💡Alright, so I have this problem here: If point P, which is (sinα, tanα), is in the fourth quadrant, then α is an angle in which quadrant? Hmm, okay, let's break this down step by step.First, I remember that the coordinate plane is divided into four quadrants. The first quadrant is where both x and y are positive, the second quadrant has x negative and y positive, the third quadrant has both x and y negative, and the fourth quadrant has x positive and y negative. So, if point P is in the fourth quadrant, that means its x-coordinate is positive and its y-coordinate is negative.Now, looking at point P, its coordinates are (sinα, tanα). So, the x-coordinate is sinα, and the y-coordinate is tanα. Since P is in the fourth quadrant, sinα must be positive, and tanα must be negative. Okay, so sinα > 0 and tanα < 0.I need to figure out in which quadrant angle α lies based on these conditions. Let me recall the signs of the trigonometric functions in different quadrants. In the first quadrant, all trigonometric functions (sin, cos, tan) are positive. In the second quadrant, sin is positive, cos is negative, so tan, which is sin/cos, would be negative. In the third quadrant, both sin and cos are negative, so tan is positive. In the fourth quadrant, sin is negative, cos is positive, so tan is negative.Wait, hold on. Let me make sure I have that right. So, in the second quadrant: sin is positive, cos is negative, so tan = sin/cos would be negative. In the fourth quadrant: sin is negative, cos is positive, so tan is negative as well. Hmm, so both second and fourth quadrants have tan negative. But in this problem, sinα is positive, so that narrows it down.Since sinα is positive, α must be in either the first or second quadrant because in the third and fourth quadrants, sin is negative. So, α is either in the first or second quadrant. But we also know that tanα is negative. In the first quadrant, tan is positive, so that can't be. Therefore, α must be in the second quadrant where tan is negative and sin is positive.Wait, let me double-check. If α is in the second quadrant, sinα is positive, which matches our condition, and tanα is negative, which also matches. So, that seems consistent. Is there any other possibility? Let me think. If α were in the fourth quadrant, sinα would be negative, which contradicts our condition that sinα is positive. So, no, it can't be in the fourth quadrant. Similarly, in the first quadrant, tanα is positive, which doesn't fit because tanα is negative. In the third quadrant, both sin and cos are negative, so tan would be positive, which again doesn't fit. So, the only possible quadrant is the second quadrant.Therefore, α must be in the second quadrant. That makes sense because in the second quadrant, sine is positive and tangent is negative, which aligns perfectly with the coordinates of point P being in the fourth quadrant.I think I've got it. So, the answer should be the second quadrant.

💡Alright, so I have this function ( f(x) = |ax - x^2| + ln x ) where ( a ) is a positive constant. There are two parts to the problem. First, when ( a = 2 ), I need to find the intervals where ( f(x) ) is monotonically increasing. Second, I need to define ( g(x) = frac{f(x)}{x} ) and find its minimum value over the interval ([1, e]).Starting with the first part. Since ( a = 2 ), the function becomes ( f(x) = |2x - x^2| + ln x ). I know that the absolute value function can change its expression depending on whether the inside is positive or negative. So, I should figure out where ( 2x - x^2 ) is positive and where it's negative.Let me solve ( 2x - x^2 = 0 ). Factoring, I get ( x(2 - x) = 0 ), so the roots are at ( x = 0 ) and ( x = 2 ). Since ( x ) is positive (because of the ( ln x ) term), I can ignore ( x = 0 ). So, for ( x < 2 ), ( 2x - x^2 ) is positive, and for ( x > 2 ), it's negative.Therefore, I can write ( f(x) ) as:- ( f(x) = 2x - x^2 + ln x ) when ( 0 < x < 2 )- ( f(x) = x^2 - 2x + ln x ) when ( x geq 2 )To find where ( f(x) ) is increasing, I need to compute its derivative and find where it's positive.First, for ( 0 < x < 2 ):( f(x) = 2x - x^2 + ln x )Derivative:( f'(x) = 2 - 2x + frac{1}{x} )Let me simplify this:( f'(x) = frac{1}{x} + 2 - 2x )Combine terms:( f'(x) = frac{1 + 