Question

設(shè)函數(shù)f(x)=(1+x)²-2ln(1+x).

(1)求函數(shù)f(x)的單調(diào)區(qū)間;

(2)當(dāng)0＜a＜2時(shí),求函數(shù)g(x)=f(x)-x²-ax-1在區(qū)間［0,3］上的最小值.

(文)已知向量a=(cosx,cosx),b=(0,sinx),c=(sinx,cosx),d=(sinx,sinx).

(1)當(dāng)x=時(shí),求向量a、b的夾角;

(2)當(dāng)x∈［0,］時(shí),求c·d的最大值;

(3)設(shè)函數(shù)f(x)=(a-b)·(c+d),將函數(shù)f(x)的圖象按向量m平移后得到函數(shù)g(x)的圖象,且g(x)=2sin2x+1,求|m|的最小值.

Answer 1

解:(1)∵f′(x)=2(x+1).                             

由f′(x)＞0,得-2＜x＜-1或x＞0;

由f′(x)＜0,得x＜-2或-1＜x＜0.

又∵f(x)的定義域?yàn)?-1,+∞),

∴函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,+∞),單調(diào)遞減區(qū)間為(-1,0).                   

(2)g(x)=2x-ax-2ln(1+x),定義域?yàn)?-1,+∞),

g′(x)=2-a-.                                           

∵0＜a＜2,∴2-a＞0且＞0.

由g′(x)＞0,得x＞,

即g(x)在(,+∞)上單調(diào)遞增;

由g′(x)＜0,得-1＜x＜,

即g(x)在(-1,)上單調(diào)遞減.                                           

①當(dāng)0＜a＜時(shí),0＜＜3,g(x)在(0,)上單調(diào)遞減,在(a,3)上單調(diào)遞增;

∴在區(qū)間［0,3］上,g(x)_min=g()=a-2ln;                           

②當(dāng)≤a＜2時(shí),≥3,g(x)在(0,3)上單調(diào)遞減,

∴在區(qū)間［0,3］上,g(x)_min=g(3)=6-3a-2ln4.                                  

綜上,可知當(dāng)0＜a＜時(shí),在區(qū)間［0,3］上,g(x)_min=g()=a-2ln;

當(dāng)≤a＜2時(shí),在區(qū)間［0,3］上,g(x)_min=g(3)=6-3a-2ln4.                         

(文)解:(1)∵x=,

∴a=(),b=(0,).                                               

則a·b=()·(0,)=,

cos〈a,b〉=.

∴向量a、b的夾角為.                                                 

(2)c·d=(sinx,cosx)·(sinx,sinx)=sin²x+sinxcosx

=

=.                                                   

∵x∈［0,］,

∴≤2x≤.                                                   

當(dāng)2x=,即x=時(shí),c·d取最大值.                             

(3)f(x)=(a-b)·(c+d)=(cosx,cosx-sinx)·(2sinx,sinx+cosx)

=sinxcosx+cos²x-sin²x

=sin2x+cos2x

=2sin(2x+).                                                            

設(shè)m=(s,t),則

g(x)=f(x-s)+t=2sin［2(x-s)+］+t

=2sin(2x-2s+)+t=2sin2x+1,

∴t=1,s=kπ+(k∈Z).

易知當(dāng)k=0時(shí),|m|_min=.