💡First, I recognize that if a quadratic equation with real coefficients has a complex root like (-1 + 2i), its complex conjugate (-1 - 2i) must also be a root.Next, I'll use Vieta's formulas to find the coefficients of the quadratic equation. According to Vieta's formulas, the sum of the roots is equal to (-b) and the product of the roots is equal to (c).Calculating the sum of the roots:[(-1 + 2i) + (-1 - 2i) = -2]This sum equals (-b), so:[-b = -2 quad Rightarrow quad b = 2]Calculating the product of the roots:[(-1 + 2i) cdot (-1 - 2i) = (-1)^2 - (2i)^2 = 1 - (-4) = 5]This product equals (c), so:[c = 5]With the values of (b) and (c) determined, the quadratic equation is:[x^2 + 2x + 5 = 0]Finally, comparing this equation with the given options, the correct choice is:[boxed{D}]

💡Okay, so I need to figure out how many sheets of paper come in a pad that Evelyn uses for her work notes. She uses this pad every week, but she takes Monday and Friday off. So, she only works on Tuesday, Wednesday, and Thursday. That means she works three days a week.She uses 12 sheets of paper each day she works. So, if I multiply the number of sheets she uses per day by the number of days she works, I should get the total number of sheets she uses in a week.Let me write that down:Number of sheets per day = 12 sheetsNumber of working days per week = 3 days (Tuesday, Wednesday, Thursday)Total sheets used per week = Number of sheets per day × Number of working days per weekTotal sheets used per week = 12 sheets/day × 3 days/weekNow, doing the multiplication:12 × 3 = 36So, Evelyn uses 36 sheets of paper each week.Therefore, a pad of paper should come with at least 36 sheets to cover her weekly usage.But wait, I should double-check if there's anything else I need to consider. Does the pad have more sheets than she uses in a week? Maybe, but the question is asking how many sheets a pad comes with based on her usage. Since she uses 36 sheets a week, it's logical to assume the pad has 36 sheets.I think that's it. So, the pad comes with 36 sheets.

💡To determine the twentieth number in the base 8 sequence, we need to convert the decimal number 20 into base 8.1. **Divide the number by 8 and record the remainder:** - (20 div 8 = 2) with a remainder of (4). This remainder is the least significant digit (rightmost digit) in base 8. - (2 div 8 = 0) with a remainder of (2). This remainder is the next digit (leftmost digit) in base 8.2. **Write the remainders in reverse order of their computation:** - The digits from the steps above are (2) and (4). - Therefore, (20_{10}) is represented as (24_8) in base 8.3. **Conclusion:** - The twentieth number in the sequence, when counting in base 8, is (24_8).Thus, the final answer is (boxed{textbf{(A)} 24}).

💡Alright, so I've got this problem here, and I need to figure out whether condition α is sufficient, necessary, both, or neither for condition β. Let me try to break this down step by step.First, let's understand what the conditions are:- α: x² = 4- β: x = 2Okay, so α is the equation x squared equals 4, and β is the equation x equals 2. I need to see how these two relate in terms of sufficiency and necessity.Let me recall what these terms mean:- **Sufficient condition**: If α is a sufficient condition for β, then whenever α is true, β must also be true. In other words, α being true guarantees that β is true.- **Necessary condition**: If α is a necessary condition for β, then β can only be true if α is true. That means that for β to hold, α must hold as well.So, I need to check both of these for α and β.Starting with α: x² = 4. Solving this equation, we get x = ±2. So, x can be either 2 or -2.Now, β is x = 2. So, β specifies that x must be exactly 2.Let's see if α is a sufficient condition for β. For α to be sufficient, whenever x² = 4, x must equal 2. But wait, x could also be -2. So, just knowing that x² = 4 doesn't necessarily mean that x is 2; it could be -2. Therefore, α is not a sufficient condition for β because α being true doesn't guarantee that β is true.Next, is α a necessary condition for β? For α to be necessary, if x = 2 (which is β), then x² must equal 4 (which is α). Let's check: If x is 2, then x² is indeed 4. So, β implies α. That means α is a necessary condition for β because β can only be true if α is true.So, putting it all together: α is a necessary condition for β because β can't be true without α being true. However, α isn't sufficient for β because α being true doesn't ensure that β is true—it could be either 2 or -2.Therefore, the answer should be that α is a necessary but not sufficient condition for β.

