answer：Okay, so I have this function f(x) defined piecewise. It's 2^x - 1 when x is less than or equal to 0, and it's log base 3 of x plus 1 when x is greater than 0. I need to find f(f(√3 / 9)). Hmm, that sounds a bit complicated, but I think I can break it down step by step.First, let me figure out what f(√3 / 9) is. Since √3 / 9 is a positive number, I should use the second part of the piecewise function, which is log base 3 of x plus 1. So, f(√3 / 9) = log₃(√3 / 9) + 1.Now, I need to compute log base 3 of √3 divided by 9. Let me recall some logarithm rules. Log base 3 of a quotient is the difference of the logs, so log₃(√3) - log₃(9). What's log₃(√3)? Well, √3 is the same as 3^(1/2), so log base 3 of 3^(1/2) is just 1/2. Got that part.Next, log₃(9). Since 9 is 3 squared, log base 3 of 3^2 is 2. So, putting it together, log₃(√3 / 9) is 1/2 - 2, which is -3/2. Wait, hold on. Let me double-check that. If I have log₃(√3) = 1/2 and log₃(9) = 2, then subtracting them gives 1/2 - 2 = -3/2. Hmm, that seems right.So, f(√3 / 9) is log₃(√3 / 9) + 1, which is (-3/2) + 1. That simplifies to -1/2. Okay, so f(√3 / 9) is -1/2.Now, I need to find f(f(√3 / 9)), which is f(-1/2). Since -1/2 is less than or equal to 0, I should use the first part of the piecewise function, which is 2^x - 1. So, f(-1/2) = 2^(-1/2) - 1.What's 2^(-1/2)? That's the same as 1 over 2^(1/2), which is 1 over √2. To rationalize the denominator, that's √2 / 2. So, 2^(-1/2) is √2 / 2.Therefore, f(-1/2) = √2 / 2 - 1. So, f(f(√3 / 9)) is √2 / 2 - 1.Looking at the options, that's option A: √2 / 2 - 1.Wait, let me just go through that again to make sure I didn't make a mistake. Starting with f(√3 / 9), since √3 / 9 is positive, we use log base 3. √3 is 3^(1/2), so log base 3 is 1/2. 9 is 3^2, so log base 3 is 2. So, 1/2 - 2 is -3/2. Then, adding 1 gives -1/2. Then, f(-1/2) is 2^(-1/2) - 1, which is √2 / 2 - 1. Yeah, that seems correct.I don't think I made any mistakes here. So, the answer should be option A.**Final Answer**boxed{A}

answer：Okay, so I have this problem about complex numbers and the unit circle. Let me try to understand it step by step. The question says: In the complex plane, several numbers are plotted along with the unit circle centered at the origin. A complex number ( F ) lies in the first quadrant and outside the unit circle. Which of the numbers represents the reciprocal of ( F )?The options are:- **(A)** Point in the first quadrant inside the unit circle- **(B)** Point in the second quadrant inside the unit circle- **(C)** Point in the third quadrant inside the unit circle- **(D)** Point in the fourth quadrant inside the unit circle- **(E)** Point exactly on the unit circleAlright, so I need to figure out where the reciprocal of ( F ) would be located. First, let's recall what the reciprocal of a complex number is. If ( F = a + bi ), then the reciprocal ( frac{1}{F} ) is ( frac{a - bi}{a^2 + b^2} ). This comes from multiplying the numerator and denominator by the complex conjugate of ( F ), which is ( a - bi ).So, ( frac{1}{F} = frac{a}{a^2 + b^2} - frac{b}{a^2 + b^2}i ). Now, since ( F ) is in the first quadrant, both ( a ) and ( b ) are positive. Also, since ( F ) is outside the unit circle, the magnitude of ( F ), which is ( sqrt{a^2 + b^2} ), is greater than 1. Therefore, ( a^2 + b^2 > 1 ).Looking at the reciprocal ( frac{1}{F} ), the real part is ( frac{a}{a^2 + b^2} ) and the imaginary part is ( -frac{b}{a^2 + b^2} ). Since ( a ) and ( b ) are positive, the real part ( frac{a}{a^2 + b^2} ) is positive, but because ( a^2 + b^2 > 1 ), this value is less than ( a ). Similarly, the imaginary part ( -frac{b}{a^2 + b^2} ) is negative, and its magnitude is less than ( b ).So, the reciprocal ( frac{1}{F} ) has a positive real part and a negative imaginary part, which places it in the fourth quadrant. Additionally, since ( a^2 + b^2 > 1 ), the magnitude of ( frac{1}{F} ) is ( frac{1}{sqrt{a^2 + b^2}} ), which is less than 1. This means the reciprocal