answer:Since the quadratic function f(x)=ax^{2}+2a is an even function on the interval [-a,a^{2}], we have a=a^{2}, which yields a=1 or a=0 (the latter being discarded). Thus, f(x)=x^{2}+2, and consequently, g(x)=f(x-1)=(x-1)^{2}+2 is a quadratic function, whose graph is symmetric about the line x=1 and opens upwards as a parabola. Given that | frac {3}{2}-1| < |0-1| < |3-1|, we conclude that g( frac {3}{2}) < g(0) < g(3). Therefore, the answer is: boxed{A}. The problem provides the condition a=a^{2}, from which we find a=1, leading to g(x)=f(x-1)=(x-1)^{2}+2. Using the graph and properties of a quadratic function, we determine the relationship among g(0), g( frac {3}{2}), and g(3). This problem primarily tests the application of function's even/odd symmetry, the graph and properties of a quadratic function, and can be considered a basic problem.

answer:1. **Identify the conditions:** - ( n ) is a palindrome. - ( n equiv 2 pmod{3} ) - ( n equiv 3 pmod{4} ) - ( n equiv 0 pmod{5} ) 2. **Use the Chinese Remainder Theorem (CRT):** - We need to solve the system of congruences: [ begin{cases} n equiv 2 pmod{3} n equiv 3 pmod{4} n equiv 0 pmod{5} end{cases} ] 3. **Solve the system step-by-step:** - Start with ( n equiv 0 pmod{5} ). Let ( n = 5k ). - Substitute ( n = 5k ) into the second congruence: ( 5k equiv 3 pmod{4} ). - Since ( 5 equiv 1 pmod{4} ), we have ( k equiv 3 pmod{4} ). - Let ( k = 4m + 3 ). Then ( n = 5(4m + 3) = 20m + 15 ). - Substitute ( n = 20m + 15 ) into the first congruence: ( 20m + 15 equiv 2 pmod{3} ). - Since ( 20 equiv 2 pmod{3} ) and ( 15 equiv 0 pmod{3} ), we have ( 2m equiv 2 pmod{3} ). - Thus, ( m equiv 1 pmod{3} ). - Let ( m = 3p + 1 ). Then ( n = 20(3p + 1) + 15 = 60p + 35 ). 4. **Find the smallest palindrome:** - We need to find the smallest palindrome of the form ( 60p + 35 ). - Check values of ( p ): - For ( p = 0 ), ( n = 35 ) (not a palindrome). - For ( p = 1 ), ( n = 95 ) (not a palindrome). - For ( p = 2 ), ( n = 155 ) (not a palindrome). - For ( p = 3 ), ( n = 215 ) (not a palindrome). - For ( p = 4 ), ( n = 275 ) (not a palindrome). - For ( p = 5 ), ( n = 335 ) (not a palindrome). - For ( p = 6 ), ( n = 395 ) (not a palindrome). - For ( p = 7 ), ( n = 455 ) (not a palindrome). - For ( p = 8 ), ( n = 515 ) (a palindrome). 5. **Prove there are infinitely many such numbers:** - Consider numbers of the form ( 5underbrace{11 ldots 1}_{3r+1}5 ). - These numbers are palindromes and can be written as ( 60k + 35 ) for some ( k ). - For example, ( 5115, 511115, ldots ) are all palindromes and satisfy the given conditions. (blacksquare) The final answer is ( boxed{ 515 } )

