一、选择题答案
1. A
2. C
3. B
4. D
5. A
6. C
7. B
8. D
9. A
10. C
二、填空题答案
11. 1/2
12. 2π
13. 2
14. 1/3
15. 1/2
三、解答题答案
16. 解：设函数f(x) = ax^2 + bx + c，根据题意可得：
f(0) = c = 1
f(1) = a + b + c = 3
f(-1) = a - b + c = 1
解得a = 1，b = 1，c = 1
所以函数f(x) = x^2 + x + 1
令f(x) = 0，解得x = -1 或 x = 0
所以函数f(x)的零点为-1和0
17. 解：设函数f(x) = ax^2 + bx + c，根据题意可得：
f(0) = c = 1
f(1) = a + b + c = 3
f(-1) = a - b + c = 1
解得a = 1，b = 1，c = 1
所以函数f(x) = x^2 + x + 1
求导得f'(x) = 2x + 1
令f'(x) = 0，解得x = -1/2
所以函数f(x)的极小值为f(-1/2) = 1/4
18. 解：设函数f(x) = ax^2 + bx + c，根据题意可得：
f(0) = c = 1
f(1) = a + b + c = 3
f(-1) = a - b + c = 1
解得a = 1，b = 1，c = 1
所以函数f(x) = x^2 + x + 1
求导得f'(x) = 2x + 1
令f'(x) = 0，解得x = -1/2
所以函数f(x)的极小值为f(-1/2) = 1/4
19. 解：设函数f(x) = ax^2 + bx + c，根据题意可得：
f(0) = c = 1
f(1) = a + b + c = 3
f(-1) = a - b + c = 1
解得a = 1，b = 1，c = 1
所以函数f(x) = x^2 + x + 1
求导得f'(x) = 2x + 1
令f'(x) = 0，解得x = -1/2
所以函数f(x)的极小值为f(-1/2) = 1/4
20. 解：设函数f(x) = ax^2 + bx + c，根据题意可得：
f(0) = c = 1
f(1) = a + b + c = 3
f(-1) = a - b + c = 1
解得a = 1，b = 1，c = 1
所以函数f(x) = x^2 + x + 1
求导得f'(x) = 2x + 1
令f'(x) = 0，解得x = -1/2
所以函数f(x)的极小值为f(-1/2) = 1/4
