answer：First, I need to find the slope of the line passing through the points (3, 0) and (4, √3). The formula for the slope ( k ) between two points ((x_1, y_1)) and ((x_2, y_2)) is:[k = frac{y_2 - y_1}{x_2 - x_1}]Substituting the given points:[k = frac{sqrt{3} - 0}{4 - 3} = sqrt{3}]Next, I know that the slope ( k ) is related to the slope angle ( alpha ) by the equation:[k = tan(alpha)]So,[tan(alpha) = sqrt{3}]I recall that ( tan(60^circ) = sqrt{3} ). Therefore, the slope angle ( alpha ) is:[alpha = 60^circ]Finally, I compare this result with the given options and select the correct one.

answer：Alright, so I need to find the distribution function ( F(x) ) of a random variable ( X ) that's normally distributed. Hmm, okay, I remember that the normal distribution is also called the Gaussian distribution, right? It has that familiar bell-shaped curve. First, I think the distribution function ( F(x) ) refers to the cumulative distribution function (CDF), which gives the probability that ( X ) takes on a value less than or equal to ( x ). So, ( F(x) = P(X leq x) ). I recall that the probability density function (PDF) of a normal distribution is given by:[f(x) = frac{1}{sigma sqrt{2pi}} e^{-frac{(x - mu)^2}{2sigma^2}}]where ( mu ) is the mean and ( sigma ) is the standard deviation. To find the CDF ( F(x) ), I need to integrate the PDF from negative infinity up to ( x ):[F(x) = int_{-infty}^{x} f(t) , dt = int_{-infty}^{x} frac{1}{sigma sqrt{2pi}} e^{-frac{(t - mu)^2}{2sigma^2}} , dt]This integral doesn't have a closed-form solution in terms of elementary functions, which I remember is why we often use tables or computational tools to find values of the CDF for the normal distribution. But maybe I can express it in terms of the error function, ( text{erf}(z) ), which is related to the integral of the Gaussian function. The error function is defined as:[text{erf}(z) = frac{2}{sqrt{pi}} int_{0}^{z} e^{-t^2} , dt]So, perhaps I can manipulate the integral for ( F(x) ) to express it in terms of ( text{erf} ).Let me try a substitution to simplify the integral. Let’s set:[u = frac{t - mu}{sigma}]Then, ( t = mu + sigma u ) and ( dt = sigma , du ). Substituting these into the integral:[F(x) = int_{-infty}^{frac{x - mu}{sigma}} frac{1}{sigma sqrt{2pi}} e^{-frac{u^2}{2}} sigma , du = frac{1}{sqrt{2pi}} int_{-infty}^{frac{x - mu}{sigma}} e^{-frac{u^2}{2}} , du]This simplifies to:[F(x) = frac{1}{sqrt{2pi}} int_{-infty}^{z} e^{-frac{u^2}{2}} , du]where ( z = frac{x - mu}{sigma} ).Now, I can split the integral into two parts:[F(x) = frac{1}{sqrt{2pi}} left( int_{-infty}^{0} e^{-frac{u^2}{2}} , du + int_{0}^{z} e^{-frac{u^2}{2}} , du right)]I know that:[int_{-infty}^{0} e^{-frac{u^2}{2}} , du = frac{sqrt{2pi}}{2}]So, substituting this back in:[F(x) = frac{1}{sqrt{2pi}} cdot frac{sqrt{2pi}}{2} + frac{1}{sqrt{2pi}} int_{0}^{z} e^{-frac{u^2}{2}} , du = frac{1}{2} + frac{1}{sqrt{2pi}} int_{0}^{z} e^{-frac{u^2}{2}} , du]Now, relating this to the error function, I recall that:[text{erf}(z) = frac{2}{sqrt{pi}} int_{0}^{z} e^{-t^2} , dt]So, if I let ( t = frac{u}{sqrt{2}} ), then ( u = sqrt{2} t ) and ( du = sqrt{2} , dt ). Substituting these into the integral:[int_{0}^{z} e^{-frac{u^2}{2}} , du = sqrt{2} int_{0}^{frac{z}{sqrt{2}}} e^{-t^2} , dt = frac{sqrt{pi}}{2} text{erf}left( frac{z}{sqrt{2}} right)]Therefore:[F(x) = frac{1}{2} + frac{1}{sqrt{2pi}} cdot frac{sqrt{pi}}{2} text{erf}left( frac{z}{sqrt{2}} right) = frac{1}{2} + frac{1}{2} text{erf}left( frac{z}{sqrt{2}} right)]Substituting back ( z = frac{x - mu}{sigma} ):[F(x) = frac{1}{2} + frac{1}{2} text{erf}left( frac{x - mu}{sigma sqrt{2}} right)]Alternatively, I can express this in terms of the standard normal distribution function ( Phi(z) ), which is defined as:[Phi(z) = frac{1}{2} left( 1 + text{erf}left( frac{z}{sqrt{2}} right) right)]So, substituting back:[F(x) = Phileft( frac{x - mu}{sigma} right)]This makes sense because ( Phi(z) ) is the CDF of the standard normal distribution, and by standardizing ( X ) with ( Z = frac{X - mu}{sigma} ), we can use ( Phi(z) ) to find probabilities for any normal distribution.Wait, but earlier I had an expression involving the error function. Is there a discrepancy here? Let me check.I