💡First, I recognize that if ( x + 1 ) and ( 4 - 2x ) are square roots of the same positive number, they must be equal in magnitude but opposite in sign.So, I set up the equation:[x + 1 = -(4 - 2x)]Next, I simplify the equation:[x + 1 = -4 + 2x][1 + 4 = 2x - x][5 = x]With ( x = 5 ), I substitute back to find the square root:[x + 1 = 5 + 1 = 6]Finally, the positive number is the square of this root:[6^2 = 36]

💡First, I recognize that the sum of two numbers is 62, and one of the numbers is 27. To find the other number, I'll set up an equation where the unknown number is represented by ( x ).So, the equation becomes:[ 27 + x = 62 ]Next, I'll solve for ( x ) by subtracting 27 from both sides of the equation:[ x = 62 - 27 ][ x = 35 ]Therefore, the other number is 35.

💡First, I need to simplify the complex number ( Z = frac{2i}{1 - i} ). To do this, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is ( 1 + i ), to eliminate the imaginary part in the denominator.[Z = frac{2i}{1 - i} times frac{1 + i}{1 + i} = frac{2i(1 + i)}{(1 - i)(1 + i)}]Next, I'll expand both the numerator and the denominator:[text{Numerator: } 2i(1 + i) = 2i + 2i^2 = 2i - 2 quad (text{since } i^2 = -1)][text{Denominator: } (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2]So, the simplified form of ( Z ) is:[Z = frac{2i - 2}{2} = -1 + i]Now, to find the modulus of ( Z ), denoted as ( |Z| ), I'll use the formula for the modulus of a complex number ( a + bi ):[|Z| = sqrt{a^2 + b^2}]Here, ( a = -1 ) and ( b = 1 ), so:[|Z| = sqrt{(-1)^2 + 1^2} = sqrt{1 + 1} = sqrt{2}]Therefore, the modulus of ( Z ) is ( sqrt{2} ).

💡Okay, so I have this problem where I need to verify if the function ( y = x^2 + x + C ) is a solution to the differential equation ( dy = (2x + 1) dx ). Hmm, let me think about how to approach this.First, I remember that to verify if a function is a solution to a differential equation, I need to compute the derivative of the function and see if it matches the differential equation given. So, the function is ( y = x^2 + x + C ). I need to find ( dy ) in terms of ( dx ) and see if it equals ( (2x + 1) dx ).Alright, let's start by finding the derivative of ( y ) with respect to ( x ). The derivative of ( x^2 ) is ( 2x ), the derivative of ( x ) is ( 1 ), and the derivative of a constant ( C ) is ( 0 ). So, putting that all together, the derivative ( frac{dy}{dx} ) should be ( 2x + 1 ).Now, if I have ( frac{dy}{dx} = 2x + 1 ), then multiplying both sides by ( dx ) should give me ( dy = (2x + 1) dx ). That seems to match exactly with the differential equation provided. So, it looks like the function ( y = x^2 + x + C ) does satisfy the differential equation.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting with ( y = x^2 + x + C ), taking the derivative term by term: ( 2x ) from ( x^2 ), ( 1 ) from ( x ), and ( 0 ) from ( C ). Yep, that gives ( 2x + 1 ). Multiplying by ( dx ) gives ( dy = (2x + 1) dx ), which is exactly what the differential equation says. Okay, that seems correct.I guess another way to think about it is integrating both sides of the differential equation. If I have ( dy = (2x + 1) dx ), integrating both sides should give me back the original function ( y ). Let's try that.Integrating ( dy ) gives ( y ) plus a constant, and integrating ( (2x + 1) dx ) should give ( x^2 + x + C ), which matches the function provided. So, that also confirms that ( y = x^2 + x + C ) is indeed a solution.Hmm, is there anything else I need to consider? Maybe check if there are any restrictions or if the constant ( C ) affects the solution. Well, since ( C ) is just a constant of integration, it doesn't affect the derivative, so it shouldn't matter in this case. The differential equation is satisfied regardless of the value of ( C ).I think I've covered all the bases here. I computed the derivative, checked it against the differential equation, and even thought about integrating back to confirm. Everything seems to line up. So, yeah, I'm pretty confident that ( y = x^2 + x + C ) is a valid solution to the differential equation ( dy = (2x + 1) dx ).**Final Answer**boxed{Yes}

