函数f(x,θ)=x2-x-xsinθ+8x-1-sinθ(x>2)的最小值为()A.42B.22C.1+42D.-1+42-高三数学

题目简介

函数f(x,θ)=x2-x-xsinθ+8x-1-sinθ(x>2)的最小值为()A.42B.22C.1+42D.-1+42-高三数学

题目详情

函数f(x,θ)=
x2-x-xsinθ+8
x-1-sinθ
(x>2)的最小值为(  )
A.4
2
B.2
2
C.1+4
2
D.-1+4
2
题型:单选题难度:偏易来源:不详

答案

∵x>2,
∴x-1-sinθ>0,
f(x,θ)=
x2-x-xsinθ+8
x-1-sinθ
=
x(x-1-sinθ)+8
x-1-sinθ
=x+class="stub"8
x-1-sinθ
=x-1-sinθ+class="stub"8
x-1-sinθ
+1+sinθ≥2
(x-1-sinθ)•class="stub"8
x-1-sinθ
+1+sinθ,
当且仅当x-1-sinθ=class="stub"8
x-1-sinθ
即x-1-sinθ=2
2
此时x=1+2
2
+sinθ取等号;
而sinθ∈[-1,1],
∴当sinθ=-1,x=2
2
时,函数f(x,θ)=
x2-x-xsinθ+8
x-1-sinθ
(x>2)取最小值为4
2

故选A.

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