设函数f(x)=ex+x?a(a∈R,e为自然对数的底数),若曲线y=sinx上存在点(x0,y0)使得f(f(y0))=y0,

\u8bbe\u51fd\u6570f\uff08x\uff09=ex+x?a\uff08a\u2208R\uff0ce\u4e3a\u81ea\u7136\u5bf9\u6570\u7684\u5e95\u6570\uff09\uff0e\u82e5\u66f2\u7ebfy=sinx\u4e0a\u5b58\u5728\u70b9\uff08x0\uff0cy0\uff09\u4f7f\u5f97f\uff08f\uff08y0\uff09\uff09=y0

\u7531\u9898\u610f\u53ef\u5f97 y0=sinx0\u2208[-1\uff0c1]\uff0cf\uff08y0\uff09=ey0+y0?a\uff0c\u2235\u66f2\u7ebfy=sinx\u4e0a\u5b58\u5728\u70b9\uff08x0\uff0cy0\uff09\u4f7f\u5f97f\uff08f\uff08y0\uff09\uff09=y0\uff0c\u2234\u5b58\u5728y0\u2208[0\uff0c1]\uff0c\u4f7ff\uff08y0\uff09=y0\u6210\u7acb\uff0c\u5373f\uff08x\uff09=x\u5728[0\uff0c1]\u4e0a\u6709\u89e3\uff0c\u5373 ex+x-x2=a \u5728[0\uff0c1]\u4e0a\u6709\u89e3\uff0e\u4ee4g\uff08x\uff09=ex+x-x2\uff0c\u5219a\u4e3ag\uff08x\uff09\u5728[0\uff0c1]\u4e0a\u7684\u503c\u57df\uff0e\u2235\u5f53x\u2208[0\uff0c1]\u65f6\uff0cg\u2032\uff08x\uff09=ex+1-2x\uff1e0\uff0c\u6545\u51fd\u6570g\uff08x\uff09\u5728[0\uff0c1]\u4e0a\u662f\u589e\u51fd\u6570\uff0c\u6545g\uff080\uff09\u2264g\uff08x\uff09\u2264g\uff081\uff09\uff0c\u53731\u2264a\u2264e\uff0c\u6545\u7b54\u6848\u4e3a\uff1a[1\uff0ce]\uff0e

\u89e3\uff1a\u7531f\uff08f\uff08b\uff09\uff09=b\uff0c\u53ef\u5f97f\uff08b\uff09=f-1\uff08b\uff09\u5176\u4e2df-1\uff08x\uff09\u662f\u51fd\u6570f\uff08x\uff09\u7684\u53cd\u51fd\u6570\u56e0\u6b64\u547d\u9898\u201c\u5b58\u5728b\u2208[0\uff0c1]\u4f7ff\uff08f\uff08b\uff09\uff09=b\u6210\u7acb\u201d\uff0c\u8f6c\u5316\u4e3a\u201c\u5b58\u5728b\u2208[0\uff0c1]\uff0c\u4f7ff\uff08b\uff09=f-1\uff08b\uff09\u201d\uff0c\u5373y=f\uff08x\uff09\u7684\u56fe\u8c61\u4e0e\u51fd\u6570y=f-1\uff08x\uff09\u7684\u56fe\u8c61\u6709\u4ea4\u70b9\uff0c\u4e14\u4ea4\u70b9\u7684\u6a2a\u5750\u6807b\u2208[0\uff0c1]\uff0c\u2235y=f\uff08x\uff09\u7684\u56fe\u8c61\u4e0ey=f-1\uff08x\uff09\u7684\u56fe\u8c61\u5173\u4e8e\u76f4\u7ebfy=x\u5bf9\u79f0\uff0c\u2234y=f\uff08x\uff09\u7684\u56fe\u8c61\u4e0e\u51fd\u6570y=f-1\uff08x\uff09\u7684\u56fe\u8c61\u7684\u4ea4\u70b9\u5fc5\u5b9a\u5728\u76f4\u7ebfy=x\u4e0a\uff0c\u7531\u6b64\u53ef\u5f97\uff0cy=f\uff08x\uff09\u7684\u56fe\u8c61\u4e0e\u76f4\u7ebfy=x\u6709\u4ea4\u70b9\uff0c\u4e14\u4ea4\u70b9\u6a2a\u5750\u6807b\u2208[0\uff0c1]\uff0c\u6839\u636eex+x?a\uff1dx\uff0c\u5316\u7b80\u6574\u7406\u5f97ex=x2-x+a\u8bb0F\uff08x\uff09=ex\uff0cG\uff08x\uff09=x2-x+a\uff0c\u5728\u540c\u4e00\u5750\u6807\u7cfb\u5185\u4f5c\u51fa\u5b83\u4eec\u7684\u56fe\u8c61\uff0c\u53ef\u5f97F(0)\u2264G(0)F(1)\u2265G(1)\uff0c\u5373e0\u226402?0+ae1\u226512?1+a\uff0c\u89e3\u4e4b\u5f971\u2264a\u2264e\u5373\u5b9e\u6570a\u7684\u53d6\u503c\u8303\u56f4\u4e3a[1\uff0ce]\u6545\u9009\uff1aA

曲线y=sinx上存在点(x0,y0)使得f(f(y0))=y0,则y0∈[-1,1]
考查四个选项,B,D两个选项中参数值都可取0,C,D两个选项中参数都可取e+1,A,B,C,D四个选项参数都可取1,由此可先验证参数为0与e+1时是否符合题意,即可得出正确选项
当a=0时,f(x)=


简单计算一下,答案如图所示



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