Question

高一数学 急求！

Answer 1

1.
sin(π-x)=sinx
要在[π/3，π]上有两个实根，两实根互不相等且互为补角。
π/3≤x≤2π/3且x≠π/2
(1-a)/2≠π/2，解得a≠1-π
π/3≤(1-a)/2≤2π/3
-4π/3≤a-1≤-2π/3
1-4π/3≤a≤1-2π/3
综上，得：1- 4π/3≤a≤1- 2π/3且a≠1-π
a的取值范围为[1-4π/3，1-π)U(1-π，1- 2π/3]
2.
k∈N*
f(k)+f(2012-k)
=cos(kπ/4)+cos[(2012-k)π/4]
=cos(kπ/4)+cos(502π+π-kπ/4)
=cos(kπ/4)-cos(kπ/4)
=0
f(1)+f(2)+...+f(2010)+f(2011)
=[f(1)+f(2011)]+[f(2)+f(2010)]+...+[f(1005)+f(1007)]+f(1006)
=0+0+...+0+cos(1006·π/4)
=cos(252π-π/2)
=cos(-π/2)
=0