answer：Okay, so I have this problem here where I need to find the maximum value of the dot product of vectors PA and PB, where P is a moving point on a circle, and A and B are fixed points. Let me try to break this down step by step.First, let me visualize the problem. There's a circle with equation x² + (y - 3)² = 1. That means the center of the circle is at (0, 3) and it has a radius of 1. Points A and B are at (2, 0) and (-2, 0) respectively, so they are on the x-axis, two units away from the origin in opposite directions.I need to find the maximum value of the dot product of vectors PA and PB as point P moves around the circle. Hmm, okay. So, PA is the vector from P to A, and PB is the vector from P to B. The dot product of two vectors is given by the product of their magnitudes and the cosine of the angle between them. But maybe there's a more straightforward way to compute this without dealing with angles.Let me recall that the dot product of two vectors (a, b) and (c, d) is ac + bd. So, if I can express vectors PA and PB in terms of coordinates, I can compute their dot product algebraically.Let me denote point P as (x, y). Then, vector PA would be A - P, which is (2 - x, 0 - y) = (2 - x, -y). Similarly, vector PB would be B - P, which is (-2 - x, 0 - y) = (-2 - x, -y).So, PA = (2 - x, -y) and PB = (-2 - x, -y). Now, let's compute their dot product:PA · PB = (2 - x)(-2 - x) + (-y)(-y)Let me compute each part:First, (2 - x)(-2 - x). Let's expand this:= 2*(-2) + 2*(-x) - x*(-2) - x*(-x)= -4 - 2x + 2x + x²= x² - 4Wait, that's interesting. The middle terms cancel out. So, (2 - x)(-2 - x) simplifies to x² - 4.Next, (-y)(-y) is y².So, putting it all together:PA · PB = (x² - 4) + y² = x² + y² - 4Okay, so the dot product simplifies to x² + y² - 4. Now, I need to find the maximum value of this expression as P moves along the circle x² + (y - 3)² = 1.Since P lies on the circle, its coordinates satisfy x² + (y - 3)² = 1. Maybe I can express x² in terms of y and substitute it into the expression for PA · PB.Let's solve for x²:x² = 1 - (y - 3)²So, substituting into PA · PB:PA · PB = [1 - (y - 3)²] + y² - 4Let me expand (y - 3)²:(y - 3)² = y² - 6y + 9So,PA · PB = 1 - (y² - 6y + 9) + y² - 4= 1 - y² + 6y - 9 + y² - 4Now, let's simplify this:The -y² and +y² cancel out.So, we have:1 + 6y - 9 - 4Combine the constants:1 - 9 - 4 = -12So,PA · PB = 6y - 12Wow, okay. So, the dot product simplifies to 6y - 12. That's a linear function in terms of y. So, to find its maximum value, I just need to find the maximum value of y on the circle, because 6y - 12 will increase as y increases.So, what's the range of y on the circle x² + (y - 3)² = 1? The circle is centered at (0, 3) with radius 1, so the y-coordinate of any point on the circle will range from 3 - 1 = 2 to 3 + 1 = 4. So, y is between 2 and 4.Therefore, the maximum value of y is 4. Plugging that into PA · PB:PA · PB = 6*4 - 12 = 24 - 12 = 12So, the maximum value of the dot product is 12.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I found vectors PA and PB correctly: (2 - x, -y) and (-2 - x, -y).2. Calculated their dot product: (2 - x)(-2 - x) + (-y)(-y) = x² - 4 + y².3. Expressed x² from the circle equation: x² = 1 - (y - 3)².4. Substituted into the dot product expression: 1 - (y² - 6y + 9) + y² - 4 = 6y - 12.5. Determined that y ranges from 2 to 4 on the circle, so maximum y is 4.6. Plugged y = 4 into 6y - 12 to get 12.Everything seems to check out. So, I think the maximum value is indeed 12.**Final Answer**The maximum value of overrightarrow{PA} cdot overrightarrow{PB} is boxed{12}.