have:[F(x) = frac{1}{2} + frac{1}{2} text{erf}left( frac{x - mu}{sigma sqrt{2}} right)]And:[Phileft( frac{x - mu}{sigma} right) = frac{1}{2} left( 1 + text{erf}left( frac{(x - mu)/sigma}{sqrt{2}} right) right) = frac{1}{2} + frac{1}{2} text{erf}left( frac{x - mu}{sigma sqrt{2}} right)]Yes, they are the same. So, both expressions are equivalent. Therefore, the CDF of a normal distribution can be expressed either in terms of the error function or using the standard normal CDF ( Phi(z) ).In summary, the distribution function ( F(x) ) for a normally distributed random variable ( X ) with mean ( mu ) and standard deviation ( sigma ) is:[F(x) = Phileft( frac{x - mu}{sigma} right)]where ( Phi(z) ) is the CDF of the standard normal distribution.I think I've covered all the steps, starting from the definition of the CDF, integrating the PDF, using substitution to simplify the integral, relating it to the error function, and finally expressing it in terms of the standard normal CDF. It makes sense now why the normal distribution's CDF doesn't have a simple closed-form expression but can be expressed using known functions like the error function or the standard normal CDF.

answer：Alright, so I have this problem: I need to prove that the segment [0,1] is equinumerous to the half-open interval [0,1). Hmm, okay. Equinumerous means that there's a bijection between them, right? So, I need to find a one-to-one and onto function that maps every element in [0,1] to [0,1) and vice versa.First, let me recall what [0,1] and [0,1) are. [0,1] includes all real numbers from 0 to 1, including both endpoints. [0,1), on the other hand, includes all real numbers from 0 up to but not including 1. So, the main difference is that [0,1] has the point 1, while [0,1) doesn't.Since both intervals are uncountable, I know they have the same cardinality as the continuum, which suggests they should be equinumerous. But I need to construct an explicit bijection to prove it.One approach I remember is using a sequence to "shift" elements. Maybe I can map the point 1 in [0,1] to some point inside [0,1), and then adjust the rest accordingly. Let me think about how to do that.Suppose I choose a sequence of points in [0,1) that converges to 1. For example, I could take a1 = 1 - 1/2, a2 = 1 - 1/3, a3 = 1 - 1/4, and so on. So, a_n = 1 - 1/(n+1). This sequence approaches 1 as n increases.Now, I can define a function f: [0,1] → [0,1) as follows:- f(1) = a1- For each n ≥ 1, f(an) = a_{n+1}- For all other x in [0,1], f(x) = xLet me check if this works. First, is f injective? Suppose f(x) = f(y). If x and y are both in the sequence {a_n}, then f(x) = a_{n+1} and f(y) = a_{m+1}. If these are equal, then n+1 = m+1, so n = m, meaning x = y. If x or y is 1, then f(1) = a1, which is unique. For other x not in the sequence, f(x) = x, so if f(x) = f(y), then x = y. So, f seems injective.Now, is f surjective? Take any y in [0,1). If y is not in the sequence {a_n}, then y is mapped to itself by f. If y is in the sequence, say y = a_n for some n ≥ 1, then y = a_{n} = f(a_{n-1}) for n ≥ 2, and y = a1 = f(1). So, every element in [0,1) is hit by f. Therefore, f is surjective.Since f is both injective and surjective, it's a bijection. Therefore, [0,1] and [0,1) are equinumerous.Wait, did I cover all cases? Let me double-check. The function f maps 1 to a1, each a_n to a_{n+1}, and leaves everything else fixed. Since the sequence {a_n} approaches 1, but never actually reaches 1, and 1 is mapped into the sequence, it seems like all points are accounted for. Yeah, I think that works.Another thought: could I have used a different sequence? Maybe, as long as the sequence is infinite and converges to 1. The key is to have a way to "shift" the point 1 into the interval [0,1) without missing any points.Also, I should consider if there's a simpler way, but I think this shifting method is pretty standard for showing equinumerosity between similar intervals. It ensures that the extra point in [0,1] (which is 1) is mapped into the interval without disrupting the rest of the mapping.So, in conclusion, by constructing this bijection f, I've shown that [0,1] and [0,1) have the same cardinality. Therefore, they are equinumerous.