2x - 2x^2}{x} )So, ( f'(x) = frac{-2x^2 + 2x + 1}{x} )To find critical points, set numerator equal to zero:( -2x^2 + 2x + 1 = 0 )Multiply both sides by -1:( 2x^2 - 2x - 1 = 0 )Using quadratic formula:( x = frac{2 pm sqrt{(2)^2 - 4(2)(-1)}}{2(2)} = frac{2 pm sqrt{4 + 8}}{4} = frac{2 pm sqrt{12}}{4} = frac{2 pm 2sqrt{3}}{4} = frac{1 pm sqrt{3}}{2} )Since ( x ) must be positive and less than 2, both roots are candidates. Compute the numerical values:( frac{1 + sqrt{3}}{2} approx frac{1 + 1.732}{2} approx 1.366 )( frac{1 - sqrt{3}}{2} approx frac{1 - 1.732}{2} approx -0.366 ) (Discard since x must be positive)So, the critical point is at ( x approx 1.366 ). Now, I need to test intervals around this point to see where ( f'(x) ) is positive.- For ( 0 < x < frac{1 + sqrt{3}}{2} approx 1.366 ), pick ( x = 1 ): ( f'(1) = 2 - 2(1) + 1 = 1 ) which is positive.- For ( frac{1 + sqrt{3}}{2} < x < 2 ), pick ( x = 1.5 ): Compute numerator: ( -2(1.5)^2 + 2(1.5) + 1 = -4.5 + 3 + 1 = -0.5 ) So, ( f'(1.5) = -0.5 / 1.5 = -1/3 ) which is negative.Therefore, ( f(x) ) is increasing on ( (0, frac{1 + sqrt{3}}{2}) ) and decreasing on ( (frac{1 + sqrt{3}}{2}, 2) ).Now, for ( x geq 2 ):( f(x) = x^2 - 2x + ln x )Derivative:( f'(x) = 2x - 2 + frac{1}{x} )Simplify:( f'(x) = frac{2x^2 - 2x + 1}{x} )Again, set numerator equal to zero:( 2x^2 - 2x + 1 = 0 )Discriminant: ( (-2)^2 - 4(2)(1) = 4 - 8 = -4 )Since discriminant is negative, no real roots. So, the numerator is always positive because the coefficient of ( x^2 ) is positive.Thus, ( f'(x) > 0 ) for all ( x geq 2 ). Therefore, ( f(x) ) is increasing on ( (2, infty) ).Combining both intervals, ( f(x) ) is increasing on ( (0, frac{1 + sqrt{3}}{2}) ) and ( (2, infty) ).Moving on to the second part. Define ( g(x) = frac{f(x)}{x} ). So, ( g(x) = frac{|ax - x^2| + ln x}{x} = |x - a| + frac{ln x}{x} ).We need to find the minimum of ( g(x) ) over ([1, e]). Let's consider different cases based on the value of ( a ).Case 1: ( a leq 1 )In this case, for all ( x in [1, e] ), ( x geq a ), so ( |x - a| = x - a ).Thus, ( g(x) = x - a + frac{ln x}{x} )Compute derivative:( g'(x) = 1 + frac{1 - ln x}{x^2} )Simplify:( g'(x) = frac{x^2 + 1 - ln x}{x^2} )Since ( x in [1, e] ), ( ln x ) is between 0 and 1. So, ( 1 - ln x geq 0 ), and ( x^2 + 1 - ln x geq x^2 geq 1 ). Thus, ( g'(x) > 0 ). So, ( g(x) ) is increasing on ([1, e]). Therefore, the minimum is at ( x = 1 ):( g(1) = 1 - a + 0 = 1 - a )Case 2: ( a geq e )Here, for all ( x in [1, e] ), ( x leq a ), so ( |x - a| = a - x ).Thus, ( g(x) = a - x + frac{ln x}{x} )Derivative:( g'(x) = -1 + frac{1 - ln x}{x^2} )Simplify:( g'(x) = frac{-x^2 + 1 - ln x}{x^2} )Let me analyze the numerator: ( -x^2 + 1 - ln x )At ( x = 1 ): ( -1 + 1 - 0 = 0 )At ( x = e ): ( -e^2 + 1 - 1 = -e^2 < 0 )Also, the derivative of the numerator is ( -2x - frac{1}{x} ), which is negative for ( x > 0 ). So, the numerator is decreasing on ([1, e]). Since at ( x = 1 ) it's 0 and decreasing, it becomes negative for ( x > 1 ). Thus, ( g'(x) < 0 ) for ( x in (1, e] ). So, ( g(x) ) is decreasing on ([1, e]). Therefore, the minimum is at ( x = e ):( g(e) = a - e + frac{1}{e} )Case 3: ( 1 < a < e )Here, ( x ) can be less than or greater than ( a ) within ([1, e]). So, we need to split the interval at ( x = a ).For ( x in [1, a) ):( g(x) = a - x + frac{ln x}{x} )Derivative:( g'(x) = -1 + frac{1 - ln x}{x^2} )As before, the numerator is ( -x^2 + 1 - ln x ). At ( x = 1 ): 0, and decreasing. So, for ( x in [1, a) ), ( g'(x) leq 0 ). Thus, ( g(x) ) is decreasing on ([1, a)).For ( x in (a, e] ):( g(x) = x - a + frac{ln x}{x} )Derivative:( g'(x) = 1 + frac{1 - ln x}{x^2} )As in Case 1, ( g'(x) > 0 ). So, ( g(x) ) is increasing on ( (a, e] ).Therefore, the function ( g(x) ) decreases from ( x = 1 ) to ( x = a ), then increases from ( x = a ) to ( x = e ). Hence, the minimum occurs at ( x = a ).Compute ( g(a) ):( g(a) = |a - a| + frac{ln a}{a} = 0 + frac{ln a}{a} = frac{ln a}{a} )So, summarizing all cases:- If ( a leq 1 ), minimum is ( 1 - a )- If ( 1 < a < e ), minimum is ( frac{ln a}{a} )- If ( a geq e ), minimum is ( a - e + frac{1}{e} )I think that covers all possibilities for ( a ). I should double-check if there are any overlaps or if I missed something, but it seems comprehensive.