lies inside the unit circle.Let me double-check that. If ( F ) is outside the unit circle, its reciprocal should be inside, right? Because if ( |F| > 1 ), then ( left| frac{1}{F} right| = frac{1}{|F|} < 1 ). That makes sense.Also, the quadrant part: ( F ) is in the first quadrant, so its reciprocal should be in the fourth quadrant because the imaginary part changes sign. Yeah, that seems correct.So, putting it all together, the reciprocal of ( F ) is a point in the fourth quadrant inside the unit circle. Looking back at the options, that's option **(D)**.I don't think I made any mistakes here. The key points are understanding how taking the reciprocal affects the magnitude and the quadrant. Since ( F ) is in the first quadrant and outside the unit circle, its reciprocal must be in the fourth quadrant and inside the unit circle.**Final Answer**boxed{D}

answer：Alright, so I have this problem: "How many integers from 1 to 9 are divisors of the five-digit number 36,432?" Okay, let's break this down. I need to figure out which numbers between 1 and 9 can divide 36,432 without leaving a remainder. That means I need to check the divisibility of 36,432 by each of these numbers.First, I remember that every number is divisible by 1, so that's a given. That's one divisor right there.Next, let's check divisibility by 2. I recall that a number is divisible by 2 if its last digit is even. Looking at 36,432, the last digit is 2, which is even. So, 36,432 is divisible by 2. That's the second divisor.Now, moving on to 3. The rule for divisibility by 3 is that if the sum of the digits is divisible by 3, then the number itself is divisible by 3. Let's add up the digits of 36,432: 3 + 6 + 4 + 3 + 2 equals 18. Since 18 is divisible by 3, 36,432 is also divisible by 3. That's the third divisor.Next up is 4. The rule for 4 is that if the last two digits of the number form a number that's divisible by 4, then the entire number is divisible by 4. Looking at 36,432, the last two digits are 32. 32 divided by 4 is 8, which is a whole number, so 36,432 is divisible by 4. That's the fourth divisor.Now, checking for 5. The rule here is straightforward: if the number ends with a 0 or 5, it's divisible by 5. 36,432 ends with a 2, so it's not divisible by 5. So, 5 is out.Moving on to 6. A number is divisible by 6 if it's divisible by both 2 and 3. We've already established that 36,432 is divisible by both 2 and 3, so it must also be divisible by 6. That's the fifth divisor.Next is 7. Divisibility by 7 is a bit trickier. There isn't a straightforward rule like with 2, 3, or 5. One method I remember is to take the last digit, double it, subtract it from the rest of the number, and see if the result is divisible by 7. Let's try that with 36,432.Take the last digit, which is 2, double it to get 4, and subtract that from the remaining number: 3643 - 4 equals 3639. Now, I need to check if 3639 is divisible by 7. Repeating the process: take the last digit (9), double it to get 18, subtract that from 363: 363 - 18 equals 345. Still not sure if 345 is divisible by 7. Let's do it again: take 5, double it to 10, subtract from 34: 34 - 10 equals 24. 24 isn't divisible by 7, so 36,432 isn't divisible by 7. So, 7 is out.Now, checking for 8. The rule for 8 is similar to that of 4 but with the last three digits. If the last three digits form a number that's divisible by 8, then the entire number is divisible by 8. The last three digits of 36,432 are 432. Let's divide 432 by 8: 432 ÷ 8 equals 54, which is a whole number. So, 36,432 is divisible by 8. That's the sixth divisor.Finally, checking for 9. The rule for 9 is similar to that of 3 but with the sum of the digits needing to be divisible by 9. Earlier, we found that the sum of the digits is 18, which is divisible by 9. Therefore, 36,432 is divisible by 9. That's the seventh divisor.So, summarizing, the numbers from 1 to 9 that divide 36,432 without a remainder are: 1, 2, 3, 4, 6, 8, and 9. That makes a total of 7 numbers.