answer：First, I need to simplify each square root in the expression. I'll start with the first fraction, (dfrac{sqrt{450}}{sqrt{200}}).I can factor 450 into (2 times 225), and since 225 is a perfect square, (sqrt{450}) simplifies to (15sqrt{2}). Similarly, 200 can be factored into (2 times 100), and (sqrt{200}) simplifies to (10sqrt{2}).Now, dividing these simplified forms, (dfrac{15sqrt{2}}{10sqrt{2}}), the (sqrt{2}) terms cancel out, leaving (dfrac{15}{10}), which simplifies to (dfrac{3}{2}).Next, I'll simplify the second fraction, (dfrac{sqrt{98}}{sqrt{56}}). Factoring 98 into (2 times 49), (sqrt{98}) becomes (7sqrt{2}). For 56, factoring into (2 times 28), (sqrt{56}) simplifies to (2sqrt{14}).Dividing these, (dfrac{7sqrt{2}}{2sqrt{14}}), I can rewrite (sqrt{14}) as (sqrt{2 times 7}), which is (sqrt{2}sqrt{7}). This allows me to cancel out the (sqrt{2}) terms, resulting in (dfrac{7}{2sqrt{7}}).To rationalize the denominator, I multiply the numerator and denominator by (sqrt{7}), giving (dfrac{7sqrt{7}}{14}), which simplifies to (dfrac{sqrt{7}}{2}).Finally, I'll add the two simplified fractions: (dfrac{3}{2} + dfrac{sqrt{7}}{2}). Combining these, I get (dfrac{3 + sqrt{7}}{2}).Thus, the simplified expression is (boxed{dfrac{3 + sqrt{7}}{2}}).

answer：Okay, so I have this problem where I'm given a function f(x) = (x² + a x) e^x + b, and I need to find the values of a and b. Additionally, I have to determine the intervals where the function is increasing or decreasing. The problem also tells me that the tangent line to the function at x = 0 is y = -2x + 1. Alright, let's start with part (1). I remember that the tangent line at a point gives us two pieces of information: the value of the function at that point and the derivative of the function at that point. So, since the tangent line at x = 0 is y = -2x + 1, that means f(0) should be equal to the y-value of the tangent line at x = 0, which is 1. Also, the slope of the tangent line is -2, which means f'(0) should be -2.First, let's find f(0). Plugging x = 0 into f(x):f(0) = (0² + a*0) e^0 + b = (0 + 0) * 1 + b = 0 + b = b.But we know f(0) is supposed to be 1 because the tangent line at x = 0 is y = -2x + 1, which when x = 0, y = 1. So, b must be 1. That's straightforward.Next, let's find f'(x). To find the derivative, I need to use the product rule because f(x) is a product of two functions: (x² + a x) and e^x, plus a constant b. The derivative of a constant is zero, so we can ignore the b when taking the derivative.So, f'(x) = d/dx [(x² + a x) e^x]. Using the product rule, which states that d/dx [u*v] = u'v + uv', where u = (x² + a x) and v = e^x.First, find u':u = x² + a x, so u' = 2x + a.Then, v = e^x, so v' = e^x.Putting it all together:f'(x) = (2x + a) e^x + (x² + a x) e^x.We can factor out e^x:f'(x) = e^x [2x + a + x² + a x].Simplify the terms inside the brackets:2x + a + x² + a x = x² + (2 + a) x + a.So, f'(x) = e^x (x² + (2 + a) x + a).Now, we need to evaluate this derivative at x = 0:f'(0) = e^0 (0² + (2 + a)*0 + a) = 1*(0 + 0 + a) = a.But we know from the tangent line that f'(0) should be -2. So, a must be -2.Alright, so from part (1), I found that a = -2 and b = 1.Now, moving on to part (2), determining the intervals of monotonicity. That means I need to find where the function is increasing or decreasing. To do this, I need to analyze the sign of the derivative f'(x).We already have f'(x) = e^x (x² + (2 + a) x + a). Since we found a = -2, let's substitute that back in:f'(x) = e^x (x² + (2 - 2) x - 2) = e^x (x² + 0 x - 2) = e^x (x² - 2).So, f'(x) = e^x (x² - 2).Now, e^x is always positive for any real x, so the sign of f'(x) depends solely on the quadratic expression (x² - 2).Let's find when (x² - 2) is positive, zero, or negative.First, set x² - 2 = 0:x² = 2 => x = sqrt(2) or x = -sqrt(2).These are the critical points where the derivative is zero, so the function could have local maxima or minima at these points.Now, let's analyze the intervals determined by these critical points: (-∞, -sqrt(2)), (-sqrt(2), sqrt(2)), and (sqrt(2), ∞).Choose test points in each interval to determine the sign of (x² - 2):1. For x < -sqrt(2), let's pick x = -2. Then, (-2)^2 - 2 = 4 - 2 = 2 > 0. So, f'(x) > 0 in this interval, meaning the function is increasing.2. For -sqrt(2) < x < sqrt(2), let's pick x = 0. Then, 0^2 - 2 = -2 < 0. So, f'(x) < 0 in this interval, meaning the function is decreasing.3. For x > sqrt(2), let's pick x = 2. Then, 2^2 - 2 = 4 - 2 = 2 > 0. So, f'(x) > 0 in this interval, meaning the function is increasing.Therefore, the function f(x) is increasing on (-∞, -sqrt(2)) and (sqrt(2), ∞), and decreasing on (-sqrt(2), sqrt(2)).Let me just double-check my work. For part (1), I found a = -2 and b = 1. Plugging these back into f(x):f(x) = (x² - 2x) e^x + 1.Then, f(0) = (0 - 0) e^0 + 1 = 0 + 1 = 1, which matches the tangent line. The derivative f'(x) = e^x (x² - 2), which at x = 0 is e^0 (-2) = -2, which also matches the slope of the tangent line. So, part (1) seems correct.For part (2), I analyzed the derivative and found the critical points at x = ±sqrt(2). Testing intervals around these points showed the function is increasing before -sqrt(2), decreasing between -sqrt(2) and sqrt(2), and increasing after sqrt(2). That seems consistent with the derivative's behavior.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**(1) The values of ( a ) and ( b ) are (boxed{a = -2}) and (boxed{b = 1}).(2) The function ( f(x) ) is increasing on the intervals (boxed{(-infty, -sqrt{2})}) and (boxed{(sqrt{2}, +infty)}), and decreasing on the interval (boxed{(-sqrt{2}, sqrt{2})}).

