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已知幂函数f(x)=(n^2-2n+1)x^(n^2-2)在(0,+∞)上是增函数,向量a=(sinθ,-2﹚,向量b=(1,cosθ),g(x)=f(.
g(x)=f(sin+cos)+2√3·cos²x.
1)当向量a⊥向量b时,求g(θ)的值;
2)求g(x)的最值以及g(x)取最值时x的最值时x的取值集合
人气:393 ℃ 时间:2020-02-06 01:00:33
解答
1)幂函数f(x),所以n^2-2n+1=1, (n-1)^2=1 n-1=±1 n=2或0
当n=2时,f(x)=x^2; 当n=0时,f(x)=x^(-2)[舍去]
g(x)=(sinx+cosx)^2+2√3cos^2x=1+sin2x+√3cos2x+√3=2sin(2x+π/3)+1+√3
a*b=(sinθ,-2﹚*(1,cosθ)=sinθ-2cosθ=0 tanθ=2,
sin2θ=2tanθ/(1+tan^2θ)=4/5, cos2θ=(1-tan^2θ)/(1+tan^2θ)= -3/5
g(θ)=2sin(2θ+π/3)+1+√3=2(sin2θcosπ/3+cos2θsinπ/3)+1+√3
=2(4/5*1/2+(-3/5) √3/2)=(9+4√3)/5
2)g(x)= 2sin(2x+π/3)+1+√3,最大值=3+√3 当2x+π/3=2kπ+π/2,x=kπ+π/12
最小值=-1+√3,当2x+π/3=2kπ-π/2 ,x=kπ-5π/12 (k∈Z)
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