Question

已知M、N兩點(diǎn)的坐標(biāo)分別是M(1+cos2x,1)、N(1,sin2x+a)(x,a∈R,a是常數(shù)),令f(x)=·(O是坐標(biāo)原點(diǎn)).

(1)求函數(shù)f(x)的解析式,并求函數(shù)f(x)在［0,π］上的單調(diào)遞增區(qū)間;

(2)當(dāng)x∈［0,］時(shí),f(x)的最大值為4,求a的值,并說(shuō)明此時(shí)f(x)的圖象可由函數(shù)y=2sin(x+)的圖象經(jīng)過(guò)怎樣的平移和伸縮變換而得到.

Answer 1

解:(1)y=·=1+cos2x+sin2x+a,

∴f(x)=1+cos2x+3sin2x+a=2sin(2x+)+a+1.

由2kπ-≤2x+≤2kπ+,得kπ≤x≤kπ+,k∈Z.

∴f(x)在［0,π］上的單調(diào)遞增區(qū)間為［0,］和［,π］.

(2)f(x)=2sin(2x+)+a+1,x∈［0, ］,2x+∈［,］,2sin(2x+)∈［-1,2］,

∴當(dāng)x=時(shí),f(x)取最大值a+3=4.解得a=1,f(x)=2sin(2x+)+2.

將y=2sin(x+)的圖象上的每一點(diǎn)的橫坐標(biāo)縮短到原來(lái)的一半,縱坐標(biāo)保持不變,再向上平移2個(gè)單位長(zhǎng)度,得f(x)=2sin(2x+)+2的圖象.