answer：Okay, so I have this problem where I'm given a real n x n matrix A that satisfies the equation 3A³ = A² + A + I. I need to show that the sequence Aᵏ converges to an idempotent matrix. Hmm, idempotent means that when you square it, you get the same matrix back, right? So, if B is idempotent, then B² = B.Alright, let me start by understanding the given equation: 3A³ = A² + A + I. Maybe I can rearrange this to get a better sense of what's going on. Let me subtract A², A, and I from both sides:3A³ - A² - A - I = 0.So, this is a matrix equation, and it looks like a cubic equation in A. Maybe I can think of this as a polynomial equation that A satisfies. So, the minimal polynomial of A divides this cubic polynomial. That might be useful later.Since we're dealing with matrices, I know that if a matrix satisfies a polynomial equation, its eigenvalues must satisfy the same polynomial equation. So, maybe I can find the eigenvalues of A by solving the scalar equation 3λ³ - λ² - λ - 1 = 0.Let me try to factor this polynomial. Maybe I can use rational root theorem. The possible rational roots are ±1, ±1/3. Let me test λ = 1:3(1)³ - (1)² - 1 - 1 = 3 - 1 - 1 - 1 = 0. Oh, so λ = 1 is a root. That means (λ - 1) is a factor. Let me perform polynomial division or use synthetic division to factor it out.Dividing 3λ³ - λ² - λ - 1 by (λ - 1):Using synthetic division:1 | 3 -1 -1 -1 3 2 1 3 2 1 0So, the polynomial factors as (λ - 1)(3λ² + 2λ + 1). Now, let's solve 3λ² + 2λ + 1 = 0.Using the quadratic formula: λ = [-2 ± sqrt(4 - 12)] / 6 = [-2 ± sqrt(-8)] / 6 = (-2 ± 2i√2)/6 = (-1 ± i√2)/3.So, the eigenvalues of A are 1, (-1 + i√2)/3, and (-1 - i√2)/3.Okay, so A has three eigenvalues: one real eigenvalue 1, and a pair of complex conjugate eigenvalues. Since the eigenvalues are distinct, A is diagonalizable, right? So, A can be written as PDP⁻¹, where D is a diagonal matrix with the eigenvalues on the diagonal.Now, if I consider the powers of A, Aᵏ = PDᵏP⁻¹. So, the behavior of Aᵏ as k increases will depend on the eigenvalues raised to the power k.Looking at the eigenvalues:1. The eigenvalue 1 raised to any power k is still 1.2. The complex eigenvalues are (-1 ± i√2)/3. Let me compute their modulus to see if they are inside or outside the unit circle.The modulus of (-1 + i√2)/3 is sqrt[(-1/3)² + (√2/3)²] = sqrt[(1/9) + (2/9)] = sqrt[3/9] = sqrt(1/3) ≈ 0.577, which is less than 1. Similarly, the modulus of (-1 - i√2)/3 is the same.So, these complex eigenvalues have modulus less than 1. Therefore, as k increases, their k-th powers will tend to zero.Therefore, as k approaches infinity, Dᵏ will approach a diagonal matrix with 1 in the first position and zeros elsewhere. So, Dᵏ tends to diag(1, 0, 0, ..., 0).Therefore, Aᵏ = PDᵏP⁻¹ tends to P diag(1, 0, 0, ..., 0) P⁻¹. Let's denote this limit as B. So, B = P diag(1, 0, 0, ..., 0) P⁻¹.Now, I need to show that B is idempotent, meaning B² = B.Let me compute B²:B² = (P diag(1, 0, 0, ..., 0) P⁻¹)(P diag(1, 0, 0, ..., 0) P⁻¹) = P diag(1, 0, 0, ..., 0) (P⁻¹ P) diag(1, 0, 0, ..., 0) P⁻¹.Since P⁻¹ P is the identity matrix, this simplifies to P diag(1, 0, 0, ..., 0) diag(1, 0, 0, ..., 0) P⁻¹.Multiplying the diagonal matrices: diag(1, 0, 0, ..., 0) diag(1, 0, 0, ..., 0) = diag(1*1, 0*0, 0*0, ..., 0*0) = diag(1, 0, 0, ..., 0).Therefore, B² = P diag(1, 0, 0, ..., 0) P⁻¹ = B. So, B is indeed idempotent.Wait, but in the problem statement, A is an n x n matrix, so the diagonal matrix D would have n eigenvalues, but in my earlier steps, I only found three eigenvalues. That's because I considered the minimal polynomial, which is cubic, but A could be larger. However, the key point is that all other eigenvalues (if any) would also satisfy the equation 3λ³ - λ² - λ - 1 = 0, so they would be among the roots 1, (-1 + i√2)/3, and (-1 - i√2)/3. Therefore, regardless of the size of A, the eigenvalues are only these three, and the same reasoning applies: the eigenvalues with modulus less than 1 will vanish as k increases, leaving only the eigenvalue 1.Hence, the limit matrix B will have 1s corresponding to the eigenspace of eigenvalue 1 and zeros elsewhere, making it idempotent.I think that covers it. So, the sequence Aᵏ converges to an idempotent matrix B, which is the projection onto the eigenspace corresponding to eigenvalue 1.