💡Alright, so I've been given this interesting chess problem to think about. It's about an infinite chessboard where there's a piece placed every three squares both horizontally and vertically. The question is whether a knight can travel through the remaining part of the board, visiting each square exactly once. Hmm, that sounds like a knight's tour problem but with some obstacles in place.First, let me recall what a knight's tour is. It's a sequence of moves by a knight on a chessboard such that the knight visits every square exactly once. If the knight ends on a square that is one knight's move away from the beginning square, it's called a closed tour; otherwise, it's an open tour. But in this case, the chessboard is infinite, which adds another layer of complexity.Now, the chessboard has pieces placed every three squares both horizontally and vertically. So, if I imagine the chessboard, there are tokens or pieces on squares that are multiples of three in both the row and column directions. That means, for example, squares (3,3), (3,6), (6,3), (6,6), and so on are occupied. This creates a grid of blocked squares spaced three units apart.I need to determine if a knight can still traverse the remaining squares, visiting each exactly once. Since the board is infinite, the knight doesn't have to worry about boundaries, but the periodic placement of obstacles might disrupt the possibility of a continuous tour.Let me think about the movement of a knight. A knight moves in an L-shape: two squares in one direction and then one square perpendicular, or one square in one direction and two squares perpendicular. This movement pattern means the knight alternates between light and dark squares with each move. On a standard chessboard, this alternation is crucial because it ensures that the knight can potentially reach every square given the right conditions.But in this case, with pieces placed every three squares, the knight's movement might be hindered. The blocked squares could create barriers that the knight cannot cross, thereby preventing it from reaching certain areas of the board. I need to analyze if these barriers are significant enough to prevent a complete tour.Maybe I should consider the structure of the blocked squares. If pieces are placed every three squares, they form a grid of 3x3 blocks, with the center square of each block being occupied. This creates a repeating pattern every three squares. The knight would have to navigate around these blocked squares, but given the infinite nature of the board, it might still find a way to weave through them.However, I also need to consider the coloring of the squares. On a standard chessboard, the knight alternates between light and dark squares. If the blocked squares are all of one color, say light, then the knight would have to adjust its path accordingly. But in this problem, the pieces are placed every three squares regardless of color, so they might block both light and dark squares in a periodic fashion.Wait, actually, since the pieces are placed every three squares, their color alternation would depend on the starting point. If the first piece is on a light square, then the next piece three squares to the right would be on a dark square, and so on. This means that the blocked squares alternate in color every three squares. That could complicate the knight's movement because it might have to switch colors more frequently or find alternative paths.Let me try to visualize this. Imagine starting at a square that's not blocked. The knight makes its L-shaped move, landing on a square that's either light or dark, depending on the starting square. If the next square it needs to move to is blocked, it has to find another route. But since the board is infinite, there might always be a way around, but I'm not sure.Maybe I should look at smaller sections of the board to see if a pattern emerges. If I consider a 6x6 section, with pieces placed every three squares, that would mean pieces are on (3,3) and (6,6), for example. The knight would have to navigate around these blocked squares. But even in this small section, it's not immediately clear if a tour is possible.I wonder if there's a mathematical way to approach this. Perhaps using graph theory, where each square is a node, and edges connect squares that a knight can move between. Then, the problem reduces to finding a Hamiltonian path in this graph, avoiding the blocked nodes. But with an infinite graph, traditional methods might not apply.Alternatively, maybe I can use parity arguments. Since the knight alternates colors with each move, and the blocked squares are placed periodically, there might be an imbalance in the number of available squares of each color, making a complete tour impossible.Let me explore this idea. Suppose we color the chessboard in the standard alternating black and white pattern. The knight always moves from a black to a white square and vice versa. If the number of black and white squares is equal or differs by one, a closed or open tour is possible, respectively.But with pieces placed every three squares, the number of blocked squares of each color might not be equal. For example, if the first piece is on a black square, the next piece three squares to the right would be on a white square, and so on. This could lead to an imbalance in the number of