answer：Alright, so I've got this problem about assigning volunteers to different tasks for the Liaoning National Games. Let me try to understand it step by step. First, there are five volunteers: Xiao Zhang, Xiao Zhao, Xiao Li, Xiao Luo, and Xiao Wang. The organizing committee needs to select four of them to undertake four different tasks: translation, tour guide, etiquette, and driver. Now, the tricky part is that Xiao Zhang and Xiao Zhao can only work in the first two positions, which I assume are translation and tour guide. The other three volunteers—Xiao Li, Xiao Luo, and Xiao Wang—can work in any of the four positions. Okay, so I need to figure out how many different assignment plans there are under these conditions. The options given are A: 36, B: 12, C: 18, and D: 48. Let me break this down. Since Xiao Zhang and Xiao Zhao are restricted to the first two tasks, I should consider two scenarios: one where both Xiao Zhang and Xiao Zhao are selected, and another where only one of them is selected. **Scenario 1: Both Xiao Zhang and Xiao Zhao are selected.**- If both are selected, they must be assigned to translation and tour guide. There are 2 ways to assign them to these two positions: Xiao Zhang as translation and Xiao Zhao as tour guide, or vice versa. - Now, we have three remaining tasks: etiquette, driver, and one more task (since we've already assigned two tasks). Wait, no, actually, since we've selected four volunteers, and two of them are already assigned, we have two more tasks to assign to the remaining two volunteers. - The remaining volunteers are Xiao Li, Xiao Luo, and Xiao Wang. We need to choose two out of these three to assign to the remaining two tasks. - The number of ways to choose two volunteers out of three is C(3,2) = 3. - For each of these choices, the two selected volunteers can be assigned to the two remaining tasks in 2! = 2 ways. - So, the total number of assignments in this scenario is 2 (for Xiao Zhang and Xiao Zhao) * 3 (choosing two volunteers) * 2 (assigning tasks) = 12 ways. **Scenario 2: Only one of Xiao Zhang or Xiao Zhao is selected.**- There are two choices here: either Xiao Zhang is selected or Xiao Zhao is selected. - Let's say we choose Xiao Zhang. He can be assigned to either translation or tour guide, so that's 2 options. - Now, we need to select three more volunteers from the remaining four: Xiao Zhao is excluded since we're only selecting one of them, so we have Xiao Li, Xiao Luo, and Xiao Wang. Wait, no, we need to select three more volunteers from the remaining four, but Xiao Zhao is still available if we choose Xiao Zhang. Wait, no, if we're only selecting one of Xiao Zhang or Xiao Zhao, then if we choose Xiao Zhang, Xiao Zhao is not selected, and vice versa. - So, if we choose Xiao Zhang, we need to select three more volunteers from Xiao Li, Xiao Luo, and Xiao Wang. But wait, we only have three volunteers left, so we need to select all three. - Similarly, if we choose Xiao Zhao, we need to select three more volunteers from Xiao Li, Xiao Luo, and Xiao Wang. - So, for each choice (Xiao Zhang or Xiao Zhao), we have 2 options for their task (translation or tour guide). - Then, we have three volunteers left to assign to the remaining three tasks. - The number of ways to assign three volunteers to three tasks is 3! = 6. - So, for each choice (Xiao Zhang or Xiao Zhao), the total number of assignments is 2 (task options) * 6 (assigning the remaining volunteers) = 12. - Since there are two choices (Xiao Zhang or Xiao Zhao), the total for this scenario is 12 * 2 = 24 ways. **Total number of assignment plans:**- Scenario 1: 12 ways- Scenario 2: 24 ways- Total: 12 + 24 = 36 waysSo, the total number of different assignment plans is 36, which corresponds to option A.