💡Okay, so I have this function f(x) = x² - (a + 2)x + a ln x. There are two parts to the problem. Let me tackle them one by one.Starting with part (1): When a = 1, find the extreme values of the function f(x).Alright, so first, I need to substitute a = 1 into the function. That gives me f(x) = x² - (1 + 2)x + 1 ln x, which simplifies to f(x) = x² - 3x + ln x.To find the extreme values, I remember that I need to take the derivative of f(x) with respect to x and set it equal to zero. So, let's compute f'(x).The derivative of x² is 2x, the derivative of -3x is -3, and the derivative of ln x is 1/x. So putting it all together, f'(x) = 2x - 3 + 1/x.Now, I need to set this derivative equal to zero to find critical points:2x - 3 + 1/x = 0.Hmm, this is a bit tricky because of the 1/x term. Maybe I can multiply both sides by x to eliminate the denominator. Let's try that:x*(2x - 3 + 1/x) = 0*x2x² - 3x + 1 = 0.Now, that's a quadratic equation in terms of x. Let me solve it using the quadratic formula. The quadratic is 2x² - 3x + 1 = 0.The quadratic formula is x = [ -b ± sqrt(b² - 4ac) ] / (2a). Here, a = 2, b = -3, c = 1.Plugging in the values:x = [ 3 ± sqrt( (-3)² - 4*2*1 ) ] / (2*2)x = [ 3 ± sqrt(9 - 8) ] / 4x = [ 3 ± sqrt(1) ] / 4x = [ 3 ± 1 ] / 4.So, the solutions are:x = (3 + 1)/4 = 4/4 = 1,x = (3 - 1)/4 = 2/4 = 1/2.So, the critical points are at x = 1/2 and x = 1.Now, I need to determine whether these critical points are maxima or minima. For that, I can use the second derivative test or analyze the sign changes of the first derivative.Let me compute the second derivative f''(x). The first derivative is f'(x) = 2x - 3 + 1/x. Taking the derivative again:f''(x) = 2 - 1/x².Now, let's evaluate f''(x) at the critical points.First, at x = 1/2:f''(1/2) = 2 - 1/( (1/2)² ) = 2 - 1/(1/4) = 2 - 4 = -2.Since f''(1/2) is negative, the function is concave down at x = 1/2, which means this is a local maximum.Next, at x = 1:f''(1) = 2 - 1/(1²) = 2 - 1 = 1.Since f''(1) is positive, the function is concave up at x = 1, which means this is a local minimum.So, now I need to find the corresponding f(x) values at these points.First, at x = 1/2:f(1/2) = (1/2)² - 3*(1/2) + ln(1/2)= 1/4 - 3/2 + ln(1/2)= (1/4 - 3/2) + ln(1/2)= (-5/4) + ln(1/2).I can leave it like that, but maybe I can write it as -5/4 + ln(1/2). Alternatively, ln(1/2) is equal to -ln 2, so it's -5/4 - ln 2. Either way is fine.Next, at x = 1:f(1) = (1)² - 3*(1) + ln(1)= 1 - 3 + 0= -2.So, summarizing, at x = 1/2, we have a local maximum of -5/4 + ln(1/2), and at x = 1, a local minimum of -2.I think that's part (1) done. Now, moving on to part (2).Part (2) is a bit more complex. Let me read it again.We have a function y = g(x) defined on domain D. The tangent line at point P(x₀, y₀) is l: y = h(x). If for x ≠ x₀, [g(x) - h(x)] / (x - x₀) > 0 always holds within D, then P is called a "turning point" of the function y = g(x).When a = 8, does the function y = f(x) have a "turning point"? If it exists, find the x-coordinate; if not, explain why.Alright, so first, let's understand what a "turning point" is in this context. It's a point where the function's graph has a tangent line such that for all other points x ≠ x₀ in the domain D, the difference [g(x) - h(x)] divided by (x - x₀) is positive. So, [g(x) - h(x)] / (x - x₀) > 0 for all x ≠ x₀ in D.This seems similar to the concept of convexity or concavity. If this expression is positive, it might mean that the function is above its tangent line on one side and below on the other, but the sign depends on the direction.Wait, actually, let's think about it. If [g(x) - h(x)] / (x - x₀) > 0, then:- If x > x₀, then (x - x₀) is positive, so [g(x) - h(x)] must be positive. So, g(x) > h(x) for x > x₀.