrt{3}} = -frac{sqrt{3}}{2} + frac{sqrt{3}}{3} = -frac{3sqrt{3}}{6} + frac{2sqrt{3}}{6} = -frac{sqrt{3}}{6}](II) Simplify (frac{sin theta}{1 - frac{1}{tan theta}} + frac{cos theta}{1 - tan theta}):1. Simplify the first term: [ frac{sin theta}{1 - frac{1}{tan theta}} = frac{sin theta}{1 - cot theta} = frac{sin theta}{1 - frac{cos theta}{sin theta}} = frac{sin theta}{frac{sin theta - cos theta}{sin theta}} = frac{sin^2 theta}{sin theta - cos theta} ]2. Simplify the second term: [ frac{cos theta}{1 - tan theta} = frac{cos theta}{1 - frac{sin theta}{cos theta}} = frac{cos theta}{frac{cos theta - sin theta}{cos theta}} = frac{cos^2 theta}{cos theta - sin theta} = -frac{cos^2 theta}{sin theta - cos theta} ]3. Combine the terms: [ frac{sin^2 theta}{sin theta - cos theta} - frac{cos^2 theta}{sin theta - cos theta} = frac{sin^2 theta - cos^2 theta}{sin theta - cos theta} = frac{(sin theta - cos theta)(sin theta + cos theta)}{sin theta - cos theta} = sin theta + cos theta ]Final answers:(I) (boxed{-dfrac{sqrt{3}}{6}});(II) (boxed{sin theta + cos theta})]

🔑:(I) To calculate the expression, we first adjust the angles within the trigonometric functions:sin frac{5pi }{3}+cos frac{11pi }{2}+tan (-frac{11pi }{6})=sin (pi -frac{pi }{3})+cos (6pi -frac{pi }{2})+tan (-2pi +frac{pi }{6})=-sin frac{pi }{3}-cos frac{pi }{2}+tan frac{pi }{6}=-frac{sqrt{3}}{2}+0+frac{sqrt{3}}{3}=-frac{sqrt{3}}{6}.So, the result of the calculation is boxed{-frac{sqrt{3}}{6}}.(II) To simplify the expression, we first convert tan theta into frac{sin theta }{cos theta }:frac{sin theta }{1-frac{1}{tan theta }}+frac{cos theta }{1-tan theta }=frac{sin theta }{1-frac{cos theta }{sin theta }}+frac{cos theta }{1-frac{sin theta }{cos theta }}=frac{sin ^{2}theta }{sin theta -cos theta }-frac{cos ^{2}theta }{sin theta -cos theta }=frac{(sin theta -cos theta )(sin theta +cos theta )}{sin theta -cos theta }=sin theta +cos theta.So, the simplified expression is boxed{sin theta +cos theta }.