几何时,往往先设出直线方程,但要注意直线的斜率是否存在,如本题中当斜率不存在时也符合题意.Р21.(1)解:当a=1时,设g(x)=f/(x)=2(ex-x-1),g/(x)=2(ex-1)0,(x0)f/(x)在[0,+ )上递增,即x0时f/(x)f/(0)=0, f(x)的增区间为[0,+),无减区间,且x0时,f(x)=2ex-2-2x-x2f(0)=04(分)Р(2)解法一:<1>当a1时f/(x)=2(ex-x-a)2(x+1-x-a)=2(1-a)0x0时f(x)f(0)=0Р即当a1时,f(x)0恒成立,x[0,+ )6(分)Р<2>当a>1时,设h(x)=f/(x)=2(ex-a-x),h/(x)=2(ex-1)0, (x0) f/(x)在[0,+ )上递增Р又f/(0)=2(1-a)<0,f/(a)=2(ea-2a)由(1)已证2ex-2-2x-x20知ex1+x+x2Р f/(a)2(1+a+a2-2a)=(a-1)2+1>0 f/(x)在(0,a)上存在唯一零点xo,即-a-x0=0,Р f(x)在(0,xo)上递减,在(xo,+)上递增8(分)Р又f(xo)= 2-2-2axo-xo2=2(-1-x0+xo2),令g(x)=ex-1-xex+x2,x(0,a),g/(x)=x(1-ex)<0,Р当x>0时g(x)<g(0)=0,即f(xo)<0,不满足f(x)0恒成立,由<1><2>可知a的取值范围为(-,1].Р 12(分)Р解法二:分离变量Рx=0时f(0)=0,x>0时f(x)0a=g(x),g/(x)=,Р令h(x)=xex-ex+1-x2,h/(x)=x(ex-1)>0x>0时h(x)>h(0)=0g/(x)>0,即g(x)在(0,+)上递增,Р由洛比达法则g(x)= (ex-x)=1(适用于参加自主招生学生)Рa的取值范围为(-,1].
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