💡Alright, so I've got this probability problem here, and I need to figure out the chance that the sum of the leftmost white point and the rightmost black point is at most 1. Let me try to unpack this step by step.First, there are 2019 points chosen uniformly at random on the interval [0,1]. That means each point has an equal chance of being anywhere between 0 and 1. Then, Tairitsu randomly picks 1000 of these points to color black, leaving the remaining 1019 points white. Hikari then looks at the leftmost white point and the rightmost black point and adds their positions together. We need to find the probability that this sum is ≤ 1.Okay, so let's break it down. Since all points are chosen uniformly, their positions are independent and identically distributed. The key here is that the coloring is done randomly after the points are chosen. So, the black and white points are randomly selected from the 2019 points.I think it might help to consider the order statistics here. Order statistics deal with the positions of ordered points. For example, the leftmost white point would be the minimum of the white points, and the rightmost black point would be the maximum of the black points.Let me denote the positions of the white points as ( W_1, W_2, ldots, W_{1019} ) and the positions of the black points as ( B_1, B_2, ldots, B_{1000} ). Then, the leftmost white point is ( W_{(1)} = min{W_1, W_2, ldots, W_{1019}} ) and the rightmost black point is ( B_{(1000)} = max{B_1, B_2, ldots, B_{1000}} ).We need to find the probability that ( W_{(1)} + B_{(1000)} leq 1 ).Hmm, this seems a bit tricky. Maybe I can think about the joint distribution of ( W_{(1)} ) and ( B_{(1000)} ). But that might be complicated because these two statistics are not independent; they're both derived from the same set of points.Wait, maybe there's a smarter way to approach this. Since all points are randomly colored, perhaps the problem has some symmetry that I can exploit.Let me consider the entire set of 2019 points. The leftmost point among all points is either white or black. Similarly, the rightmost point is either white or black. But in our case, we're dealing with the leftmost white and the rightmost black.I wonder if there's a way to relate ( W_{(1)} ) and ( B_{(1000)} ) to the overall order statistics of all 2019 points.Let me denote the order statistics of all 2019 points as ( X_{(1)}, X_{(2)}, ldots, X_{(2019)} ), where ( X_{(1)} ) is the leftmost point and ( X_{(2019)} ) is the rightmost point.Now, ( W_{(1)} ) is the minimum of the white points, which is essentially the minimum of a subset of these 2019 points. Similarly, ( B_{(1000)} ) is the maximum of the black points, which is the maximum of another subset.But how does this help me? Maybe I can think about the positions of these minima and maxima relative to each other.Another thought: since the coloring is random, the probability that a particular point is white or black is independent of its position. So, the color of each point doesn't depend on where it is on the interval [0,1].Wait, that might not be entirely true. The coloring is done after selecting the points, so the positions are fixed, and then colors are assigned randomly. So, the color assignment is indeed independent of the positions.Given that, perhaps I can model the problem as follows: for each point, it's white with probability ( frac{1019}{2019} ) and black with probability ( frac{1000}{2019} ). But actually, since the coloring is done without replacement, it's more like a hypergeometric distribution rather than independent Bernoulli trials.But maybe for large numbers, the difference isn't too significant. Hmm, not sure if that's helpful.Let me think about the event ( W_{(1)} + B_{(1000)} leq 1 ). What does this mean geometrically? It means that the leftmost white point is to the left of ( 1 - B_{(1000)} ). Or, equivalently, the rightmost black point is to the left of ( 1 - W_{(1)} ).Wait, that