- If x < x₀, then (x - x₀) is negative, so [g(x) - h(x)] must be negative to make the whole expression positive. So, g(x) < h(x) for x < x₀.Therefore, the function lies above its tangent line to the right of x₀ and below its tangent line to the left of x₀. This is characteristic of a point where the function changes its concavity, i.e., an inflection point. But wait, an inflection point is where the concavity changes, but here it's defined in terms of the tangent line.Wait, actually, if the function is above the tangent line on one side and below on the other, it's called a "point of inflection" in some contexts, but the definition here is slightly different. It's called a "turning point" if [g(x) - h(x)] / (x - x₀) > 0 for all x ≠ x₀.So, in other words, the function is above the tangent line for x > x₀ and below for x < x₀. So, the tangent line is crossed only once at x₀, and the function doesn't cross it again. So, in terms of convexity, if the function is convex on one side and concave on the other, then it's an inflection point. But here, it's defined as a "turning point".So, in this case, we need to check whether such a point exists for f(x) when a = 8.First, let's write down f(x) when a = 8:f(x) = x² - (8 + 2)x + 8 ln x = x² - 10x + 8 ln x.So, f(x) = x² - 10x + 8 ln x.We need to find if there's a point x₀ such that for all x ≠ x₀, [f(x) - h(x)] / (x - x₀) > 0, where h(x) is the tangent line at x₀.So, let's denote h(x) as the tangent line at x₀. The equation of the tangent line at x₀ is:h(x) = f(x₀) + f'(x₀)(x - x₀).So, h(x) = f(x₀) + f'(x₀)(x - x₀).Then, the expression [f(x) - h(x)] / (x - x₀) simplifies to [f(x) - f(x₀) - f'(x₀)(x - x₀)] / (x - x₀).This is essentially the difference quotient for the derivative at x₀, but in the limit as x approaches x₀, it tends to zero. However, here we're considering x ≠ x₀.So, let's define F(x) = f(x) - h(x). Then, F(x) = f(x) - [f(x₀) + f'(x₀)(x - x₀)].So, F(x) = f(x) - f(x₀) - f'(x₀)(x - x₀).We need to analyze the sign of F(x)/(x - x₀) for x ≠ x₀.Given that [F(x)] / (x - x₀) > 0 for all x ≠ x₀ in D.So, for x > x₀, F(x) > 0, and for x < x₀, F(x) < 0.Which implies that x₀ is a point where the function f(x) crosses its tangent line from below to above as x increases through x₀.This is similar to the concept of a point where the function has a horizontal tangent and is convex on one side and concave on the other. But more precisely, it's about the function being above the tangent line on one side and below on the other.So, to find such a point, we can analyze the behavior of F(x) around x₀.Alternatively, we can consider the derivative of F(x). Let's compute F'(x):F'(x) = f'(x) - f'(x₀).Because h(x) is linear, its derivative is constant, equal to f'(x₀). So, F'(x) = f'(x) - f'(x₀).So, F'(x) = f'(x) - f'(x₀).Therefore, the sign of F'(x) tells us about the increasing or decreasing nature of F(x).If F'(x) is positive, F(x) is increasing; if negative, decreasing.Given that F(x₀) = 0, and we need F(x) to be positive for x > x₀ and negative for x < x₀, the function F(x) must have a minimum at x₀.Wait, no. If F(x) is negative for x < x₀ and positive for x > x₀, then x₀ is a minimum point for F(x). So, F'(x) must be negative before x₀ and positive after x₀, meaning that F'(x) changes sign from negative to positive at x₀, which implies that x₀ is a local minimum of F(x).But F'(x) = f'(x) - f'(x₀). So, for F'(x) to change sign from negative to positive at x₀, f'(x) must be increasing through x₀.In other words, f'(x) must be increasing at x₀, meaning that f''(x₀) > 0.Wait, let me think again.If F'(x) = f'(x) - f'(x₀), then at x = x₀, F'(x₀) = 0.If F'(x) changes sign from negative to positive at x₀, then F(x) has a minimum at x₀, which is consistent with F(x) being negative before x₀ and positive after x₀.Therefore, for F(x) to satisfy [F(x)] / (x - x₀) > 0 for all x ≠ x₀, it's necessary that F(x) has a minimum at x₀, which requires that F'(x) changes sign from negative to positive at x₀.But F'(x) = f'(x) - f'(x₀). So, for F'(x) to change sign from negative to positive at x₀, f'(x) must be increasing through x₀.Which means that f''(x₀) > 0.So, x₀ must be a point where f''(x₀) > 