answer：Alright, so I have this probability problem here where I need to find the probability that when I randomly select a number x in the interval [0, π], the value of tan x is between -1 and √3. The options are A. 7/12, B. 2/3, C. 1/3, D. 1/4. Hmm, okay, let me try to figure this out step by step.First, I know that tan x is the tangent function, which has some specific properties. I remember that tan x is periodic with a period of π, and it has vertical asymptotes where cos x is zero, which happens at x = π/2. So, in the interval [0, π], tan x is defined everywhere except at x = π/2, where it goes to infinity.Now, the problem is asking for the probability that tan x is between -1 and √3. Probability, in this case, would be the length of the interval(s) where tan x satisfies this condition divided by the total length of the interval [0, π], which is π. So, I need to find all x in [0, π] such that -1 ≤ tan x ≤ √3 and then calculate the measure of those x's.Let me break this down into two parts: when tan x is between -1 and √3. Since tan x can be both positive and negative in the interval [0, π], I need to consider both cases.First, let's consider where tan x is positive. In the interval [0, π], tan x is positive in [0, π/2) and negative in (π/2, π]. So, for tan x to be between -1 and √3, I need to look at both the positive and negative parts.Starting with the positive part: tan x ≤ √3. I know that tan(π/3) = √3, so tan x = √3 at x = π/3. Since tan x increases from 0 to infinity as x approaches π/2, tan x will be less than or equal to √3 for x in [0, π/3]. So, that's one interval: [0, π/3].Now, for the negative part: tan x ≥ -1. In the interval (π/2, π], tan x is negative and approaches negative infinity as x approaches π/2 from the right. It increases from negative infinity to 0 as x approaches π. So, tan x = -1 at x = 3π/4 because tan(3π/4) = -1. Therefore, tan x is greater than or equal to -1 for x in [3π/4, π].So, combining these two intervals where tan x is between -1 and √3, we have [0, π/3] and [3π/4, π]. Now, I need to find the total length of these intervals.Calculating the length of [0, π/3]: that's simply π/3 - 0 = π/3.Calculating the length of [3π/4, π]: that's π - 3π/4 = π/4.So, the total length where tan x satisfies the condition is π/3 + π/4. To add these, I need a common denominator, which is 12. So, π/3 is 4π/12 and π/4 is 3π/12. Adding them together gives 7π/12.Now, the total length of the interval [0, π] is π. Therefore, the probability is the total favorable length divided by the total length, which is (7π/12) / π = 7/12.Wait, let me double-check that. So, the favorable intervals are [0, π/3] and [3π/4, π]. Their lengths are π/3 and π/4, which add up to 7π/12. Divided by π, that's 7/12. That seems correct.But just to make sure I didn't miss anything, let me visualize the graph of tan x between 0 and π. From 0 to π/2, tan x goes from 0 to infinity. At π/3, it's √3, so up to π/3, it's less than or equal to √3. Then, from π/2 to π, tan x goes from negative infinity to 0. At 3π/4, it's -1, so from 3π/4 to π, tan x is between -1 and 0. So, yes, those are the intervals where tan x is between -1 and √3.Therefore, the probability is 7/12, which is option A.**Final Answer**boxed{A}

answer：First, I need to find the derivative of the function ( y = x^3 - 4x^2 + 4 ) to determine the slope of the tangent line at the point ( (1, 1) ).Calculating the derivative:[y' = 3x^2 - 8x]Next, I'll evaluate the derivative at ( x = 1 ) to find the slope at that specific point:[y'|_{x=1} = 3(1)^2 - 8(1) = 3 - 8 = -5]Now that I have the slope (-5), I can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is:[y - y_1 = m(x - x_1)]where ( m ) is the slope and ( (x_1, y_1) ) is the point of tangency.Plugging in the values:[y - 1 = -5(x - 1)]Simplifying this equation:[y - 1 = -5x + 5][y = -5x + 6]Finally, I'll compare this equation with the given options to identify the correct one.