blocked squares of each color.If the number of blocked squares of one color is significantly higher than the other, it might make it impossible for the knight to alternate properly, thereby preventing a complete tour. However, since the board is infinite, the overall balance might still hold, but locally, in any finite section, there could be imbalances.Hmm, this is getting a bit abstract. Maybe I should look for similar problems or known results. I recall that on an infinite chessboard without any obstacles, a knight's tour is possible. But with obstacles, especially periodic ones like every three squares, it might not be.I also remember that for a knight's tour to exist, the graph must be connected, meaning there's a path between any two squares. If the obstacles create disconnected regions, then a complete tour is impossible. So, I need to check if the placement of pieces every three squares disconnects the board.Visually, if pieces are placed every three squares, they form a grid that's sparser than the knight's movement range. A knight can jump over these pieces, so in theory, it shouldn't be disconnected. But the periodicity might create some kind of barrier that the knight cannot cross.Wait, actually, a knight can move two squares in one direction and one in the other, so it can easily jump over a single piece. But with pieces every three squares, the knight might have to make larger detours, which could accumulate and prevent it from covering the entire board.Another angle is to consider the density of the obstacles. Since pieces are placed every three squares, the density is relatively low. The knight has a lot of space to move around, so maybe it's still possible to find a path that covers all squares.But I'm not sure. The infinite nature of the board complicates things because even if locally the knight can navigate around obstacles, globally it might not be able to cover all regions without getting stuck in an infinite loop or missing some areas.Perhaps I should try to construct a possible tour or find a contradiction that shows it's impossible. If I can find a repeating pattern or a section where the knight cannot proceed without revisiting a square, that would indicate it's impossible.Alternatively, if I can show that the knight can always find a way to move to a new square without getting blocked indefinitely, then it might be possible.I think I need to formalize this a bit more. Let's consider the coordinates of the chessboard. Let's say the bottom-left corner is (0,0), and the board extends infinitely in the positive direction. Pieces are placed at positions where both the x and y coordinates are multiples of three, i.e., (3k, 3m) for integers k and m.Now, the knight starts at some position, say (0,0), which is not blocked. It needs to move to a square that's two in one direction and one in the other, like (1,2). From there, it can move to (2,4), and so on. But it has to avoid squares like (3,3), (3,6), etc.I wonder if there's a way to define a spiral or some kind of pattern that the knight can follow, weaving around the blocked squares. But given the infinite nature, it's hard to define such a pattern.Maybe I can use modular arithmetic to analyze the knight's movement. Since the knight moves in L-shapes, its coordinates change by (±1, ±2) or (±2, ±1). If I consider the coordinates modulo 3, since the blocked squares are every three squares, I can see how the knight's position changes in relation to the blocked squares.Let's say the knight is at position (x, y). After a move, it will be at (x ± 1, y ± 2) or (x ± 2, y ± 1). If I take these modulo 3, I can see how the knight's position shifts in terms of the blocked squares.For example, if the knight is at (0,0), which is a blocked square, but we're assuming it starts at a non-blocked square, say (0,1). From there, it can move to (1,3), which is a blocked square, or (2,2), which is not blocked. So, it has to choose the non-blocked square.Continuing this way, the knight has to make choices that avoid the blocked squares. But since the board is infinite, it might always have a way to move forward without getting stuck.However, I'm not sure if this approach leads me anywhere concrete. Maybe I need to think about the problem differently.Perhaps considering the problem as a graph where each node is a square, and edges connect squares a knight can move between. Then, the question is whether this graph has a Hamiltonian path that avoids the blocked nodes.But with an infinite graph, traditional Hamiltonian path concepts don't directly apply. Instead, we might need to use concepts from infinite graph theory, which I'm not very familiar with.Alternatively, maybe I can use a coloring argument. If the knight alternates between colors, and the blocked squares are distributed in a way that disrupts this alternation, it might make a complete tour impossible.For instance, if the number of blocked squares of one color is significantly higher than the other, the knight might not be able to alternate properly, leading to an imbalance that prevents visiting all squares.But since the board is infinite, the overall balance of colors remains equal, even with periodic blocking. So, maybe this isn't an issue.Wait, but locally, in any finite section, the number of blocked squares of each color might differ, which could cause problems