已知函数f(x)=sinx+tanx,项数为27的等差数列An满足An∈(-π/2,π/2),且公差d≠0.若f(a1)+f(a2)+f(a3)+.+f(a27)=0,则当k=?时,f(Ak)=0
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![已知函数f(x)=sinx+tanx,项数为27的等差数列An满足An∈(-π/2,π/2),且公差d≠0.若f(a1)+f(a2)+f(a3)+.+f(a27)=0,则当k=?时,f(Ak)=0](/uploads/image/z/1324576-64-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsinx%2Btanx%2C%E9%A1%B9%E6%95%B0%E4%B8%BA27%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97An%E6%BB%A1%E8%B6%B3An%E2%88%88%28-%CF%80%2F2%2C%CF%80%2F2%29%2C%E4%B8%94%E5%85%AC%E5%B7%AEd%E2%89%A00.%E8%8B%A5f%EF%BC%88a1%EF%BC%89%2Bf%EF%BC%88a2%EF%BC%89%2Bf%EF%BC%88a3%EF%BC%89%2B.%2Bf%28a27%29%3D0%2C%E5%88%99%E5%BD%93k%3D%3F%E6%97%B6%2Cf%28Ak%29%3D0)
已知函数f(x)=sinx+tanx,项数为27的等差数列An满足An∈(-π/2,π/2),且公差d≠0.若f(a1)+f(a2)+f(a3)+.+f(a27)=0,则当k=?时,f(Ak)=0
已知函数f(x)=sinx+tanx,项数为27的等差数列An满足An∈(-π/2,π/2),且公差d≠0.
若f(a1)+f(a2)+f(a3)+.+f(a27)=0,则当k=?时,f(Ak)=0
已知函数f(x)=sinx+tanx,项数为27的等差数列An满足An∈(-π/2,π/2),且公差d≠0.若f(a1)+f(a2)+f(a3)+.+f(a27)=0,则当k=?时,f(Ak)=0
①{an}是等差数列,容易看出,当a14=0时,a1+a27=a2+a26=.=a13+a15=0,且f(a14)=f(0)=0,
又易f(x)是x∈(-π/2,π/2)奇函数且单调递增,所以
S=f(a1)+f(a2)+.+f(a13)+f(a14)+f(15)+.+f(a27)
=f(a1)+f(a2)+.+f(a13)+0+f(-a13)+f(-a12)+.+f(-a1)=0
②由a1与a27,a2与a26,.,a13与a15关于a14对称,且a14=0时,S=0
再由f(x)是增函数知,当a14>0时,27个点整体向右移动(与a14=0时比较),所得函数值比a14=0时大,所以 S>0;
③同理,当a14
因为f(x)是奇函数,且f(a1)+f(a2)+f(a3)+....+f(a27)=0,可知a1到a27是一组对称且互为相反数,又因为An是项数为27的等差数列,所以An不是递增就是递减,综上所述可得
a1=-a27,a2=-a26......a13=-a15,所以f(a14)=0
即k=14时,f(Ak)=0
sinx与tanx在[-90,90]度内关于原点对称,所以此数列应该在此区间内
x在0度时f(x)=0,即ak=0
a1+a27=a2+a26=......=2a14=0
k=14时,f(Ak)=0