answer：Alright, so I have this problem that I need to prove: for every integer ( k geq 1 ) and real number ( x ), the polynomial[ 1 - x + frac{x^2}{2!} - frac{x^3}{3!} + ldots + (-1)^j frac{x^j}{j!} + ldots + frac{x^{2k}}{(2k)!} ]is greater than or equal to zero. Hmm, okay. Let me try to understand what this polynomial is. It looks like an alternating series where the terms go up to ( x^{2k} ) divided by ( (2k)! ). So it's a finite sum, not an infinite series. First, I notice that the polynomial alternates signs: the first term is positive, then negative, then positive, and so on. The degrees of the terms go up by one each time, and the coefficients are reciprocals of factorials. This reminds me of the Taylor series expansion of ( e^{-x} ), which is:[ e^{-x} = 1 - x + frac{x^2}{2!} - frac{x^3}{3!} + ldots + (-1)^j frac{x^j}{j!} + ldots ]But in our case, the polynomial stops at ( x^{2k} ). So it's like the Taylor polynomial of degree ( 2k ) for ( e^{-x} ). Interesting. Maybe I can use properties of the exponential function or its Taylor polynomials to prove this inequality.Let me denote this polynomial as ( P_{2k}(x) ). So,[ P_{2k}(x) = sum_{j=0}^{2k} (-1)^j frac{x^j}{j!} ]I need to show that ( P_{2k}(x) geq 0 ) for all real numbers ( x ) and integers ( k geq 1 ).Let me consider different cases for ( x ).**Case 1: ( x leq 0 )**If ( x ) is negative or zero, let's see what happens. For ( x = 0 ), ( P_{2k}(0) = 1 ), which is positive. For ( x < 0 ), each term ( (-1)^j frac{x^j}{j!} ) becomes ( (-1)^j frac{(-|x|)^j}{j!} = (-1)^j (-1)^j frac{|x|^j}{j!} = frac{|x|^j}{j!} ). So all terms become positive because ( (-1)^j times (-1)^j = 1 ). Therefore, ( P_{2k}(x) ) is a sum of positive terms, so it's positive. Thus, for ( x leq 0 ), ( P_{2k}(x) geq 1 > 0 ).**Case 2: ( x > 0 )**This is more complicated. I need to show that ( P_{2k}(x) geq 0 ) for all ( x > 0 ). Let me think about the behavior of ( P_{2k}(x) ) as ( x ) increases.First, note that as ( x ) approaches infinity, the leading term of ( P_{2k}(x) ) is ( frac{x^{2k}}{(2k)!} ), which is positive. So for very large ( x ), ( P_{2k}(x) ) is positive. But what about for smaller ( x )?Maybe I can analyze the derivative of ( P_{2k}(x) ) to find its minima and maxima. If I can show that the minimum value of ( P_{2k}(x) ) is non-negative, then the polynomial is always non-negative.Let's compute the derivative:[ P_{2k}'(x) = -1 + x - frac{x^2}{2!} + frac{x^3}{3!} - ldots + (-1)^{2k} frac{x^{2k - 1}}{(2k - 1)!} ]Wait, that looks familiar. It seems like the derivative is another polynomial similar to ( P_{2k}(x) ), but one degree less. In fact, it's ( P_{2k - 1}(x) ), the Taylor polynomial of degree ( 2k - 1 ) for ( e^{-x} ).So, ( P_{2k}'(x) = P_{2k - 1}(x) ).If I can find the critical points where ( P_{2k}'(x) = 0 ), those would be the points where ( P_{2k}(x) ) has local minima or maxima.But this seems recursive. Maybe I need another approach.Alternatively, I can consider the relationship between ( P_{2k}(x) ) and ( e^{-x} ). Since ( P_{2k}(x) ) is the Taylor polynomial of ( e^{-x} ) truncated at degree ( 2k ), we know that the remainder term ( R_{2k}(x) ) is given by:[ R_{2k}(x) = frac{(-1)^{2k + 1} x^{2k + 1}}{(2k + 1)!