answer:Given: [ f(x) = x^2 + a x + b quad text{where} ; a, b in mathbb{R} ] Set ( A = { x mid f(x) = x, x in mathbb{R} } ) and ( B = { x mid f(f(x)) = x, x in mathbb{R} } ). We know ( A = {-1, 3} ). 1. **Solve for a and b using A**: Since A = {-1, 3}, we know that ( f(-1) = -1 ) and ( f(3) = 3 ). Let's start with: [ f(x) = x ] This implies: [ x^2 + a x + b = x ] Simplify this to: [ x^2 + (a-1)x + b = 0 ] Since the roots are given by (-1) and (3), we use these to form the factorized quadratic equation: [ (x + 1)(x - 3) = x^2 - 2 x - 3 ] Thus, we compare: [ x^2 + (a-1)x + b = x^2 - 2 x - 3 ] This gives us: [ a - 1 = -2 ] [ b = -3 ] Solving for (a): [ a = -2 + 1 = -1 ] Thus, we have: [ a = -1 quad text{and} quad b = -3 ] 2. **Rewrite the function f(x)**: Now we have: [ f(x) = x^2 - x - 3 ] 3. **Find B for f(f(x)) = x**: We need to solve ( f(f(x)) = x ). First, apply (f(x)) to itself: [ f(f(x)) = f(x^2 - x - 3) ] Substitute (x^2 - x - 3) into (f): [ f(x^2 - x - 3) = (x^2 - x - 3)^2 - (x^2 - x - 3) - 3 ] Expand ((x^2 - x - 3)^2): [ (x^2 - x - 3)^2 = x^4 - 2x^3 + x^2 - 6x^2 + 6x + 9 = x^4 - 2x^3 - 5x^2 + 6x + 9 ] Thus: [ f(x^2 - x - 3) = x^4 - 2x^3 - 5x^2 + 6x + 9 - x^2 + x + 3 - 3 ] Simplify this to: [ f(x^2 - x - 3) = x^4 - 2x^3 - 6x^2 + 7x + 9 ] Now we set ( f(f(x)) = x ): [ x^4 - 2x^3 - 6x^2 + 7x + 9 = x ] Simplifying gives: [ x^4 - 2x^3 - 6x^2 + 6x + 9 = 0 ] We need to find the real roots. Through factorization and solving, we get the roots: [ x = -1, 3, sqrt{3}, -sqrt{3} ] Therefore: [ B = {-1, sqrt{3}, -sqrt{3}, 3} ] # Conclusion: [ boxed{{-1, sqrt{3}, -sqrt{3}, 3}} ]

answer:To understand the relationship between the sets M and N, let's first clearly define each set based on the given conditions: For M={x|0 lt xleqslant 3}, this means M includes all numbers greater than 0 and up to and including 3. For N={x|0 lt xleqslant 2}, this indicates N includes all numbers greater than 0 and up to and including 2. Given these definitions, we observe that every element in N is also in M, because all numbers greater than 0 and up to 2 are also included in the range up to 3. This observation can be mathematically expressed as Nsubset M. Now, regarding the statement "ain M" as a condition for "ain N": - Because Nsubset M, if ain N, then ain M. This shows that being in M is a **necessary condition** for being in N, because without being in M, it is impossible to be in N. - However, not all elements of M are in N, since M includes numbers between 2 and 3 that are not included in N. Thus, ain M does not guarantee ain N, making it **not a sufficient condition**. Therefore, we conclude that "ain M" is a **necessary but not sufficient condition** for "ain N". The correct answer to the question is thus: boxed{text{B}}: Necessary but not sufficient condition.

answer:First, we need to determine the formulas for S_n when n is odd and when n is even, as defined from previous knowledge. 1. **When n is even**, the sequence pairs up completely, with each pair summing to -1. There are n/2 such pairs, so S_n = -n/2. 2. **When n is odd**, the sequence pairs up leaving one unpaired number at the end. Each pair sums to -1, and there are (n-1)/2 pairs. The last term, a positive n, adds to the total, so S_n = -frac{n-1}{2} + n = frac{n+1}{2}. Calculating the specific values: - S_{15}: Since 15 is odd, S_{15} = frac{15+1}{2} = frac{16}{2} = 8. - S_{20}: Since 20 is even, S_{20} = -frac{20}{2} = -10. - S_{35}: Since 35 is odd, S_{35} = frac{35+1}{2} = frac{36}{2} = 18. Combine these values: S_{15} + S_{20} + S_{35} = 8 - 10 + 18 = 16. Conclusion: The final computed sum is 16. The final answer is boxed{text{(D) } 16}.

answer:Let's set (AB=BC=1). Then, since (cos B=-dfrac{7}{18}), we have (AC^2=AB^2+BC^2-2ABcdot BCcdotcos B=dfrac{25}{9}), thus (AC=dfrac{5}{3}), so (2a=1+dfrac{5}{3}=dfrac{8}{3}), since (2c=1), we get (e=dfrac{2c}{2a}=dfrac{3}{8}). Therefore, the eccentricity of the ellipse is boxed{dfrac{3}{8}}.