已知函数 f(x)=x2−2x+a(ex−1+e−x+1)f\left( x\right) = {x}^{2} - {2x} + a\left( {{e}^{x - 1} + {e}^{-x + 1}}\right)f(x)=x2−2x+a(ex−1+e−x+1) 有唯一零点,则 a=a =a=A. −12- \frac{1}{2}−21 B. 13\frac{1}{3}31 C. 12\frac{1}{2}21 D. 1