for the knight's alternation.I'm getting stuck here. Maybe I should look for patterns or try to construct a possible tour.Let me try to imagine the knight moving in a specific pattern. Suppose it moves in a way that it always skips over the blocked squares. For example, if it encounters a blocked square, it takes a detour around it.But with an infinite number of blocked squares, the knight would have to make an infinite number of detours, which might not be feasible in a single tour.Alternatively, maybe the knight can follow a spiral pattern that naturally avoids the blocked squares. But I'm not sure how to formalize that.Another idea is to consider the problem in terms of tiling the board with knight's moves, avoiding the blocked squares. If the knight can tile the board in such a way that it covers every square exactly once, then a tour is possible.But tiling an infinite board is non-trivial, especially with obstacles.I think I need to approach this more systematically. Let's consider the coordinates again. The knight moves from (x, y) to (x ± 1, y ± 2) or (x ± 2, y ± 1). The blocked squares are at (3k, 3m).If I can find a way for the knight to move such that it never lands on a blocked square and can reach any other square, then a tour is possible.But how can I ensure that? Maybe by defining a path that systematically covers all squares while avoiding the blocked ones.Alternatively, perhaps I can use a recursive approach, where the knight covers a section of the board, then moves to the next section, and so on, ensuring it doesn't get trapped.But I'm not sure how to implement this idea.Wait, maybe I can use the fact that the knight can move in both positive and negative directions. So, even if it encounters a blocked square in one direction, it can turn around and go the other way.But with an infinite board, the knight has to keep moving forward without getting stuck in a loop.I'm starting to think that it might not be possible for the knight to tour the entire board under these conditions. The periodic placement of blocked squares might create insurmountable barriers that prevent the knight from reaching certain areas.But I need to find a concrete argument or proof to support this intuition.Perhaps I can consider the problem in terms of graph connectivity. If the graph of available squares is disconnected due to the blocked squares, then a complete tour is impossible.But with the knight's ability to jump over squares, it's not clear if the graph is disconnected. The knight can jump over blocked squares, so it might still be connected.Wait, but the knight can't land on blocked squares, but it can jump over them. So, the graph of available squares is still connected because the knight can move around the blocked squares.Hmm, that suggests that a tour might still be possible.But I'm not sure. Maybe the periodicity of the blocked squares creates some kind of periodicity in the graph that makes a complete tour impossible.I think I need to look for a contradiction or a specific property that makes a complete tour impossible.Perhaps considering the number of available squares modulo some number. For example, if the number of available squares is not compatible with the knight's movement, it might prevent a complete tour.But since the board is infinite, the number of available squares is also infinite, so modular arguments might not apply.Alternatively, maybe I can use a parity argument. If the knight alternates between two types of squares, and the blocked squares disrupt this alternation in a way that makes it impossible to cover all squares, then a tour is impossible.But I'm not sure how to formalize this.Wait, let's think about the coordinates again. If I color the squares based on their coordinates modulo 3, I can see how the knight's movement affects this.For example, if a square is at (x, y), its color modulo 3 would be (x mod 3, y mod 3). The knight's move changes the coordinates by (±1, ±2) or (±2, ±1), so the modulo 3 coordinates change accordingly.If I can show that the knight is confined to certain modulo classes, it might prevent it from reaching all squares.But I'm not sure. The knight can reach various modulo classes through its moves, so it might still be able to cover all squares.I'm going in circles here. Maybe I should try to find a specific example or a pattern that shows it's impossible.Alternatively, maybe I can use the fact that the knight's graph is bipartite, with two sets of squares (like black and white), and the blocked squares might disrupt this bipartition in a way that makes a complete tour impossible.But since the blocked squares are placed periodically, they might not disrupt the bipartition globally, even if they do locally.I think I need to conclude that it's impossible for the knight to tour the entire board under these conditions, but I'm not entirely sure. Maybe I can look for similar problems or research to back this up.After some thought, I believe that the periodic placement of blocked squares creates insurmountable barriers for the knight, making a complete tour impossible. The knight cannot navigate around all the blocked squares without getting stuck or missing some areas of the board.So, my final answer is that it is not possible for the knight to travel through the remaining part of the board, visiting each square exactly once.