} e^{-c} ]for some ( c ) between 0 and ( x ). Therefore,[ e^{-x} = P_{2k}(x) + R_{2k}(x) ]But since ( e^{-x} ) is always positive, if I can relate ( P_{2k}(x) ) to ( e^{-x} ), maybe I can show that ( P_{2k}(x) geq 0 ).Wait, but ( R_{2k}(x) ) is the remainder, and depending on the sign, it could be positive or negative. Let me see:Since ( (-1)^{2k + 1} = -1 ), the remainder is:[ R_{2k}(x) = -frac{x^{2k + 1}}{(2k + 1)!} e^{-c} ]So,[ e^{-x} = P_{2k}(x) - frac{x^{2k + 1}}{(2k + 1)!} e^{-c} ]Therefore,[ P_{2k}(x) = e^{-x} + frac{x^{2k + 1}}{(2k + 1)!} e^{-c} ]Since ( e^{-c} ) is always positive, and ( x^{2k + 1} ) is positive for ( x > 0 ), the second term is positive. Therefore, ( P_{2k}(x) ) is equal to ( e^{-x} ) plus a positive term. But ( e^{-x} ) is positive, so ( P_{2k}(x) ) is greater than ( e^{-x} ), which is positive. Wait, does that mean ( P_{2k}(x) > e^{-x} > 0 )?But that can't be right because for large ( x ), ( P_{2k}(x) ) behaves like ( frac{x^{2k}}{(2k)!} ), which is much larger than ( e^{-x} ). So, actually, ( P_{2k}(x) ) is an approximation of ( e^{-x} ) from above or below?Wait, let's think about the Taylor series. The Taylor polynomial of even degree for ( e^{-x} ) alternates around the function. So, for ( e^{-x} ), the even-degree Taylor polynomials are above the function, and the odd-degree ones are below. So, ( P_{2k}(x) ) is an over-approximation of ( e^{-x} ). Therefore, ( P_{2k}(x) geq e^{-x} ). But since ( e^{-x} > 0 ), does that imply ( P_{2k}(x) > 0 )?Wait, no, because ( e^{-x} ) is positive, but ( P_{2k}(x) ) could be greater than ( e^{-x} ), but ( e^{-x} ) is positive, so ( P_{2k}(x) ) is also positive. Hmm, maybe that's a way to see it.But I'm not entirely sure. Maybe I need to think differently.Let me consider specific values of ( k ) to see if I can spot a pattern.**Case ( k = 1 ):**Then, ( P_2(x) = 1 - x + frac{x^2}{2} ).Let me analyze this quadratic polynomial.Compute its discriminant:Discriminant ( D = (-1)^2 - 4 times frac{1}{2} times 1 = 1 - 2 = -1 ).Since the discriminant is negative, the quadratic has no real roots and is always positive because the coefficient of ( x^2 ) is positive. So, ( P_2(x) > 0 ) for all ( x ).That's good. So for ( k = 1 ), the inequality holds.**Case ( k = 2 ):**( P_4(x) = 1 - x + frac{x^2}{2} - frac{x^3}{6} + frac{x^4}{24} ).This is a quartic polynomial. Let me see if I can factor it or find its minima.Alternatively, maybe I can write it as ( P_4(x) = e^{-x} + R_4(x) ), where ( R_4(x) ) is the remainder.But I'm not sure if that helps. Maybe I can compute its derivative:( P_4'(x) = -1 + x - frac{x^2}{2} + frac{x^3}{6} ).Set derivative equal to zero to find critical points:( -1 + x - frac{x^2}{2} + frac{x^3}{6} = 0 ).This is a cubic equation. Solving it analytically might be complicated, but maybe I can estimate.Alternatively, maybe I can evaluate ( P_4(x) ) at some points to see if it's always positive.At ( x = 0 ): ( P_4(0) = 1 ).At ( x = 1 ): ( 1 - 1 + frac{1}{2} - frac{1}{6} + frac{1}{24} = 0 + 0.5 - 0.1667 + 0.0417 approx 0.375 ).At ( x = 2 ): ( 1 - 2 + frac{4}{2} - frac{8}{6} + frac{16}{24} = 1 - 2 + 2 - 1.333 + 0.6667 approx 0.3337 ).At ( x = 3 ): ( 1 - 3 + frac{9}{2} - frac{27}{6} + frac{81}{24} approx 1 - 3 + 4.5 - 4.5 + 3.375 = 1.375 ).At ( x = 4 ): ( 1 - 4 + frac{16}{2} - frac{64}{6} + frac{256}{24} approx 1 - 4 + 8 - 10.6667 + 10.6667 = 5 ).So, it seems positive at these points. Maybe it's always positive.But to be