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已知函数 ${\displaystyle f(x)={\mathrm{e}}^x}$.

1. 若函数 ${\displaystyle G(x)=f(x)-p\sqrt{x}+q\sin x}$ 存在零点. 求证: ${\displaystyle p^2+q^2>\mathrm{e}}$.
2. 令 ${\displaystyle h(x)=xf{\displaystyle \left(\frac{3x}{2}\right)-\lambda }}$ 有两不同零点 ${\displaystyle a,b(a<b)}$,
   1. 求实数 ${\displaystyle \lambda }$ 的取值范围;
   2. 求证: ${\displaystyle {\mathrm{e}}^b-{\mathrm{e}}^a<2\lambda +1}$.

1. 证明 设 ${\displaystyle G(x)}$ 零点为 ${\displaystyle x_0>0}$, 则消去 ${\displaystyle p}$ 后利用柯西不等式：$$\begin{array}{rl}{p}^2+q^2& =\frac{({\mathrm{e}}^{x_0}+q\sin x_0{)}^2}{x_0}+q^2\ge \frac{({\mathrm{e}}^{x_0}+q\sin x_0-q\sin x_0{)}^2}{x_0+{\sin }^2{x}_0}\\ & =\frac{{\mathrm{e}}^{2x_0}}{x_0+{\sin }^2{x}_0}\ge \frac{{\mathrm{e}}^{2x_0}}{x_0+|\sin x_0|}\ge \frac{{\mathrm{e}}^{2x_0}}{2x_0}\ge \mathrm{e}.\end{array}$$这样不等式得证.
2. 1. 解 求导得 ${\displaystyle h^{\prime }(x)={\displaystyle (\frac{3x}{2}+1){\mathrm{e}}^{\frac{3x}{2}}}}$, 有零点 ${\displaystyle x={\displaystyle -\frac{2}{3}}}$, 因此至少要求 $$\begin{array}{r}h(x{)}_{min}=h(-\frac{2}{3})=-\frac{2}{3\mathrm{e}}-\lambda <0\Rightarrow \lambda >-\frac{2}{3\mathrm{e}}.\end{array}$$
      而 ${\displaystyle x{\displaystyle {\mathrm{e}}^{\frac{3x}{2}}>0⟺x>0}}$, 且 ${\displaystyle x\mathrm{e}{\displaystyle {}^{\frac{3x}{2}}}}$ 在 ${\displaystyle (0,+\mathrm{\infty })}$ 上增，因此若 ${\displaystyle \lambda \ge 0}$, ${\displaystyle h(x)}$ 至多只有一个零点，这样得到 ${\displaystyle \lambda \in {\displaystyle (-\frac{2}{3\mathrm{e}},0)}}$.
      此时，${\displaystyle h(0)=-\lambda >0,h{\displaystyle (-\frac{2}{3})<0}}$, 而利用 ${\displaystyle {\mathrm{e}}^x>{\displaystyle \frac{1}{2}{x}^2,x>0}}$, 有$$\begin{array}{r}h(x_1)=\frac{x_1}{{\mathrm{e}}^{\frac{3}{2}(-x_1)}}-\lambda >\frac{x_1}{\frac{1}{2}\cdot \frac{9}{4}{x}_1^2}-\lambda \ge 0,x_1=min\{\frac{8}{9\lambda },-\frac{2}{3}\},\end{array}$$
      故 ${\displaystyle h(x)}$ 在 ${\displaystyle {\displaystyle (x_1,-\frac{2}{3}),(-\frac{2}{3},0)}}$ 上各有一零点. 综上，${\displaystyle \lambda \in {\displaystyle (-\frac{2}{3\mathrm{e}},0)}}$.
   2. 证明 记 ${\displaystyle {\mathrm{e}}^a=m,{\mathrm{e}}^b=n}$, 则 ${\displaystyle m,n\in (0,1)}$. 原问题等价于: ${\displaystyle \ln m\cdot m{\displaystyle {}^{\frac{3}{2}}=\ln n\cdot n^{\frac{3}{2}}=\lambda }}$，证明 ${\displaystyle n-m<2\lambda +1}$.
      一方面，$$\begin{array}{rl}& \lambda -(n-1)=\ln n\cdot n^{\frac{3}{2}}-(n-1)\\ & (1-\frac{1}{n})n^{\frac{3}{2}}+(1-n)=(1-n)(1-n^{\frac{1}{2}})>0,\end{array}$$
      所以 ${\displaystyle n<\lambda +1}$.
      另一方面 $$\begin{array}{r}\frac{\lambda }{m}=\ln m\cdot m^{\frac{1}{2}}=-2\ln \frac{1}{m^{\frac{1}{2}}}\cdot m^{\frac{1}{2}}>-2\frac{1}{\mathrm{e}{m}^{\frac{1}{2}}}\cdot m^{\frac{1}{2}}>-1\Rightarrow m>-\lambda ,\end{array}$$
      故 ${\displaystyle n-m<2\lambda +1}$.