💡Okay, so I have this function f(x) defined piecewise. For x less than or equal to 0, it's x squared plus 2x. But for x greater than 0, it's defined recursively as f(x - 1) plus 1. Hmm, that's interesting. So, for positive x, the function depends on its value at x - 1, which is one unit to the left. I think this means the function is built up incrementally for each interval [n, n+1) where n is a non-negative integer.The problem is asking me to find the sum of all solutions to the equation f(x) = x - 1/5 for x in the interval [0, 100]. So, I need to figure out where f(x) intersects the line y = x - 1/5 between 0 and 100 and then add up all those x-values.Let me start by understanding how f(x) behaves for positive x. Since it's defined recursively, maybe I can find a pattern or an explicit formula for f(x) in each interval [n, n+1). Let's try to compute f(x) for x in [0,1), [1,2), [2,3), and so on.Starting with x in [0,1): Here, x is greater than 0, so f(x) = f(x - 1) + 1. But x - 1 is in [-1,0), which is less than or equal to 0. So, f(x - 1) is (x - 1)^2 + 2(x - 1). Therefore, f(x) = (x - 1)^2 + 2(x - 1) + 1.Let me simplify that: (x - 1)^2 is x^2 - 2x + 1, and 2(x - 1) is 2x - 2. Adding them together: x^2 - 2x + 1 + 2x - 2 + 1. The -2x and +2x cancel out, and 1 - 2 + 1 is 0. So, f(x) = x^2 for x in [0,1). That's nice, it simplifies to x squared.Now, moving on to x in [1,2): Here, f(x) = f(x - 1) + 1. Since x - 1 is in [0,1), we already know f(x - 1) is (x - 1)^2. So, f(x) = (x - 1)^2 + 1.Similarly, for x in [2,3): f(x) = f(x - 1) + 1. x - 1 is in [1,2), so f(x - 1) is (x - 2)^2 + 1. Therefore, f(x) = (x - 2)^2 + 1 + 1 = (x - 2)^2 + 2.Wait, I see a pattern here. For each interval [n, n+1), where n is a non-negative integer, f(x) seems to be (x - n)^2 + n. Let me test this for n=3. For x in [3,4), f(x) = f(x - 1) + 1. x - 1 is in [2,3), so f(x - 1) is (x - 3)^2 + 2. Therefore, f(x) = (x - 3)^2 + 2 + 1 = (x - 3)^2 + 3. Yep, that fits the pattern.So, in general, for x in [n, n+1), f(x) = (x - n)^2 + n. That makes sense because each time we increment n, we add 1 to the previous function's value.Now, I need to solve f(x) = x - 1/5 for each interval [n, n+1) where n ranges from 0 to 99 (since x is in [0,100]). For each interval, I'll set up the equation (x - n)^2 + n = x - 1/5 and solve for x, then sum all the solutions.Let me write the equation for a general n:(x - n)^2 + n = x - 1/5Expanding the left side: (x^2 - 2nx + n^2) + n = x - 1/5Simplify: x^2 - 2nx + n^2 + n = x - 1/5Bring all terms to one side: x^2 - 2nx + n^2 + n - x + 1/5 = 0Combine like terms: x^2 - (2n + 1)x + (n^2 + n + 1/5) = 0This is a quadratic equation in x. Let me write it as:x^2 - (2n + 1)x + (n^2 + n + 1/5) = 0To find the solutions, I can use the quadratic formula:x = [ (2n + 1) ± sqrt( (2n + 1)^2 - 4*(n^2 + n + 1/5) ) ] / 2Let me compute the discriminant D:D = (2n + 1)^2 - 4*(n^2 + n + 1/5)Expand (2n + 1)^2: 4n^2 + 4n + 1Subtract 4*(n^2 + n + 1/5): 4n^2 + 4n + 1 - 4n^2 - 4n - 4/5Simplify: (4n^2 - 4n^2) + (4n - 4n) + (1 - 4/5) = 0 + 0 + 1/5 = 1/5So, D = 1/5, which is positive, meaning two real solutions for each n.Therefore, the solutions are:x = [ (2n + 1) ± sqrt(1/5) ] / 2Simplify sqrt(1/5): that's 1/sqrt(5) or sqrt(5)/5.So,x = [ (2n + 1) ± sqrt(5)/5 ] / 2Which can be written as:x = (2n + 1)/2 ± sqrt(5)/10Simplify further:x = n + 1/2 ± sqrt(5)/10So, the two solutions in each interval [n, n+1) are:x1 = n + 1/2 + sqrt(5)/10x2 = n + 1/2 - sqrt(5)/10Now, I need to check if these solutions lie within the interval [n, n+1). Let's compute the numerical values of sqrt(5)/10 and 1/2.sqrt(5) is approximately 2.236, so sqrt(5)/10 ≈ 0.2236.1/2 is 0.5.So, x1 = n + 0.5 + 0.2236 ≈ n + 0.7236x2 = n + 0.5 - 0.2236 ≈ n + 0.2764Since n is an integer, x1 is approximately n + 0.7236, which is less than n + 1, so it's within [n, n+1).Similarly, x2 is approximately n + 0.2764, which is greater than n, so it's also within [n, n+1).Therefore, both solutions are valid in each interval.Now, the sum of the solutions for each n is x1 + x2 = [n + 1/2 + sqrt(5)/10] + [n + 1/2 - sqrt(5)/10] = 2n + 1.So, for each n from 0 to 99, the sum of solutions in [n, n+1) is 2n + 1.Therefore, the total sum over all intervals [0,1), [1,2), ..., [99,100] is the sum from n=0 to n=99 of (2n + 1).Let me compute this sum.First, note that the sum of (2n + 1) from n=0 to n=99 is equal to 2*sum(n from 0 to 99) + sum(1 from n=0 to 99).Compute sum(n from 0 to 99): that's (99*100)/2 = 4950.Compute sum(1 from n=0 to 99): that's 100 terms of 1, so 100.Therefore, the total sum is 2*4950 + 100 = 9900 + 100 = 10,000.Wait, but let me double-check that.Sum from n=0 to n=99 of (2n + 1) = 2*sum(n=0 to 99 n) + sum(n=0 to 99 1)sum(n=0 to 99 n) = (99)(100)/2 = 4950sum(n=0 to 99 1) = 100So, 2*4950 = 99009900 + 100 = 10,000Yes, that seems correct.Therefore, the sum of all solutions in [0,100] is 10,000.I think that's the answer. Let me just recap to make sure I didn't miss anything.- For each interval [n, n+1), f(x) = (x - n)^2 + n.- Set this equal to x - 1/5, solve quadratic, get two solutions per interval.- Sum of solutions per interval is 2n + 1.- Sum over n=0 to 99: sum(2n + 1) = 10,000.Yes, that seems consistent.

💡First, I notice that both terms in the expression ( x^3 - x ) have a common factor of ( x ). I'll factor out the ( x ) to simplify the expression.After factoring out ( x ), the expression becomes ( x(x^2 - 1) ). Now, I observe that ( x^2 - 1 ) is a difference of squares, which can be factored further.Finally, I factor ( x^2 - 1 ) into ( (x + 1)(x - 1) ), resulting in the complete factorization of the original expression.

💡Alright, so I've got this problem about parametric equations and converting them to Cartesian form, and then finding intersection points with another curve given in polar coordinates. Let me try to break it down step by step.First, part (1) asks for the Cartesian equation of curve ( C_1 ). The parametric equations are given as:[begin{cases}x = dfrac{2 + t}{6}, y = sqrt{t}end{cases}]where ( t ) is the parameter. Okay, so I need to eliminate the parameter ( t ) to get a relation directly between ( x ) and ( y ).Looking at the first equation, I can solve for ( t ):[x = dfrac{2 + t}{6} implies 6x = 2 + t implies t = 6x - 2]So, ( t ) is expressed in terms of ( x ). Now, looking at the second equation, ( y = sqrt{t} ). Since ( t = 6x - 2 ), I can substitute that into the equation for ( y ):[y = sqrt{6x - 2}]To get rid of the square root, I'll square both sides:[y^2 = 6x - 2]But wait, since ( y = sqrt{t} ), ( y ) must be non-negative. So, the Cartesian equation is:[y^2 = 6x - 2 quad text{with} quad y geq 0]Alright, that seems straightforward. So, part (1) is done.Moving on to part (2). We need to find the intersection points of ( C_3 ) with both ( C_1 ) and ( C_2 ). The polar equation of ( C_3 ) is given as:[2costheta - sintheta = 0]First, I should convert this polar equation to Cartesian coordinates because the other curves are already in Cartesian form.Recall that in polar coordinates, ( x = rcostheta ) and ( y = rsintheta ). So, let's rewrite the equation:[2costheta - sintheta = 0]Multiply both sides by ( r ) to convert to Cartesian:[2rcostheta - rsintheta = 0 implies 2x - y = 0]So, the Cartesian equation of ( C_3 ) is ( 2x - y = 0 ) or ( y = 2x ).Now, to find the intersection points with ( C_1 ), I need to solve the system:[begin{cases}y = 2x, y^2 = 6x - 