Question

已知向量a=(x,1),b=(1,-sinx),函數(shù)f(x)=a·b.

(1)若x∈［0,π］,試求函數(shù)f(x)的值域；

(2)若θ為常數(shù)，且θ∈(0,π),設(shè)g(x)=,x∈［0,π］,請討論g(x)的單調(diào)性，并判斷g(x)的符號.

Answer 1

解：(1)f(x)=a·b=x-sinx,∴f′(x)=1-cosx,x∈［0,π］.

∴f′(x)≥0.∴f(x)在［0，π］上單調(diào)遞增.                                     

于是f(0)≤f(x)≤f(π),即0≤f(x)≤π,

∴f(x)的值域?yàn)椋?，π］.                                                  

(2)g(x)=,

∴g′(x)=cosx+cos.                                            

∵x∈［0,π］,θ∈(0,π),∴∈(0,π).而y=cosx在［0,π］內(nèi)單調(diào)遞減,

∴由g′(x)=0,得x=,即x=θ.

因此，當(dāng)0≤x＜θ時，g′(x)＜0,g(x)單調(diào)遞減;

當(dāng)θ＜x≤π時，g′(x)＞0,g(x)單調(diào)遞增.                                       

由g(x)的單調(diào)性，知g(θ)是g(x)在［0,π］上的最小值,

∴當(dāng)x=θ時，g(x)=g(θ)=0；當(dāng)x≠θ時，g(x)＞g(θ)=0.                          

綜上,知當(dāng)x∈［0,θ)時，g(x)單調(diào)遞減;

當(dāng)x∈(θ,π］時，g(x)單調(diào)遞增.

當(dāng)x=θ時，g(x)=0,

當(dāng)x≠θ時，g(x)＞0.