answer:Solution: (Ⅰ) The domain of f(x) is (0,+infty), therefore f'(x)= frac{{e}^{x}cdot{x}^{2}-{e}^{x}cdot2x}{{x}^{4}}-kleft( frac{1}{x}- frac{2}{{x}^{2}}right)= frac{(x-2)({e}^{x}-kx)}{{x}^{3}}(x > 0), When kleqslant 0, kxleqslant 0, therefore e^{x}-kx > 0, Let f′(x)=0, then x=2, therefore When 0 < x < 2, f′(x) < 0, f(x) is monotonically decreasing; When x > 2, f′(x) > 0, f(x) is monotonically increasing, therefore f(x)'s monotonically decreasing interval is (0,2), and the monotonically increasing interval is (2,+infty). (Ⅱ) From (Ⅰ), we know that when kleqslant 0, the function f(x) is monotonically decreasing in (0,2), Therefore, f(x) does not have extreme points in (0,2); When k > 0, let the function g(x)=e^{x}-kx, xin(0,+infty), because g′(x)=e^{x}-k=e^{x}-e^{ln k}, When 0 < kleqslant 1, for xin(0,2), g′(x)=e^{x}-k > 0, y=g(x) is monotonically increasing, Therefore, f(x) does not have two extreme points in (0,2); When k > 1, we get for xin(0,ln k), g′(x) < 0, the function y=g(x) is monotonically decreasing, For xin(ln k,+infty), g′(x) > 0, the function y=g(x) is monotonically increasing, therefore The function y=g(x)'s minimum value is g(ln k)=k(1-ln k), the function f(x) has two extreme points in (0,2),   if and only if begin{cases}g(0) > 0 g(ln k) < 0 g(2) > 0 0 < ln k < 2end{cases},  Solving, we get: e < k < frac{{e}^{2}}{2}, In summary, when the function f(x) has two extreme points in (0,2), the range of values for k is boxed{(e, frac{{e}^{2}}{2})}.

answer:Let's denote the ending number as ( n ). We are looking for numbers between 10 and ( n ) that are divisible by 9. The first number divisible by 9 after 10 is 18, and the last number divisible by 9 before ( n ) will be ( n ) itself if ( n ) is divisible by 9, or the largest number less than ( n ) that is divisible by 9. The numbers divisible by 9 form an arithmetic sequence with the first term ( a_1 = 18 ) and the common difference ( d = 9 ). If ( n ) is the last term of this sequence, then ( n ) must be of the form ( 18 + 9k ), where ( k ) is the number of terms after the first one. The average of an arithmetic sequence is given by the formula: [ text{Average} = frac{text{First term} + text{Last term}}{2} ] Given that the average is 49.5, we have: [ 49.5 = frac{18 + n}{2} ] Now, let's solve for ( n ): [ 99 = 18 + n ] [ n = 99 - 18 ] [ n = 81 ] Since 81 is divisible by 9, it is the last term of the sequence. Therefore, the ending number is boxed{81} .

answer:Nous devons trouver le cardinal moyen d'un sous-ensemble de {1, 2, ldots, n}. Première méthode : En regroupant chaque ensemble avec son complémentaire, nous pouvons calculer le cardinal moyen de la manière suivante : 1. Considérons un sous-ensemble (A subset {1, 2, ldots, n}) et son complémentaire (A^c), qui est l'ensemble des éléments de ({1, 2, ldots, n}) qui ne sont pas dans (A). 2. Pour tous les sous-ensembles (A), nous avons la propriété que la somme des cardinaux de (A) et de son complémentaire (A^c) est égale à (n) : [ operatorname{Card}(A) + operatorname{Card}(A^c) = n ] 3. Calculons le cardinal moyen (N) : [ N = frac{1}{2^n} sum_{A subset {1, 2, ldots, n}} operatorname{Card}(A) ] 4. Remarquons que pour chaque paire ((A, A^c)), on a : [ frac{operatorname{Card}(A) + operatorname{Card}(A^c)}{2} = frac{n}{2} ] 5. Donc la somme totale est : [ sum_{A subset {1, 2, ldots, n}} operatorname{Card}(A) = sum_{A subset {1, 2, ldots, n}} frac{n}{2^n} = 2^n cdot frac{n}{2^n} = n ] [ N = frac{n}{2} ] Conclusion : Le cardinal moyen d'un sous-ensemble de ({1, 2, ldots, n}) est boxed{frac{n}{2}}. Seconde méthode : Nous devons évaluer la somme : [ S_n = frac{1}{2^n} sum_{k=0}^n k binom{n}{k} ] 1. Observe that the summation is equivalent to counting the occurrences of elements in all subsets. 2. Each element appears in exactly half of the subsets, hence reducing the summation to: [ sum_{k=0}^n k binom{n}{k} = n cdot 2^{n-1} ] 3. Dividing by (2^n): [ S_n = frac{n cdot 2^{n-1}}{2^n} = frac{n}{2} ] Conclusion : Le cardinal moyen est donc boxed{frac{n}{2}}. Troisième méthode : Utilisons le polynôme générateur (P_n(x)=sum_{k=0}^n binom{n}{k} x^k). 1. D'après la formule du binôme de Newton : [ P_n(x) = (1+x)^n ] 2. Dérivons cette égalité par rapport à (x): [ frac{d}{dx} P_n(x) = sum_{k=1}^n k binom{n}{k} x^{k-1} = n(1+x)^{n-1} ] 3. En évaluant cette dérivée en (x = 1): [ sum_{k=1}^n k binom{n}{k} = n cdot 2^{n-1} ] 4. En divisant par (2^n) pour obtenir la moyenne : [ S_n = frac{n cdot 2^{n-1}}{2^n} = frac{n}{2} ] Conclusion : Le cardinal moyen est donc boxed{frac{n}{2}}.

answer:To achieve a digit sum of 23, we must consider a 3-digit prime or a small 4-digit prime, since the largest possible digit sum for a 2-digit number is 18. Let's consider the smallest 3-digit numbers whose digits sum up to 23: - If the first digit is 1, an impossible situation arises as the other two digits would need to sum to 22, which is not possible. - If the first digit is 2, the smallest possible number is 299 (since 2 + 9 + 9 = 20, and 2 + 9 + 10 is not possible due to 10 being a two-digit number). - If the first digit is 8, the smallest possible number is 887 (since 8 + 8 + 7 = 23). Now we check if 887 is prime. Since the square root of 887 is approximately 29.8, we only need to check divisibility by prime numbers less than or equal to 29: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. - 887 is odd, so it is not divisible by 2. - The sum of the digits 8 + 8 + 7 = 23, which is not divisible by 3. - The number does not end in 0 or 5, so it is not divisible by 5. - By manually checking, 887 is not divisible by 7, 11, 13, 17, 19, 23, or 29. Since 887 is not divisible by any prime number up to 29, it is a prime number. Thus, the smallest prime number whose digits sum to 23 is boxed{887}.