answer：Alright, so I have this inequality to prove for any natural number ( n ). The inequality is:[frac{sqrt{2}}{2} cdot frac{1}{sqrt{2n}} leq frac{1 cdot 3 cdot ldots cdot (2n-1)}{2 cdot 4 cdot ldots cdot 2n} < frac{sqrt{3}}{2} cdot frac{1}{sqrt{2n}}]First, I need to understand what the middle term represents. It looks like a product of odd numbers divided by a product of even numbers. Let me denote this middle term as ( Q_n ). So,[Q_n = frac{1 cdot 3 cdot 5 cdot ldots cdot (2n-1)}{2 cdot 4 cdot 6 cdot ldots cdot 2n}]I recall that this kind of product can be related to binomial coefficients or perhaps something involving factorials. Let me see if I can express ( Q_n ) in terms of factorials.The numerator is the product of the first ( n ) odd numbers, which can be written as ( (2n)! ) divided by the product of the first ( n ) even numbers. Wait, actually, the product of the first ( n ) odd numbers is ( frac{(2n)!}{2^n n!} ). Let me verify that.Yes, because:[(2n)! = 2n cdot (2n-1) cdot 2(n-1) cdot (2n-3) cdot ldots cdot 2 cdot 1]So, if I factor out the 2s from each even term, I get:[(2n)! = 2^n n! cdot (1 cdot 3 cdot 5 cdot ldots cdot (2n-1))]Therefore, the product of the first ( n ) odd numbers is:[1 cdot 3 cdot 5 cdot ldots cdot (2n-1) = frac{(2n)!}{2^n n!}]Similarly, the denominator of ( Q_n ) is the product of the first ( n ) even numbers, which is ( 2 cdot 4 cdot 6 cdot ldots cdot 2n = 2^n n! ).So, putting it all together:[Q_n = frac{frac{(2n)!}{2^n n!}}{2^n n!} = frac{(2n)!}{(2^n n!)^2}]That simplifies to:[Q_n = frac{(2n)!}{4^n (n!)^2}]Hmm, I also know that ( frac{(2n)!}{(n!)^2} ) is the central binomial coefficient, often denoted as ( binom{2n}{n} ). So,[Q_n = frac{binom{2n}{n}}{4^n}]I remember that the central binomial coefficient has an approximation involving ( frac{4^n}{sqrt{pi n}} ) for large ( n ), but I'm not sure if that's directly helpful here. Maybe I can use some inequalities related to the central binomial coefficient or perhaps use induction to prove the given bounds.Let me consider the lower bound first:[frac{sqrt{2}}{2} cdot frac{1}{sqrt{2n}} leq Q_n]Substituting ( Q_n ) with the expression I found:[frac{sqrt{2}}{2} cdot frac{1}{sqrt{2n}} leq frac{binom{2n}{n}}{4^n}]Simplifying the left side:[frac{sqrt{2}}{2} cdot frac{1}{sqrt{2n}} = frac{sqrt{2}}{2sqrt{2n}} = frac{1}{2sqrt{n}}]So, the inequality becomes:[frac{1}{2sqrt{n}} leq frac{binom{2n}{n}}{4^n}]I wonder if I can find an expression or inequality that relates ( binom{2n}{n} ) to ( frac{4^n}{sqrt{pi n}} ). I recall that:[binom{2n}{n} approx frac{4^n}{sqrt{pi n}}]But this is an approximation for large ( n ). However, I need an exact inequality. Maybe I can use the fact that ( binom{2n}{n} geq frac{4^n}{sqrt{pi n}} ) for all ( n geq 1 ). If that's true, then:[frac{binom{2n}{n}}{4^n} geq frac{1}{sqrt{pi n}}]But ( frac{1}{sqrt{pi n}} ) is approximately ( frac{1}{1.772sqrt{n}} ), which is larger than ( frac{1}{2sqrt{n}} ) since ( 1.772 < 2 ). So, this suggests that:[frac{binom{2n}{n}}{4^n} geq frac{1}{sqrt{pi n}} > frac{1}{2sqrt{n}}]Therefore, the lower bound holds. But I need to make sure that this inequality is valid for all natural numbers ( n ), not just asymptotically. Maybe I can test it for small ( n ).For ( n = 1 ):[Q_1 = frac{1}{2} = 0.5][frac{sqrt{2}}{2} cdot frac{1}{sqrt{2 cdot 1}} = frac{sqrt{2}}{2} cdot frac{1}{sqrt{2}} = frac{sqrt{2}}{2sqrt{2}} = frac{1}{2}]So, equality holds for ( n = 1 ).For ( n = 2 ):[Q_2 = frac{1 cdot 3}{2 cdot 4} = frac{3}{8} = 0.375][frac{sqrt{2}}{2} cdot frac{1}{sqrt{4}} = frac{sqrt{2}}{2} cdot frac{1}{2} = frac{sqrt{2}}{4} approx 0.3535]So, ( 0.375 > 0.3535 ), which satisfies the inequality.For ( n = 3 ):[Q_3 = frac{1 cdot 3 cdot 5}{2 cdot 4 cdot 6} = frac{15}{48} = 0.3125][frac{sqrt{2}}{2} cdot frac{1}{sqrt{6}} approx frac{1.4142}{2} cdot frac{1}{2.4495} approx 0.7071 cdot 0.4082 approx 0.289]Again, ( 0.3125 > 0.289 ), so the inequality holds.Okay, so empirically, it seems the lower bound holds. Now, for the upper bound:[Q_n < frac{sqrt{3}}{2} cdot frac{1}{sqrt{2n}}]Again, substituting ( Q_n ):[frac{binom{2n}{n}}{4^n} < frac{sqrt{3}}{2} cdot frac{1}{sqrt{2n}}]Simplifying the right side:[frac{sqrt{3}}{2} cdot frac{1}{sqrt{2n}} = frac{sqrt{3}}{2sqrt{2n}} = frac{sqrt{3}}{2sqrt{2}sqrt{n}} = frac{sqrt{3}}{2sqrt{2}} cdot frac{1}{sqrt{n}} approx 0.6124 cdot frac{1}{sqrt{n}}]I need to show that ( frac{binom{2n}{n}}{4^n} < frac{sqrt{3}}{2sqrt{2}} cdot frac{1}{sqrt{n}} ).Again, using the approximation ( binom{2n}{n} approx frac{4^n}{sqrt{pi n}} ), we have:[frac{binom{2n}{n}}{4^n} approx frac{1}{sqrt{pi n}} approx 0.5642 cdot frac{1}{sqrt{n}}]Which is less than ( 0.6124 cdot frac{1}{sqrt{n}} ). So, the upper bound seems to hold asymptotically. But I need to ensure it's valid for all ( n geq 1 ).Testing for small ( n ):For ( n = 1 ):[Q_1 = 0.5][frac{sqrt{3}}{2} cdot frac{1}{sqrt{2}} approx 0.8660 cdot 0.7071 approx 0.6124]So, ( 0.5 < 0.6124 ), which holds.For ( n = 2 ):[Q_2 = 0.375][frac{sqrt{3}}{2} cdot frac{1}{sqrt{4}} = frac{sqrt{3}}{2} cdot frac{1}{2} approx 0.8660 cdot 0.5 approx 0.4330]So, ( 0.375 < 0.4330 ), which holds.For ( n = 3 ):[Q_3 = 0.3125][frac{sqrt{3}}{2} cdot frac{1}{sqrt{6}} approx 0.8660 cdot 0.4082 approx 0.3535]So, ( 0.3125 < 0.3535 ), which holds.For ( n = 4 ):[Q_4 = frac{1 cdot 3 cdot 5 cdot 7}{2 cdot 4 cdot 6 cdot 8} = frac{105}{384} approx 0.2734][frac{sqrt{3}}{2} cdot frac{1}{sqrt{8}} approx 0.8660 cdot 0.3536 approx 0.306]So, ( 0.2734 < 0.306 ), which holds.It seems the upper bound holds for these small values. To generalize, maybe I can use induction or find a recursive relation for ( Q_n ).Let me consider the ratio ( frac{Q_{n+1}}{Q_n} ):[frac{Q_{n+1}}{Q_n} = frac{frac{1 cdot 3 cdot ldots cdot (2n+1)}{2 cdot 4 cdot ldots cdot (2n+2)}}{frac{1 cdot 3 cdot ldots cdot (2n-1)}{2 cdot 4 cdot ldots cdot 2n}} = frac{2n+1}{2n+2} = frac{2n+1}{2(n+1)}]So, ( Q_{n+1} = Q_n