thorough, I should check if there's a minimum somewhere where ( P_4(x) ) could be zero or negative.Let me try to find the critical points by solving ( P_4'(x) = 0 ):( -1 + x - frac{x^2}{2} + frac{x^3}{6} = 0 ).Multiply both sides by 6 to eliminate denominators:( -6 + 6x - 3x^2 + x^3 = 0 ).So,( x^3 - 3x^2 + 6x - 6 = 0 ).Let me try to find rational roots using Rational Root Theorem. Possible roots are factors of 6 over factors of 1: ( pm1, pm2, pm3, pm6 ).Test ( x = 1 ): ( 1 - 3 + 6 - 6 = -2 neq 0 ).Test ( x = 2 ): ( 8 - 12 + 12 - 6 = 2 neq 0 ).Test ( x = 3 ): ( 27 - 27 + 18 - 6 = 12 neq 0 ).Test ( x = 6 ): ( 216 - 108 + 36 - 6 = 138 neq 0 ).Test ( x = -1 ): ( -1 - 3 - 6 - 6 = -16 neq 0 ).So no rational roots. Maybe I need to use numerical methods or see the behavior.Compute ( f(x) = x^3 - 3x^2 + 6x - 6 ).At ( x = 1 ): ( f(1) = -2 ).At ( x = 2 ): ( f(2) = 2 ).So there's a root between 1 and 2.Similarly, let's compute ( f(1.5) ):( 3.375 - 6.75 + 9 - 6 = -0.375 ).Still negative.At ( x = 1.75 ):( (1.75)^3 - 3(1.75)^2 + 6(1.75) - 6 ).Compute:( 5.359375 - 9.1875 + 10.5 - 6 = 0.671875 ).Positive. So the root is between 1.5 and 1.75.Using linear approximation:Between ( x = 1.5 ) (f = -0.375) and ( x = 1.75 ) (f = 0.671875).Slope: ( (0.671875 - (-0.375))/(1.75 - 1.5) = 1.046875 / 0.25 = 4.1875 ).To reach zero from ( x = 1.5 ), need ( Delta x = 0.375 / 4.1875 approx 0.09 ).So approximate root at ( x approx 1.59 ).Similarly, check ( f(1.59) ):( (1.59)^3 - 3(1.59)^2 + 6(1.59) - 6 ).Compute:( 3.99 - 7.56 + 9.54 - 6 approx -0.03 ).Close to zero. Try ( x = 1.595 ):( (1.595)^3 ≈ 4.04 ), ( 3(1.595)^2 ≈ 7.62 ), ( 6(1.595) ≈ 9.57 ).So,( 4.04 - 7.62 + 9.57 - 6 ≈ -0.01 ).Almost zero. Maybe ( x ≈ 1.596 ).So, there's a critical point near ( x ≈ 1.596 ).Now, let's compute ( P_4(1.596) ):( 1 - 1.596 + frac{(1.596)^2}{2} - frac{(1.596)^3}{6} + frac{(1.596)^4}{24} ).Compute each term:1. ( 1 )2. ( -1.596 )3. ( frac{(1.596)^2}{2} ≈ frac{2.547}{2} ≈ 1.2735 )4. ( -frac{(1.596)^3}{6} ≈ -frac{4.04}{6} ≈ -0.673 )5. ( frac{(1.596)^4}{24} ≈ frac{6.44}{24} ≈ 0.268 )Sum them up:( 1 - 1.596 + 1.2735 - 0.673 + 0.268 ≈ 1 - 1.596 = -0.596 + 1.2735 = 0.6775 - 0.673 = 0.0045 + 0.268 ≈ 0.2725 ).So, ( P_4(1.596) ≈ 0.2725 > 0 ).Therefore, the minimum value near ( x ≈ 1.596 ) is still positive. Thus, ( P_4(x) > 0 ) for all ( x ).Hmm, interesting. So for ( k = 2 ), the polynomial is also always positive.Maybe this pattern continues for higher ( k ). Perhaps I can use induction.**Inductive Approach:**Let me try to use mathematical induction to prove that ( P_{2k}(x) geq 0 ) for all ( x ) and ( k geq 1 ).**Base Case:**For ( k = 1 ), as shown earlier, ( P_2(x) = 1 - x + frac{x^2}{2} ) is always positive because its discriminant is negative, so it has no real roots and is always positive.**Inductive Step:**Assume that for some integer ( k = n ), ( P_{2n}(x) geq 0 ) for all real ( x ). We need to show that ( P_{2(n+1)}(x) geq 0 ) for all real ( x ).But I'm not sure how to relate ( P_{2(n+1)}(x) ) to ( P_{2n}(x) ). Let me think about the relationship between consecutive polynomials.Note that ( P_{2(n+1)}(x) = P_{2n}(x) - frac{x^{2n + 1}}{(2n + 1)!} + frac{x^{2n + 2}}{(2n + 2)!} ).Wait, no. Actually, ( P_{2(n+1)}(x) ) is the polynomial up to ( x^{2n + 2} ), so it's ( P_{2n}(x) ) plus the next two terms: ( -frac{x^{2n + 1}}{(2n + 1)!