(2013·成都理) 设函数f\left( x \right)=\sqrt{{{\text{e}}^{x}}+x-a}（a\in \mathbb{R}，{\text{e}}为自然对数的底数）.若曲线y=\sin x上存在\left( {{x}_{0}},{{y}_{0}} \right)使得f\left( f\left( {{y}_{0}} \right) \right)={{y}_{0}}，则a的取值范围是
A. [1, e]
B. [e^{−1} − 1, 1]
C. [1, e+1]
D. [e^{−1} − 1, e+1]
由已知{{y}_{0}}\in \left[ -1,1 \right]，先证明：f\left( {{y}_{0}} \right)={{y}_{0}}.
假设f\left( {{y}_{0}} \right)\ne {{y}_{0}}，由已知f\left( x \right)在定义域内单调递增，则：
①f\left( {{y}_{0}} \right)>{{y}_{0}}\Leftrightarrow f\left( f\left( {{y}_{0}} \right) \right)>f\left( {{y}_{0}} \right)>{{y}_{0}}，这与f\left( f\left( {{y}_{0}} \right) \right)={{y}_{0}}矛盾；
②f\left( {{y}_{0}} \right)<{{y}_{0}}\Leftrightarrow f\left( f\left( {{y}_{0}} \right) \right)<f\left( {{y}_{0}} \right)<{{y}_{0}}，这与f\left( f\left( {{y}_{0}} \right) \right)={{y}_{0}}矛盾.
综上，f\left( {{y}_{0}} \right)={{y}_{0}}.
问题等价于方程f\left( {{y}_{0}} \right)={{y}_{0}}在{{y}_{0}}\in \left[ -1,1 \right]上有解，显然{{y}_{0}}\in \left[ -1,0 \right)时无解；
{{y}_{0}}\in \left[ 0,1 \right]时，等价于a={{e}^{x}}+x-{{x}^{2}}在x\in \left[ 0,1 \right]上有解.
设g\left( x \right)={{\text{e}}^{x}}-{{x}^{2}}+x\left( x\in \left[ 0,1 \right] \right)，g'\left( x \right)={{\text{e}}^{x}}-2x+1，设h\left( x \right)=g'\left( x \right)，
h'\left( x \right)={{\text{e}}^{x}}-2，列表，得
| x | \left(0,\ln2\right) | \ln2 | \left(\ln2,1\right) |
|---|
| h'(x) | - | 0 | + |
| h(x) | 单调递减 | 极小值 | 单调递增 |
故g'\left( x \right)=h\left( x \right)\ge h\left( \ln 2 \right)=3-\ln 4>0，即g\left( x \right)在x\in \left[ 0,1 \right] 上单调递增.
故a\in \left[ g\left( 0 \right),g\left( 1 \right) \right]=\left[ 1,e \right]，经检验知此时定义域内存在符合题意的y_0，故答案选A.