2end{cases}]Substitute ( y = 2x ) into the second equation:[(2x)^2 = 6x - 2 implies 4x^2 = 6x - 2]Bring all terms to one side:[4x^2 - 6x + 2 = 0]This is a quadratic equation in ( x ). Let's solve it using the quadratic formula:[x = dfrac{6 pm sqrt{(-6)^2 - 4 cdot 4 cdot 2}}{2 cdot 4} = dfrac{6 pm sqrt{36 - 32}}{8} = dfrac{6 pm sqrt{4}}{8} = dfrac{6 pm 2}{8}]So, the solutions are:[x = dfrac{6 + 2}{8} = 1 quad text{and} quad x = dfrac{6 - 2}{8} = dfrac{1}{2}]Now, find the corresponding ( y ) values using ( y = 2x ):- For ( x = 1 ), ( y = 2 cdot 1 = 2 )- For ( x = dfrac{1}{2} ), ( y = 2 cdot dfrac{1}{2} = 1 )So, the intersection points with ( C_1 ) are ( (1, 2) ) and ( left( dfrac{1}{2}, 1 right) ).Now, onto finding the intersection points with ( C_2 ). The parametric equations for ( C_2 ) are:[begin{cases}x = -dfrac{2 + s}{6}, y = -sqrt{s}end{cases}]where ( s ) is the parameter. Similar to part (1), let's eliminate the parameter ( s ).From the first equation:[x = -dfrac{2 + s}{6} implies 6x = -2 - s implies s = -6x - 2]From the second equation:[y = -sqrt{s} implies y = -sqrt{-6x - 2}]Again, squaring both sides to eliminate the square root:[y^2 = (-6x - 2) implies y^2 = -6x - 2]But since ( y = -sqrt{s} ), ( y ) must be non-positive. So, the Cartesian equation of ( C_2 ) is:[y^2 = -6x - 2 quad text{with} quad y leq 0]Now, to find the intersection points with ( C_3 ), which is ( y = 2x ), we set up the system:[begin{cases}y = 2x, y^2 = -6x - 2end{cases}]Substitute ( y = 2x ) into the second equation:[(2x)^2 = -6x - 2 implies 4x^2 = -6x - 2]Bring all terms to one side:[4x^2 + 6x + 2 = 0]Again, solving this quadratic equation using the quadratic formula:[x = dfrac{-6 pm sqrt{6^2 - 4 cdot 4 cdot 2}}{2 cdot 4} = dfrac{-6 pm sqrt{36 - 32}}{8} = dfrac{-6 pm sqrt{4}}{8} = dfrac{-6 pm 2}{8}]So, the solutions are:[x = dfrac{-6 + 2}{8} = -dfrac{1}{2} quad text{and} quad x = dfrac{-6 - 2}{8} = -1]Now, find the corresponding ( y ) values using ( y = 2x ):- For ( x = -dfrac{1}{2} ), ( y = 2 cdot left( -dfrac{1}{2} right) = -1 )- For ( x = -1 ), ( y = 2 cdot (-1) = -2 )So, the intersection points with ( C_2 ) are ( left( -dfrac{1}{2}, -1 right) ) and ( (-1, -2) ).Wait a second, let me double-check these results. For ( C_2 ), the parametric equations give ( y = -sqrt{s} ), so ( y ) should be non-positive, which matches our solutions. Also, substituting back into the equations seems to hold. For ( C_1 ), ( y ) is non-negative, which also matches. So, I think these are correct.Just to make sure, let me plug the points back into the original equations.For ( C_1 ) and ( C_3 ):- Point ( (1, 2) ): - ( y = 2x ) gives ( 2 = 2 cdot 1 ), which is true. - ( y^2 = 6x - 2 ) gives ( 4 = 6 cdot 1 - 2 = 4 ), which is true.- Point ( left( dfrac{1}{2}, 1 right) ): - ( y = 2x ) gives ( 1 = 2 cdot dfrac{1}{2} = 1 ), which is true. - ( y^2 = 6x - 2 ) gives ( 1 = 6 cdot dfrac{1}{2} - 2 = 3 - 2 = 1 ), which is true.For ( C_2 ) and ( C_3 ):- Point ( left( -dfrac{1}{2}, -1 right) ): - ( y = 2x ) gives ( -1 = 2 cdot left( -dfrac{1}{2} right) = -1 ), which is true. - ( y^2 = -6x - 2 ) gives ( 1 = -6 cdot left( -dfrac{1}{2} right) - 2 = 3 - 2 = 1 ), which is true.- Point ( (-1, -2) ): - ( y = 2x ) gives ( -2 = 2 cdot (-1) = -2 ), which is true. - ( y^2 = -6x - 2 ) gives ( 4 = -6 cdot (-1) - 2 = 6 - 2 = 4 ), which is true.Everything checks out. So, I think I've got the correct intersection points.**Final Answer**(1) The Cartesian equation of ( C_1 ) is (boxed{y^2 = 6x - 2 quad (y geq 0)}).(2) The Cartesian coordinates of the intersection points are:- With ( C_1 ): (boxed{left( dfrac{1}{2}, 1 right)}) and (boxed{(1, 2)}).- With ( C_2 ): (boxed{left( -dfrac{1}{2}, -1 right)}) and (boxed{(-1, -2)}).