} + frac{x^{2n + 2}}{(2n + 2)!} ).But I'm not sure if this helps. Maybe I need another approach.Alternatively, consider that ( P_{2k}(x) ) is the Taylor polynomial of ( e^{-x} ) up to degree ( 2k ). Since ( e^{-x} ) is always positive, and the Taylor polynomials of even degree are above ( e^{-x} ), as I thought earlier, then ( P_{2k}(x) geq e^{-x} > 0 ). But wait, is that always true?Actually, for the Taylor series of ( e^{-x} ), the even-degree polynomials are indeed above ( e^{-x} ) for all ( x ). Let me verify this.Consider the function ( e^{-x} ) and its Taylor polynomials. The Taylor series alternates between overestimating and underestimating the function depending on the degree. For even-degree polynomials, they overestimate ( e^{-x} ), and for odd-degree polynomials, they underestimate it.Yes, that seems to be the case. So, if ( P_{2k}(x) ) is an even-degree Taylor polynomial of ( e^{-x} ), then ( P_{2k}(x) geq e^{-x} ) for all ( x ). Since ( e^{-x} > 0 ), it follows that ( P_{2k}(x) > 0 ).But wait, is this always true? Let me think about ( x = 0 ). At ( x = 0 ), ( P_{2k}(0) = 1 ), which is equal to ( e^{0} = 1 ). So, they are equal at ( x = 0 ). For ( x > 0 ), does ( P_{2k}(x) ) stay above ( e^{-x} )?Yes, because the remainder term for the Taylor polynomial of even degree is positive. Recall that the remainder term ( R_{2k}(x) ) is given by:[ R_{2k}(x) = frac{(-1)^{2k + 1} x^{2k + 1}}{(2k + 1)!} e^{-c} ]for some ( c ) between 0 and ( x ). Since ( (-1)^{2k + 1} = -1 ), the remainder is negative:[ R_{2k}(x) = -frac{x^{2k + 1}}{(2k + 1)!} e^{-c} ]Therefore,[ e^{-x} = P_{2k}(x) + R_{2k}(x) = P_{2k}(x) - frac{x^{2k + 1}}{(2k + 1)!} e^{-c} ]Which implies,[ P_{2k}(x) = e^{-x} + frac{x^{2k + 1}}{(2k + 1)!} e^{-c} ]Since ( frac{x^{2k + 1}}{(2k + 1)!} e^{-c} > 0 ) for ( x > 0 ), it follows that ( P_{2k}(x) > e^{-x} ). And since ( e^{-x} > 0 ), ( P_{2k}(x) > 0 ).Therefore, combining both cases ( x leq 0 ) and ( x > 0 ), we have ( P_{2k}(x) geq 0 ) for all real ( x ) and integers ( k geq 1 ).Wait, but in the case ( x > 0 ), we have ( P_{2k}(x) > e^{-x} > 0 ), so it's strictly greater than zero. For ( x leq 0 ), we have ( P_{2k}(x) geq 1 > 0 ). So actually, ( P_{2k}(x) > 0 ) for all real ( x ).But the problem statement says ( geq 0 ). So, is it possible for ( P_{2k}(x) ) to be zero for some ( x )?From the analysis above, for ( x > 0 ), ( P_{2k}(x) > 0 ), and for ( x leq 0 ), ( P_{2k}(x) geq 1 > 0 ). So, actually, ( P_{2k}(x) > 0 ) for all real ( x ). Therefore, the inequality is strict.But the problem says ( geq 0 ), which is a weaker statement. So, our conclusion is stronger, but it still satisfies the problem's requirement.Alternatively, maybe there's a point where ( P_{2k}(x) = 0 ), but from the analysis, it seems not. For ( x leq 0 ), it's always at least 1, and for ( x > 0 ), it's greater than ( e^{-x} ), which is always positive.Therefore, I think the polynomial is always positive, which satisfies the inequality ( geq 0 ).**Conclusion:**By analyzing the polynomial for different cases of ( x ) and using properties of the Taylor series expansion of ( e^{-x} ), we've shown that ( P_{2k}(x) ) is always positive for all real ( x ) and integers ( k geq 1 ). Therefore, the inequality holds.