cdot frac{2n+1}{2n+2} ).This recursive relation might help in establishing the bounds. Let me see.Suppose I assume that ( Q_n ) is decreasing, which seems to be the case from the examples above. So, each subsequent term is smaller than the previous one. That might help in setting up the inequalities.Alternatively, maybe I can use the Wallis formula or some integral approximation to bound ( Q_n ).Wait, Wallis formula relates to the product ( frac{pi}{2} = prod_{k=1}^{infty} frac{4k^2}{4k^2 - 1} ), which is similar to the terms in ( Q_n ). Maybe I can use that.But I'm not sure if that's the right approach. Maybe I can use the AM-GM inequality or some other inequality to bound the product.Alternatively, I can consider taking the logarithm of ( Q_n ) and then approximating the sum.Let me try that. Taking the natural logarithm:[ln Q_n = sum_{k=1}^{n} lnleft(frac{2k-1}{2k}right) = sum_{k=1}^{n} lnleft(1 - frac{1}{2k}right)]Using the approximation ( ln(1 - x) approx -x - frac{x^2}{2} ) for small ( x ), but I'm not sure if that's helpful here.Alternatively, maybe I can use the inequality ( ln(1 - x) leq -x ) for ( 0 < x < 1 ). So,[ln Q_n = sum_{k=1}^{n} lnleft(1 - frac{1}{2k}right) leq -sum_{k=1}^{n} frac{1}{2k} = -frac{1}{2} H_n]Where ( H_n ) is the ( n )-th harmonic number. So,[Q_n leq expleft(-frac{1}{2} H_nright)]But I'm not sure if this helps in establishing the desired bounds. Maybe I need a different approach.Let me think about the product ( Q_n ) again. It can be written as:[Q_n = prod_{k=1}^{n} frac{2k - 1}{2k} = prod_{k=1}^{n} left(1 - frac{1}{2k}right)]I recall that the infinite product ( prod_{k=1}^{infty} left(1 - frac{1}{2k}right) ) converges to zero, which makes sense because ( Q_n ) decreases to zero as ( n ) increases.But I need to find bounds for finite ( n ). Maybe I can use the inequality for the product of terms less than 1.Alternatively, I can use the fact that for ( 0 < x < 1 ), ( ln(1 - x) leq -x - frac{x^2}{2} ). So,[ln Q_n = sum_{k=1}^{n} lnleft(1 - frac{1}{2k}right) leq -sum_{k=1}^{n} left(frac{1}{2k} + frac{1}{8k^2}right)]Which gives:[ln Q_n leq -frac{1}{2} H_n - frac{1}{8} sum_{k=1}^{n} frac{1}{k^2}]But again, I'm not sure if this leads me anywhere useful.Maybe I should try to find a telescoping product or relate ( Q_n ) to some known series or product.Wait, another idea: perhaps I can compare ( Q_n ) to an integral. For example, using the integral test or approximating the sum with an integral.But I'm not sure how to set that up. Alternatively, maybe I can use the fact that ( Q_n ) is related to the beta function or gamma function.I know that:[binom{2n}{n} = frac{4^n}{sqrt{pi n}} cdot frac{Gamma(n + frac{1}{2})}{Gamma(n + 1)}]But I'm not sure if that helps in establishing the inequalities.Wait, maybe I can use the recursive relation I found earlier:[Q_{n+1} = Q_n cdot frac{2n + 1}{2n + 2}]If I can show that ( Q_n ) is bounded below and above by certain expressions