0.Therefore, to find such a point x₀, we need to find a critical point of f(x) where the second derivative is positive.Wait, but hold on. Let me make sure.Wait, actually, the function F(x) is f(x) - h(x). So, F(x) is the difference between f(x) and its tangent line at x₀.If F(x) is positive for x > x₀ and negative for x < x₀, then the graph of f(x) lies above the tangent line to the right of x₀ and below to the left of x₀.This is similar to the concept of convexity. If a function is convex (f''(x) > 0), then the function lies above its tangent lines. If it's concave (f''(x) < 0), it lies below its tangent lines.But in this case, we need the function to lie above the tangent line on one side and below on the other. So, that would require that the function changes its concavity at x₀, i.e., x₀ is an inflection point.Wait, but an inflection point is where the concavity changes. So, if x₀ is an inflection point, then f''(x) changes sign at x₀.But in our case, we need F(x) to be positive for x > x₀ and negative for x < x₀. So, if x₀ is an inflection point, then to the left of x₀, the function is concave (f''(x) < 0), so it lies above its tangent lines, and to the right, it's convex (f''(x) > 0), so it lies below its tangent lines. Wait, that would be the opposite of what we need.Wait, no. Let me think again.If f''(x) > 0, the function is convex, meaning it lies above its tangent lines.If f''(x) < 0, the function is concave, meaning it lies below its tangent lines.So, if x₀ is an inflection point, then to the left of x₀, f''(x) < 0, so the function lies below its tangent lines, and to the right, f''(x) > 0, so it lies above its tangent lines.But in our problem, we need the function to lie above the tangent line for x > x₀ and below for x < x₀. So, that would require that to the right of x₀, the function is convex (f''(x) > 0), and to the left, it's concave (f''(x) < 0). So, x₀ is an inflection point where concavity changes from concave to convex.Therefore, x₀ must be an inflection point of f(x). So, to find such a point, we need to find where f''(x) = 0 and f''(x) changes sign.So, let's compute f''(x) for a = 8.First, f(x) = x² - 10x + 8 ln x.Compute f'(x):f'(x) = 2x - 10 + 8/x.Then, f''(x) is the derivative of f'(x):f''(x) = 2 - 8/x².Set f''(x) = 0 to find potential inflection points:2 - 8/x² = 02 = 8/x²Multiply both sides by x²:2x² = 8x² = 4x = ±2.But since the domain of f(x) is x > 0 (because of the ln x term), we only consider x = 2.So, x = 2 is a potential inflection point.Now, we need to check if f''(x) changes sign at x = 2.Let's test values around x = 2.For x < 2, say x = 1:f''(1) = 2 - 8/1 = 2 - 8 = -6 < 0.For x > 2, say x = 3:f''(3) = 2 - 8/9 ≈ 2 - 0.89 ≈ 1.11 > 0.So, f''(x) changes from negative to positive at x = 2, meaning that x = 2 is indeed an inflection point.Therefore, at x = 2, the function changes from concave to convex. So, to the left of x = 2, the function is concave (f''(x) < 0), meaning it lies below its tangent lines, and to the right, it's convex (f''(x) > 0), meaning it lies above its tangent lines.But in our problem, we need the function to lie above the tangent line for x > x₀ and below for x < x₀. So, if x₀ is an inflection point where the concavity changes from concave to convex, then the function lies below the tangent line for x < x₀ and above for x > x₀, which is exactly what we need.Therefore, x₀ = 2 is a "turning point" as defined.Wait, but let me confirm this with the definition.Given that [f(x) - h(x)] / (x - x₀) > 0 for all x ≠ x₀.So, for x > x₀, [f(x) - h(x)] / (x - x₀) > 0 implies f(x) - h(x) > 0, since x - x₀ > 0.Similarly, for x < x₀, [f(x) - h(x)] / (x - x₀) > 0 implies f(x) - h(x) < 0, since x - x₀ < 0.So, f(x) > h(x) for x > x₀ and f(x) < h(x) for x < x₀.Which is exactly the case when x₀ is an inflection point where the function changes from concave to convex. Because to the left of x₀, the function is concave, so it lies below its tangent line, and to the