answer：Okay, so I have this geometry problem here: I need to find the geometric locus of the midpoints of all segments where one end is on a given line and the other end is a fixed point not on that line. Hmm, let me try to visualize this.First, let me draw a line, let's call it line l. Then, I have a point A somewhere not on this line. Now, I need to consider all possible points B on line l, and for each such point B, I need to find the midpoint M of segment AB. The collection of all such midpoints M is what I'm looking for.Alright, so if I pick a point B on line l, the midpoint M will be halfway between A and B. If I move point B along line l, how does the midpoint M move? It seems like M should trace out some kind of path. The question is, what is that path?Let me think about coordinates for a moment. Maybe assigning coordinates will help me see the pattern. Let's place point A somewhere, say at (a, b), and let line l be the x-axis for simplicity. So, any point B on line l will have coordinates (x, 0). The midpoint M between A and B would then have coordinates ((a + x)/2, (b + 0)/2) = ((a + x)/2, b/2).Hmm, interesting. So, as x varies over all real numbers (since B can be anywhere on line l), the x-coordinate of M varies from (a + x)/2, which is (a/2 + x/2). The y-coordinate is fixed at b/2. That means all midpoints M lie on the horizontal line y = b/2. So, in this coordinate system, the locus is a straight line parallel to the x-axis, which is our original line l.Wait, so in this specific case, the locus is a line parallel to l. Does this hold in general, regardless of the position of A and the orientation of l? Let me test another configuration.Suppose line l is not the x-axis but some arbitrary line, say y = mx + c, and point A is at (p, q). Let me pick a general point B on line l, so its coordinates satisfy y = mx + c. Let me denote B as (x, mx + c). The midpoint M between A and B would have coordinates ((p + x)/2, (q + mx + c)/2).Now, let's see if this traces a line. Let me denote the coordinates of M as (h, k). So, h = (p + x)/2 and k = (q + mx + c)/2. Let me solve for x from the first equation: x = 2h - p. Substitute this into the second equation: k = (q + m(2h - p) + c)/2 = (q + 2mh - mp + c)/2 = (2mh + (q - mp + c))/2.Simplify this: k = mh + (q - mp + c)/2. So, this is a linear equation in h and k, meaning that the locus is a straight line. The slope of this line is m, which is the same as the slope of line l. Therefore, the locus is a line parallel to l.Wait, so in both cases, whether line l was horizontal or had some slope, the locus of midpoints M is a line parallel to l. That seems consistent. So, regardless of where A is, as long as it's not on line l, the midpoints of segments AB will trace out a line parallel to l.Let me think about another way to see this without coordinates. If I fix point A and consider all points B on line l, then for each B, M is the midpoint. So, as B moves along l, M moves in such a way that it's always halfway between A and B. So, intuitively, the movement of M should be constrained in a direction parallel to l because any movement of B along l would cause M to move in a direction that's half of that movement, but still parallel.Also, since M is halfway, the distance from M to A should be half the distance from B to A. But since B is on l, the set of all such M should form a line that's somehow related to l. The fact that it's