involving ( frac{1}{sqrt{n}} ), then I can establish the desired inequalities.Let me consider the lower bound first. Suppose I can show that:[Q_n geq frac{sqrt{2}}{2} cdot frac{1}{sqrt{2n}}]Which simplifies to:[Q_n geq frac{1}{2sqrt{n}}]I can try to prove this by induction. Let's see.Base case: ( n = 1 )[Q_1 = frac{1}{2} = 0.5][frac{1}{2sqrt{1}} = 0.5]So, equality holds.Inductive step: Assume that for some ( k geq 1 ), ( Q_k geq frac{1}{2sqrt{k}} ). Then, we need to show that ( Q_{k+1} geq frac{1}{2sqrt{k+1}} ).Using the recursive relation:[Q_{k+1} = Q_k cdot frac{2k + 1}{2k + 2}]By the inductive hypothesis:[Q_{k+1} geq frac{1}{2sqrt{k}} cdot frac{2k + 1}{2k + 2}]We need to show that:[frac{1}{2sqrt{k}} cdot frac{2k + 1}{2k + 2} geq frac{1}{2sqrt{k+1}}]Simplifying:[frac{2k + 1}{2k + 2} geq frac{sqrt{k}}{sqrt{k+1}}]Cross-multiplying (since all terms are positive):[(2k + 1)sqrt{k+1} geq (2k + 2)sqrt{k}]Let me square both sides to eliminate the square roots:[(2k + 1)^2 (k + 1) geq (2k + 2)^2 k]Expanding both sides:Left side:[(4k^2 + 4k + 1)(k + 1) = 4k^3 + 4k^2 + k + 4k^2 + 4k + 1 = 4k^3 + 8k^2 + 5k + 1]Right side:[(4k^2 + 8k + 4)k = 4k^3 + 8k^2 + 4k]Subtracting the right side from the left side:[(4k^3 + 8k^2 + 5k + 1) - (4k^3 + 8k^2 + 4k) = k + 1 geq 0]Which is always true for ( k geq 1 ). Therefore, the inequality holds, and by induction, the lower bound is established.Now, for the upper bound:[Q_n < frac{sqrt{3}}{2} cdot frac{1}{sqrt{2n}}]Simplifying:[Q_n < frac{sqrt{3}}{2sqrt{2n}} = frac{sqrt{3}}{2sqrt{2}} cdot frac{1}{sqrt{n}} approx 0.6124 cdot frac{1}{sqrt{n}}]Again, I can try induction. Let's see.Base case: ( n = 1 )[Q_1 = frac{1}{2} = 0.5][frac{sqrt{3}}{2sqrt{2}} approx 0.6124]So, ( 0.5 < 0.6124 ), which holds.Inductive step: Assume that for some ( k geq 1 ), ( Q_k < frac{sqrt{3}}{2sqrt{2k}} ). Then, we need to show that ( Q_{k+1} < frac{sqrt{3}}{2sqrt{2(k+1)}} ).Using the recursive relation:[Q_{k+1} = Q_k cdot frac{2k + 1}{2k + 2}]By the inductive hypothesis:[Q_{k+1} < frac{sqrt{3}}{2sqrt{2k}} cdot frac{2k + 1}{2k + 2}]We need to show that:[frac{sqrt{3}}{2sqrt{2k}} cdot frac{2k + 1}{2k + 2} < frac{sqrt{3}}{2sqrt{2(k+1)}}]Simplifying:[frac{2k + 1}{2k + 2} < frac{sqrt{k}}{sqrt{k+1}}]Cross-multiplying (since all terms are positive):[(2k + 1)sqrt{k+1} < (2k + 2)sqrt{k}]Squaring both sides:[(2k + 1)^2 (k + 1) < (2k + 2)^2 k]Expanding both sides:Left side:[(4k^2 + 4k + 1)(k + 1) = 4k^3 + 4k^2 + k + 4k^2 + 4k + 1 = 4k^3 + 8k^2 + 5k + 1]Right side:[(4k^2 + 8k + 4)k = 4k^3 + 8k^2 + 4k]Subtracting the left side from the right side:[(4k^3 + 8k^2 + 4k) - (4k^3 + 8k^2 + 5k + 1) = -k - 1 < 0]Which is always true for ( k geq 1 ). Therefore, the inequality holds, and by induction, the upper bound is established.So, combining both bounds, we have:[frac{sqrt{2}}{2} cdot frac{1}{sqrt{2n}} leq Q_n < frac{sqrt{3}}{2} cdot frac{1}{sqrt{2n}}]Which is the desired inequality. Therefore, the proof is complete.

