一到高考数学题)设函数f(x)=x/2+sinx的所有正的极小值点从小到大排成的数列为{xn}.(Ⅰ)求数列{xn}的通项公式; (Ⅱ)设的前项和为Sn,求sinSn.标准答案f(x)=x/2 + sin(x),f'(x) = 1/2 + cos(x),令0=f'(x),
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![一到高考数学题)设函数f(x)=x/2+sinx的所有正的极小值点从小到大排成的数列为{xn}.(Ⅰ)求数列{xn}的通项公式; (Ⅱ)设的前项和为Sn,求sinSn.标准答案f(x)=x/2 + sin(x),f'(x) = 1/2 + cos(x),令0=f'(x),](/uploads/image/z/13401844-52-4.jpg?t=%E4%B8%80%E5%88%B0%E9%AB%98%E8%80%83%E6%95%B0%E5%AD%A6%E9%A2%98%EF%BC%89%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dx%2F2%2Bsinx%E7%9A%84%E6%89%80%E6%9C%89%E6%AD%A3%E7%9A%84%E6%9E%81%E5%B0%8F%E5%80%BC%E7%82%B9%E4%BB%8E%E5%B0%8F%E5%88%B0%E5%A4%A7%E6%8E%92%E6%88%90%E7%9A%84%E6%95%B0%E5%88%97%E4%B8%BA%7Bxn%7D.%EF%BC%88%E2%85%A0%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Bxn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B+%EF%BC%88%E2%85%A1%EF%BC%89%E8%AE%BE%E7%9A%84%E5%89%8D%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82sinSn.%E6%A0%87%E5%87%86%E7%AD%94%E6%A1%88f%28x%29%3Dx%2F2+%2B+sin%28x%29%2Cf%27%28x%29+%3D+1%2F2+%2B+cos%28x%29%2C%E4%BB%A40%3Df%27%28x%29%2C)
一到高考数学题)设函数f(x)=x/2+sinx的所有正的极小值点从小到大排成的数列为{xn}.(Ⅰ)求数列{xn}的通项公式; (Ⅱ)设的前项和为Sn,求sinSn.标准答案f(x)=x/2 + sin(x),f'(x) = 1/2 + cos(x),令0=f'(x),
一到高考数学题)设函数f(x)=x/2+sinx的所有正的极小值点从小到大排成的数列为{xn}.(Ⅰ)求数
列{xn}的通项公式;
(Ⅱ)设的前项和为Sn,求sinSn.
标准答案f(x)=x/2 +
sin(x),
f'(x) = 1/2 + cos(x),令0=f'(x),有,-1/2 = cos(x),
x=2kπ+4π/3或x=2kπ-4π/3,k=0,1,-1,2,-2,...,n,-n,...
f''(x) = - sin(x),
x为极小值点,则-sin(x)=f''(x)>=0,因此,x=2kπ+4π/3,
k=0,1,-1,2,-2,...,n,-n,...
x(n)=2(n-1)π+4π/3,
s(n)=n(n-1)π+4nπ/3,
n=3m,
s(n)=3m(3m-1)π+4mπ,sin[s(n)] = sin(0)=0,n=3m,m=1,2,...
n=3m-1,
s(n)=(3m-1)(3m-2)π + 4(3m-1)π/3 = (3m-1)(3m-2)π + 4mπ - 4π/3,
sin[s(n)] =
sin(-4π/3)=sin(2π/3)=sin(π/3)=3^(1/2)/2,n=3m-1,m=1,2,...
n=3m-2,s(n) =
(3m-2)(3m-3)π +4(3m-2)π/3 = (3m-2)(3m-3)π + 4mπ - 8π/3,
sin[s(n)] =
sin(-8π/3)=sin(-2π/3)=-sin(2π/3)=-sin(π/3)=-3^(1/2)/2,
n=3m-2,m=1,2,...
综合,有,
n=3m-2时,sin[s(n)]=-3^(1/2)/2,
n=3m-1时,sin[s(n)]=
3^(1/2)/2,
n=3m时,sin[s(n)]=0,
m=1,2,...
为什么要设3m,3m-1……
一到高考数学题)设函数f(x)=x/2+sinx的所有正的极小值点从小到大排成的数列为{xn}.(Ⅰ)求数列{xn}的通项公式; (Ⅱ)设的前项和为Sn,求sinSn.标准答案f(x)=x/2 + sin(x),f'(x) = 1/2 + cos(x),令0=f'(x),
这是分类讨论,把n分成三类,即3m,3m-1和3m-2,因为s(n)=n(n-1)π+4nπ/3,n若为3的倍数,则4nπ/3为π的偶数倍,其sin值为0,所以为了方便计算把n按除以3得到的余数不同分成三类.