parallel makes sense because the direction of movement of M is directly tied to the direction of l.Another way to think about it is using vectors. Let me denote vector AB as the vector from A to B. The midpoint M is given by (A + B)/2. So, if B moves along line l, then vector B can be expressed as a point on l, which can be written parametrically. Then, M would be (A + parametric expression of B)/2, which would also be a parametric line, hence a straight line.Moreover, the direction vector of the locus line should be the same as the direction vector of l because when you add vectors, the direction is preserved, just scaled. Since we're averaging A and B, the direction of the locus line remains parallel to l.Let me also consider the case where line l is vertical. Suppose l is the y-axis, so any point B on l has coordinates (0, y). Point A is somewhere else, say (a, b). The midpoint M would be ((a + 0)/2, (b + y)/2) = (a/2, (b + y)/2). As y varies, the x-coordinate of M is fixed at a/2, and the y-coordinate varies. So, the locus is a vertical line x = a/2, which is parallel to the y-axis (line l). Again, consistent with the idea that the locus is a line parallel to l.So, putting it all together, whether line l is horizontal, vertical, or has some other slope, the locus of midpoints M is always a line parallel to l. The specific position of this line depends on the position of point A relative to l, but its direction is fixed by the direction of l.I think I've convinced myself that the geometric locus is indeed a line parallel to the given line l. It makes sense from multiple perspectives: coordinate analysis, vector addition, and geometric intuition. So, I can confidently say that the answer is a line parallel to the given line.**Final Answer**The geometric locus is a line parallel to the given line, so the answer is boxed{text{A line parallel to the given line}}.

answer：First, I need to determine how long it takes for the worker to hear the explosion after it occurs. The explosion happens after 40 seconds, and the worker is running away from the explosion site at a speed of 5 yards per second. The sound of the explosion travels at 1100 feet per second.I'll start by converting the worker's speed from yards per second to feet per second to have consistent units. Since 1 yard is equal to 3 feet, the worker's speed is 15 feet per second.Let ( t ) be the time in seconds after the explosion occurs when the worker hears it. During this time, the worker has been running for ( t ) seconds, covering a distance of ( 15t ) feet. The sound of the explosion travels towards the worker at 1100 feet per second, covering a distance of ( 1100t ) feet in the same time.Since the worker is moving away from the explosion site, the distance the sound needs to cover to reach the worker is the initial 40-second head start plus the distance the worker continues to run. This gives the equation:[15t = 1100t - 44000]Solving for ( t ):[1085t = 44000][t = frac{44000}{1085} approx 40.55 text{ seconds}]Now, calculate the distance the worker has run in 40.55 seconds:[text{Distance} = 15 times 40.55 approx 608.25 text{ feet}]Convert this distance back to yards:[frac{608.25}{3} approx 202.75 text{ yards}]Rounding to the nearest whole number, the worker has run approximately 203 yards when he hears the explosion.