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已知函数 ${\displaystyle f(x)={\mathrm{e}}^{2x}-2a(x-2){\mathrm{e}}^x-a^2{x}^2(a>0)}$ 恰有两个零点，求 ${\displaystyle a}$ 的值.

${\displaystyle f^{\prime }(x)=2({\mathrm{e}}^x+a)({\mathrm{e}}^x-ax)}$.如果 ${\displaystyle a\le \mathrm{e}}$，则 ${\displaystyle {\mathrm{e}}^x\ge \mathrm{e}x\ge ax\Rightarrow f^{\prime }(x)\ge 0}$，这样不存在两个零点；则 ${\displaystyle a>\mathrm{e}}$，记 ${\displaystyle g(x)={\mathrm{e}}^x-ax,g^{\prime }(x)={\mathrm{e}}^x-a}$，得 ${\displaystyle g(x)}$ 有极小值点 ${\displaystyle x=\ln a>1}$，进而可知 ${\displaystyle g(x)}$ 在 ${\displaystyle (0,\ln a)}$ 上有零点 ${\displaystyle x_1}$, ${\displaystyle (\ln a,2a)}$ 上有零点 ${\displaystyle x_2}$，从而 ${\displaystyle f(x)}$ 在 ${\displaystyle (-\mathrm{\infty },x_1),(x_2,+\mathrm{\infty })}$ 上增，在 ${\displaystyle (x_1,x_2)}$ 上减.
对于 ${\displaystyle i=1,2}$, ${\displaystyle {\mathrm{e}}^{x_i}=a{x}_i}$，从而 ${\displaystyle f(x_i)=-2a^2{x}_i(x_i-2)}$.由于 ${\displaystyle {\displaystyle \underset{x\to -\mathrm{\infty }}{lim}f(x)=-\mathrm{\infty },\underset{x\to +\mathrm{\infty }}{lim}f(x)=+\mathrm{\infty }}}$，(可以进行如下放缩：当 ${\displaystyle x>6a}$ 时依据 ${\displaystyle {\mathrm{e}}^x>{\displaystyle \frac{1}{2}{x}^2}}$ 有 $$\begin{array}{rl}f(x)& =({\mathrm{e}}^x-ax{)}^2-2a^2{x}^2+4a{\mathrm{e}}^x>{(\frac{1}{2}{x}^2-ax)}^2-2a^2{x}^2\\ & >(3ax-ax{)}^2-2a^2{x}^2=2a^2{x}^2\to +\mathrm{\infty }.\end{array}$$
当 ${\displaystyle x<\ln a}$ 时 $$\begin{array}{r}f(x)<{\mathrm{e}}^{2x}+4a{\mathrm{e}}^x-a^2{x}^2<5a^2-a^2{x}^2\to -\mathrm{\infty }.\end{array}$$ 故只能是某个极值为 ${\displaystyle 0}$，即 ${\displaystyle x_1=2}$ 或 ${\displaystyle x_2=2}$，从而 ${\displaystyle {\displaystyle a=\frac{{\mathrm{e}}^2}{2}}}$.