right, it's convex, so it lies above its tangent line.Therefore, x₀ = 2 is indeed a "turning point".So, the function y = f(x) when a = 8 does have a "turning point" at x = 2.Wait, just to make sure, let me compute F(x) = f(x) - h(x) and see its behavior around x = 2.Compute h(x), the tangent line at x = 2.First, compute f(2):f(2) = (2)² - 10*(2) + 8 ln 2 = 4 - 20 + 8 ln 2 = -16 + 8 ln 2.Compute f'(2):f'(2) = 2*(2) - 10 + 8/(2) = 4 - 10 + 4 = -2.So, the tangent line at x = 2 is:h(x) = f(2) + f'(2)(x - 2) = (-16 + 8 ln 2) + (-2)(x - 2).Simplify h(x):h(x) = -16 + 8 ln 2 - 2x + 4 = (-16 + 4) + 8 ln 2 - 2x = -12 + 8 ln 2 - 2x.So, h(x) = -2x + (-12 + 8 ln 2).Now, F(x) = f(x) - h(x) = [x² - 10x + 8 ln x] - [-2x + (-12 + 8 ln 2)].Simplify F(x):F(x) = x² - 10x + 8 ln x + 2x - (-12 + 8 ln 2)= x² - 8x + 8 ln x + 12 - 8 ln 2.Wait, let me double-check:f(x) = x² - 10x + 8 ln xh(x) = -2x -12 + 8 ln 2So, F(x) = f(x) - h(x) = x² - 10x + 8 ln x - (-2x -12 + 8 ln 2)= x² - 10x + 8 ln x + 2x + 12 - 8 ln 2= x² - 8x + 8 ln x + 12 - 8 ln 2.Yes, that's correct.Now, let's analyze F(x):F(x) = x² - 8x + 8 ln x + 12 - 8 ln 2.We can factor out 8 ln x - 8 ln 2 as 8 ln(x/2).So, F(x) = x² - 8x + 12 + 8 ln(x/2).Now, let's analyze the behavior of F(x) around x = 2.First, compute F(2):F(2) = (2)² - 8*(2) + 12 + 8 ln(2/2)= 4 - 16 + 12 + 8 ln(1)= 0 + 0 = 0.Which makes sense because F(x₀) = 0.Now, let's compute F'(x):F'(x) = derivative of F(x) = 2x - 8 + 8*(1/x).So, F'(x) = 2x - 8 + 8/x.Wait, that's interesting. So, F'(x) = 2x - 8 + 8/x.But earlier, we had F'(x) = f'(x) - f'(x₀). Let's check:f'(x) = 2x - 10 + 8/x.f'(2) = 4 - 10 + 4 = -2.So, F'(x) = f'(x) - (-2) = 2x - 10 + 8/x + 2 = 2x - 8 + 8/x.Yes, that's consistent.So, F'(x) = 2x - 8 + 8/x.We can factor this as:F'(x) = 2x - 8 + 8/x = (2x² - 8x + 8)/x.Wait, let me see:Multiply numerator and denominator by x:F'(x) = (2x² - 8x + 8)/x.So, F'(x) = (2x² - 8x + 8)/x.We can factor the numerator:2x² - 8x + 8 = 2(x² - 4x + 4) = 2(x - 2)².So, F'(x) = 2(x - 2)² / x.Therefore, F'(x) = 2(x - 2)² / x.Since x > 0, the denominator is positive. The numerator is 2(x - 2)², which is always non-negative, and zero only when x = 2.Therefore, F'(x) is always non-negative, and zero only at x = 2.So, F'(x) ≥ 0 for all x > 0, with equality only at x = 2.This means that F(x) is increasing on (0, ∞), with a stationary point at x = 2.Wait, but if F'(x) is always non-negative, then F(x) is non-decreasing on its entire domain.But we know that F(x₀) = F(2) = 0.So, for x < 2, since F'(x) > 0, F(x) is increasing towards F(2) = 0. Therefore, for x < 2, F(x) < 0.For x > 2, F'(x) > 0, so F(x) is increasing away from F(2) = 0. Therefore, for x > 2, F(x) > 0.Therefore, [F(x)] / (x - x₀) > 0 for all x ≠ x₀.Because:- For x > 2, F(x) > 0 and x - 2 > 0, so their ratio is positive.- For x < 2, F(x) < 0 and x - 2 < 0, so their ratio is positive.Therefore, the point x₀ = 2 satisfies the condition of being a "turning point".Hence, the function y = f(x) when a = 8 does have a "turning point" at x = 2.So, summarizing:(1) When a = 1, the function f(x) has a local maximum at x = 1/2 with value -5/4 + ln(1/2) and a local minimum at x = 1 with value -2.(2) When a = 8, the function f(x) has a "turning point" at x = 2.**Final Answer**(1) The extreme values are a maximum at boxed{-dfrac{5}{4} + ln dfrac{1}{2}} and a minimum at boxed{-2}.(2) The function has a "turning point" at boxed{2}.

💡First, I need to understand the relationships between the variables. I know that ( y = 3x ) and ( z = 3y ). To find the sum ( 2x + 3y + z ), I should express all variables in terms of ( x ).Starting with ( y = 3x ), I can substitute this into the equation for ( z ):[z = 3y = 3(3x) = 9x]Now, I have both ( y ) and ( z ) expressed in terms of ( x ). Next, I substitute these values into the expression ( 2x + 3y + z ):[2x + 3y + z = 2x + 3(3x) + 9x]Simplifying the expression step by step:[2x + 9x + 9x = 20x]Therefore, the sum ( 2x + 3y + z ) equals ( 20x ).