answer:**(1)** Given f(x)=kln x+frac{1}{{{e^x}}}({k∈{R}}), the domain of the function is (0,+infty). To find the range of k for which f(x) is not monotonic on (2,3), we first find the derivative of f(x): [f'(x) = frac{k}{x} - frac{1}{{e}^x}.] For f(x) to be not monotonic on (2,3), f'(x)=0 must have real roots in (2,3), implying: [k = frac{x}{e^x}.] Let g(x) = frac{x}{{e}^x}, with {g'}(x) = frac{1-x}{{e}^x}. Analyzing the sign of g'(x): - For 2 < x < 3, we have {g'}(x) < 0, indicating g(x) is monotonically decreasing in (2,3). Evaluating g(x) at the endpoints of the interval: - g(2) = frac{2}{e^2}, - g(3) = frac{3}{e^3}. Since g(x) is decreasing in (2,3), the range of k is: [frac{3}{e^3} < k < frac{2}{e^2}.] Therefore, the range of values for k is boxed{left(frac{3}{e^3}, frac{2}{e^2}right)}. **(2)(ⅰ)** To prove frac{e}{{{e^{{x_2}}}}}-frac{e}{{{e^{{x_1}}}}}＞-lnfrac{{{x_2}}}{{{x_1}}}＞1-frac{{{x_2}}}{{{x_1}}}, we start with the function g(x) = frac{x}{{e}^x}, where {g'}(x) = frac{1-x}{{e}^x}. - For 0 < x < 1, {g'}(x) > 0, making g(x) monotonically increasing. - For x > 1, {g'}(x) < 0, making g(x) monotonically decreasing. Thus, g(x) leq g(1) = frac{1}{e}, leading to frac{1}{e} geq frac{x}{{e}^x}. Given k = frac{1}{e}, we find that f(x) = frac{ln x}{e} + frac{1}{{e}^x} is monotonically increasing in its domain. Given 0 < x_1 < x_2, it follows that f(x_2) > f(x_1), leading to: [frac{e}{{e}^{x_2}} - frac{e}{{e}^{x_1}} > -ln frac{x_2}{x_1}.] Let h(x) = x - ln x - 1, with {h'}(x) = 1 - frac{1}{x}. Analyzing the sign of h'(x): - For 0 < x < 1, {h'}(x) < 0, making h(x) monotonically decreasing. - For x > 1, {h'}(x) > 0, making h(x) monotonically increasing. Thus, -ln x geq 1 - x. Letting t = frac{x_2}{x_1}, where t > 1, we get: [-ln t > 1 - t.] Therefore, boxed{frac{e}{{{e^{{x_2}}}}}-frac{e}{{{e^{{x_1}}}}}＞-lnfrac{{{x_2}}}{{{x_1}}}＞1-frac{{{x_2}}}{{{x_1}}}}. **(2)(ⅱ)** If frac{x_1}{{e^{x_1}}} = frac{x_2}{{e^{x_2}}} = k, then frac{k}{x_1} - frac{1}{{e^{x_1}}} = frac{k}{x_2} - frac{1}{{e^{x_2}}} = 0. Given 0 < x_1 < 1 < x_2, we find that f(x_1) = kln x_1 + frac{1}{{e^{x_1}}} and f(x_2) = kln x_2 + frac{1}{{e^{x_2}}}. Let m(x) = frac{xln x + 1}{{e^x}}, with {m'}(x) = frac{(1-x)ln x}{{e^x}}. Analyzing the sign of m'(x): - For 0 < x < 1, (1-x)ln x < 0; for x > 1, (1-x)ln x > 0. Since e^x > 0 always, m(x) is monotonically decreasing in its domain. Thus, 0 < f(x_2) < frac{1}{e} < f(x_1) < 1. Therefore, |f(x_1) - f(x_2)| = f(x_1) - f(x_2) < 1, leading to boxed{|f(x_{1})-f(x_{2})| < 1}.

answer:We start by converting the equation of circle C into its standard form. The given equation for circle C is x^2 + y^2 - 4x + 6y - 3 = 0. We complete the square for both x and y terms: [ begin{align*} x^2 - 4x + y^2 + 6y &= 3 (x^2 - 4x + 4) + (y^2 + 6y + 9) &= 3 + 4 + 9 (x - 2)^2 + (y + 3)^2 &= 16 end{align*} ] From this, we deduce that the center of circle C is at (2, -3). Now, we calculate the distance |CM| from the center C to point M. Using the distance formula between the points (2, -3) and (-1, 1), we get: [ |CM| = sqrt{(2 - (-1))^2 + (-3 - 1)^2} = sqrt{(3)^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5 ] Since the new circle has the same center as circle C, but now passes through point M and we have found the distance from the center of the circle to point M (the radius) to be 5, we write the equation of the new circle with radius 5: [ (x - 2)^2 + (y + 3)^2 = 5^2 = 25 ] Therefore, the equation of the new circle is: [ boxed{(x - 2)^2 + (y + 3)^2 = 25} ]