might not be the most straightforward way to think about it. Maybe it's better to consider the positions of all points and see how the white and black points are distributed.Another approach: consider that the sum ( W_{(1)} + B_{(1000)} leq 1 ) implies that the leftmost white point is less than or equal to ( 1 - B_{(1000)} ). So, for this to happen, the leftmost white point must be less than or equal to ( 1 - B_{(1000)} ).But how do I relate ( W_{(1)} ) and ( B_{(1000)} )?Maybe I can think about the positions in terms of spacings. The interval [0,1] is divided into segments by the points. The spacings between consecutive points are important in order statistics.But I'm not sure if that directly helps here.Wait, maybe I can use the concept of inclusion. The event ( W_{(1)} + B_{(1000)} leq 1 ) can be thought of as the union of two events: either the leftmost white point is very far to the left, or the rightmost black point is very far to the left, or both.But that might not be precise.Alternatively, perhaps I can model this as a race between the leftmost white and the rightmost black. If the leftmost white is to the left of ( 1 - B_{(1000)} ), then the sum is ≤1.Wait, maybe it's better to consider the joint distribution of ( W_{(1)} ) and ( B_{(1000)} ). Since all points are independent, the joint distribution can be expressed as the product of their individual distributions, but I need to account for the fact that they are dependent because they are from the same set.Alternatively, maybe I can use the fact that the leftmost white and the rightmost black are both order statistics from the same sample.Wait, another idea: since the coloring is random, the probability that a particular point is white or black is independent of its position. So, perhaps I can think of the entire process as randomly assigning colors to the points and then looking at the leftmost white and rightmost black.In that case, the probability that the leftmost white is at position ( w ) and the rightmost black is at position ( b ) is some joint probability, and we need to integrate over all ( w ) and ( b ) such that ( w + b leq 1 ).But that seems complicated.Wait, maybe I can use the fact that the leftmost white is the minimum of 1019 points, and the rightmost black is the maximum of 1000 points. So, their distributions are known.The distribution of the minimum of ( n ) uniform points is ( f_{W_{(1)}}(w) = n(1 - w)^{n - 1} ).Similarly, the distribution of the maximum of ( m ) uniform points is ( f_{B_{(1000)}}(b) = m b^{m - 1} ).But since ( W_{(1)} ) and ( B_{(1000)} ) are dependent, their joint distribution isn't just the product of their marginals.Hmm, this is getting tricky. Maybe there's a smarter way.Wait, perhaps I can think about the problem in terms of the relative ordering of the leftmost white and the rightmost black.If the leftmost white is to the left of the rightmost black, then their sum could be ≤1 depending on their positions.But actually, the sum being ≤1 doesn't directly correspond to their relative order, but rather to their specific positions.Wait, another thought: if I consider the entire interval [0,1], and I want ( W_{(1)} + B_{(1000)} leq 1 ), that's equivalent to ( W_{(1)} leq 1 - B_{(1000)} ).So, the leftmost white must be to the left of ( 1 - B_{(1000)} ).But ( 1 - B_{(1000)} ) is like a reflection of the rightmost black point.Wait, maybe I can use a symmetry argument here.Suppose I reflect the entire interval [0,1] around the point 0.5, so that 0 becomes 1, 1 becomes 0, and so on. Then, the leftmost white point would correspond to the rightmost white point in the reflected interval, and the rightmost black point would correspond to the leftmost black point in the reflected interval.But I'm not sure if that helps.Alternatively, maybe I can think about the problem as a competition between the white and black points. The leftmost white and the rightmost black are competing to see who is closer to certain points.Wait, perhaps it's better to think about the probability that ( W_{(1)} leq 1 - B_{(1000)} ).So, I need to find ( P(W_{(1)} leq 1 - B_{(1000)}) ).But how do I compute this probability?Maybe I can use the law of total probability. For a fixed ( B_{(1000)} = b ), the probability that ( W_{(1)} leq 1 - b ) is equal to the probability that at least one white point is ≤ ( 1 - b ).Since ( W_{(1)} ) is the minimum of 1019 points, the probability that ( W_{(1)} leq 1 - b ) is ( 1 - (1 - (1 - b))^{1019} = 1 - b^{1019} ).Wait, that seems right. Because the probability that all white points are > ( 1 - b ) is ( (1 - (1 - b))^{1019} = b^{1019} ), so the probability that at least one is ≤ ( 1 - b ) is ( 1 - b^{1019} ).So, then, the overall probability is the expectation over ( B_{(1000)} ) of ( 1 - B_{(1000)}^{1019} ).Mathematically, that would be:( P = E[1 - B_{(1000)}^{1019}] = 1 - E[B_{(1000)}^{1019}] ).Now, I need to compute ( E[B_{(1000)}^{1019}] ).The expectation of the maximum of ( m ) uniform points raised to the power ( k ) is given by:( E[B_{(m)}^k] = frac{m}{m + k} ).Wait, is that correct? Let me recall. For the maximum of ( m ) uniform variables, the PDF is ( f_{B_{(m)}}(b) = m b^{m - 1} ).So, ( E[B_{(m)}^k] = int_0^1 b^k m b^{m - 1} db = m int_0^1 b^{m + k - 1} db = m cdot frac{1}{m + k} = frac{m}{m + k} ).Yes, that's correct.So, in our case, ( m = 1000 ) and ( k = 1019 ).Therefore,( E[B_{(1000)}^{1019}] = frac{1000}{1000 + 1019} = frac{1000}{2019} ).Therefore, the probability ( P ) is:( P = 1 - frac{1000}{2019} = frac{2019 - 1000}{2019} = frac{1019}{2019} ).So, the probability that ( W_{(1)} + B_{(1000)} leq 1 ) is ( frac{1019}{2019} ).Wait, that seems surprisingly straightforward. Let me double-check my steps.1. I defined ( W_{(1)} ) and ( B_{(1000)} ) correctly as the minimum of white points and maximum of black points, respectively.2. I recognized that ( P(W_{(1)} + B_{(1000)} leq 1) = P(W_{(1)} leq 1 - B_{(1000)}) ).3. For a fixed ( B_{(1000)} = b ), ( P(W_{(1)} leq 1 - b) = 1 - b^{1019} ).4. Then, I took the expectation over ( B_{(1000)} ), leading to ( 1 - E[B_{(1000)}^{1019}] ).5. Using the formula for the expectation of the maximum raised to a power, I found ( E[B_{(1000)}^{1019}] = frac{1000}{2019} ).6. Subtracting this from 1 gives ( frac{1019}{2019} ).Yes, that seems correct. So, the probability is ( frac{1019}{2019} ).I think that's the answer. It makes sense because there are more white points than black points, so the leftmost white is more likely to be to the left of the rightmost black, making their sum more likely to be ≤1.

💡Okay, so I have this quadratic equation: (k - 1)x² + 2x + 1 = 0. The question is asking about the range of values for k such that the equation has real roots. The options are A, B, C, D, with different ranges involving k ≤ 2, k < 2, and whether k can be 1 or not.First, I remember that for a quadratic equation ax² + bx + c = 0 to have real roots, the discriminant must be greater than or equal to zero. The discriminant is b² - 4ac. So, I need to calculate the discriminant for this equation.In this case, a is (k - 1), b is 2, and c is 1. So, plugging into the discriminant formula: discriminant = (2)² - 4*(k - 1)*(1) = 4 - 4(k - 1). Let me simplify that: 4 - 4k + 4, which is 8 - 4k.Now, for the equation to have real roots, the discriminant must be ≥ 0. So, 8 - 4k ≥ 0. Let's solve for k: 8 ≥ 4k, which simplifies to 2 ≥ k, or k ≤ 2.But wait, there's another condition. Since this is a quadratic equation, the coefficient of x², which is (k - 1), must not be zero. If k - 1 = 0, then k = 1, and the equation becomes linear, not quadratic. So, k cannot be 1.Putting these together, k must be less than or equal to 2, but k cannot be 1. So, the range is k ≤ 2 and k ≠ 1.Looking at the options, option A says k ≤ 2 and k ≠ 1, which matches what I found. Option B says k < 2 and k ≠ 1, but since k can be equal to 2, B is too restrictive. Option C says k ≤ 2, but it doesn't exclude k = 1, which is necessary. Option D says k < 2, which is also too restrictive and doesn't exclude k = 1.So, the correct answer should be A.