answer:Since we know that f(x) = 2sinalphacos x - 2sinalpha = 2sinalpha(cos x - 1), and given f(x) geq 0 for all x in mathbb{R}, Since cos x - 1 leq 0 for all x, it must then follow that sinalpha leq 0. Therefore, we have -frac{pi}{2} leq alpha leq 0. With tan(2alpha) = frac{3}{4}, we have the equation frac{2tanalpha}{1 - tan^2alpha} = frac{3}{4}. Solving this equation yields tanalpha = -3 (the solution frac{1}{3} is discarded because it doesn't satisfy the sign of sinalpha). Thus, sinalpha = -frac{3}{sqrt{10}} and cosalpha = frac{1}{sqrt{10}}. So, sinleft(alpha - frac{pi}{4}right) = frac{sqrt{2}}{2}(sinalpha - cosalpha) = frac{sqrt{2}}{2} cdot left(-frac{3}{sqrt{10}} - frac{1}{sqrt{10}}right) = -frac{sqrt{2}}{2} cdot frac{4}{sqrt{10}}. Finally, we simplify to get sinleft(alpha - frac{pi}{4}right) = -frac{2sqrt{5}}{5}. boxed{sinleft(alpha - frac{pi}{4}right) = -frac{2sqrt{5}}{5}}.

answer:1. Given the sequence defined by ( a_{0} = 0 ) and the recurrence relation: [ a_{n+1} = (k+1) a_{n} + kleft(a_{n}+1right) + 2sqrt{k(k+1) a_{n}left(a_{n}+1right)}, ] where ( k ) is a given natural number, we seek to prove that ( a_{n} ) is an integer for all ( n geq 0 ). 2. From the recurrence relation, we first simplify it: [ a_{n+1} = (k+1)a_{n} + k a_{n} + k + 2sqrt{k(k+1) a_{n}(a_{n} + 1)} ] [ = (2k+1) a_{n} + k + 2sqrt{k(k+1) a_{n}(a_{n}+1)}. ] 3. Next, square both sides to eliminate the square root: [ [a_{n+1} - (2k+1) a_{n} - k]^2 = [2sqrt{k(k+1) a_{n}(a_{n}+1)}]^2 ] [ a_{n+1}^2 - 2a_{n+1}((2k+1) a_{n} + k) + [(2k+1) a_{n} + k]^2 = 4k(k+1) a_{n}(a_{n} + 1) ] [ a_{n+1}^2 - 2(2k+1) a_{n} a_{n+1} - 2k a_{n+1} + (2k+1)^2 a_{n}^2 + 2k(2k+1) a_{n} + k^2 = 4k(k+1) a_{n}(a_{n} + 1). ] 4. Simplify and rearrange terms: [ a_{n+1}^2 + (2k^2 + 2k + 1)a_{n}^2 + 2k^2 a_{n} + k^2 - (4k+2) a_{n} a_{n+1} - 2k a_{n+1}(a_{n}+1) = 4k(k+1) a_{n}(a_{n}+1). ] 5. Notice that simplifying further results in the following equation: [ a_{n}^{2} + a_{n+1}^{2} - 4(k+2) a_{n+1} a_{n} - 2k (a_{n+1} + a_{n}) + k^2 = 0. ] 6. Now, substitute ( a_{n} ) with ( a_{n-1} ) in the previous equation: [ a_{n-1}^{2} + a_{n}^{2} - (4k+2) a_{n} a_{n-1} - 2k(a_{n} + a_{n-1}) + k^2 = 0. ] 7. Subtract the equation for ( n-1 ) from the equation for ( n ): [ a_{n+1}^2 - a_{n-1}^2 - (4k+2) a_{n}(a_{n+1} - a_{n-1}) - 2k(a_{n+1} - a_{n-1}) = 0. ] 8. Since ( a_{n+1} - a_{n-1} > 0 ) (as the sequence is strictly increasing by earlier analysis): [ a_{n+1} + a_{n-1} - (4k+2) a_{n} - 2k = 0. ] 9. This simplifies to the second recurrence relation: [ a_{n+1} = (4k + 2) a_{n} - a_{n-1} + 2k, quad n = 1, 2, ldots ] 10. Using the initial condition ( a_{0} = 0 ) and ( a_{1} = k ), and knowing ( k ) is a natural number, we prove by induction that ( a_{n} ) is an integer for all ( n ). # Conclusion: [ boxed{a_{n} text{ is an integer for all } n geq 0} ]

answer:Let x+1=t, then x=t-1, The equation is equivalent to [(t-1)^{2}+1](2t+1)^{9}=a_0+a_1t+a_2t^{2}+...+a_{11}t^{11}, That is, (t^{2}-2t+2)(2t+1)^{9}=a_0+a_1t+a_2t^{2}+...+a_{11}t^{11}, a_{11} is the coefficient of t^{11} in the expanded form, so a_{11}=2^{9}=512 a_1 is the coefficient of the first degree term, so a_1=-2×1+1× C_{ 9 }^{ 1 }×2=18-2=16. a_2 is the coefficient of the second degree term, so a_2=1×1-2× C_{ 9 }^{ 1 }×2+2× C_{ 9 }^{ 2 }×2^{2}=1-36+288=253. Thus, a_1+a_2+a_{11}=16+253+512=781, Therefore, the answer is: boxed{781}. By using the substitution method, setting x+1=t, the equation is transformed into a polynomial in terms of t, and the solution can be found based on the relationship between the coefficients. This problem primarily tests the application of the binomial theorem, and using the substitution method to transform the equation into a polynomial in terms of t is the key to solving this problem.

answer:Dhoni spent 25% of his earnings on rent. He then spent 10% less than that amount on a new dishwasher. First, let's find out what 10% of the rent amount is: 10% of 25% = 0.10 * 25 = 2.5% Now, let's find out how much he spent on the dishwasher: 25% - 2.5% = 22.5% So, Dhoni spent 22.5% of his earnings on the dishwasher. Now, let's add up the total percentage of his earnings that he spent: 25% (rent) + 22.5% (dishwasher) = 47.5% Finally, let's find out what percentage he had left over: 100% (total earnings) - 47.5% (spent) = 52.5% Dhoni had boxed{52.5%} of his last month's earnings left over.