设函数\(f(x)=\sqrt{\mathrm{e}^x+x-a},a\in\mathbb{R},\mathrm{e}\)为自然对数的底数,若曲线\(y=\sin x\)上存在点\(\left(x_0,y_0\right)\)使得\(f\left(f\left(y_0\right)\right)=y_0\)成立,则\(a\)的取值范围是\(\left(\qquad\right)\)
\(\mathrm{A}.\left[1,\mathrm{e}\right]\) \(\qquad\mathrm{B}.\left[\mathrm{e}^{-1}-1,1\right]\) \(\qquad\mathrm{C}.\left[1,\mathrm{e}+1\right]\) \(\qquad\mathrm{D}.\left[\mathrm{e}^{-1}-1,\mathrm{e}+1\right]\)
解析:
根据复合函数的单调性,可以判断\(f(x)\)为增函数,又因为$$\exists y_0\in\left[-1,1\right],f\left(f\left(y_0\right)\right)=y_0.$$所以\(y_0\)是函数\(f(x)\)的稳定点,又因为\(f(x)\)单调递增,因此该稳定点即不动点,也即有$$\exists y_0\in\left[-1,1\right], f(y_0)=y_0.$$此时\(y_0\in\left[0,1\right]\),记\(x=y_0\),可得$$a=\mathrm{e}x+x-x2,x\in\left[0,1\right].$$于是可得\(a\)的取值范围为\([1,\mathrm e]\).