answer：Okay, so I have this equation here: 7 equals (x³ - 3x² - 10x) divided by (x + 2). I need to find the sum of the x-values that satisfy this equation. Hmm, let's see. I think I should start by simplifying the equation to make it easier to solve.First, I notice that the numerator is a cubic polynomial, and the denominator is a linear polynomial. Maybe I can factor the numerator to see if anything cancels out with the denominator. Let me try factoring the numerator: x³ - 3x² - 10x. I can factor out an x from all the terms, so that gives me x(x² - 3x - 10). Now, I need to factor the quadratic part: x² - 3x - 10.To factor x² - 3x - 10, I'm looking for two numbers that multiply to -10 and add up to -3. Let me think... 2 and -5? Yeah, because 2 times -5 is -10, and 2 plus (-5) is -3. So, that factors to (x + 2)(x - 5). So, putting it all together, the numerator factors to x(x + 2)(x - 5).Now, the equation becomes 7 = [x(x + 2)(x - 5)] / (x + 2). I can cancel out the (x + 2) terms in the numerator and denominator, but I have to remember that x cannot be -2 because that would make the denominator zero, which is undefined. So, after canceling, the equation simplifies to 7 = x(x - 5).Now, let's write that out: 7 = x(x - 5). Expanding the right side, that's 7 = x² - 5x. To solve for x, I'll bring the 7 to the other side: x² - 5x - 7 = 0.This is a quadratic equation, and I can use the quadratic formula to find the roots. The quadratic formula is x = [-b ± √(b² - 4ac)] / (2a), where a = 1, b = -5, and c = -7. Plugging those values in, I get x = [5 ± √(25 + 28)] / 2, which simplifies to x = [5 ± √53] / 2.So, the two solutions are (5 + √53)/2 and (5 - √53)/2. To find the sum of these x-values, I can add them together. When I add (5 + √53)/2 and (5 - √53)/2, the √53 terms cancel out, leaving me with (5 + 5)/2, which is 10/2, or 5.Wait a second, I remember that for any quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a. In this case, a is 1 and b is -5, so the sum should be -(-5)/1, which is 5. That matches what I got when I added the two solutions. So, that makes sense.But just to make sure I didn't make any mistakes earlier, let me double-check my factoring. The original numerator was x³ - 3x² - 10x, which I factored as x(x + 2)(x - 5). If I multiply those factors back out, I should get the original numerator. Let's see: x times (x + 2) is x² + 2x, and then times (x - 5) is x³ - 5x² + 2x² - 10x, which simplifies to x³ - 3x² - 10x. Yep, that's correct.Also, when I canceled out the (x + 2) terms, I remembered that x cannot be -2, so I don't have to worry about that solution. The quadratic equation I ended up with doesn't include x = -2 as a root, so everything checks out.So, putting it all together, the sum of the x-values that satisfy the original equation is 5.