$f\left(x\right)=e^{-\sqrt{x}}+e^{\frac{-1}{{ }^{x^2}}}.{\text{\hspace{0.33em}If f}}^{"}\left(x\right)=\alpha .\frac{e^{-\sqrt{x}}}{x}\left(1+\frac{1}{\sqrt{x}}\right)+\beta \frac{e^{\frac{-1}{x^2}}}{x^4}\left(3-\frac{2}{x^2}\right)\text{ then }\left(\alpha ,\beta \right)=$

$$\begin{gathered} f^1\left(x\right)=e^{-\sqrt{x}}\left(\frac{-1}{2\sqrt{x}}\right)+e^{\frac{-1}{x^2}}.\frac{2}{x^3} \\ {f}^{11}\left(x\right)=e^{-\sqrt{x}}{\left(\frac{-1}{2\sqrt{x}}\right)}^2+e^{\sqrt{x}}\left(\frac{-1}{2}\right)\left(\frac{-1}{2}{x}^{\frac{-3}{2}}\right)+e^{\frac{-1}{{ }^{x^2}}}{\left(\frac{2}{x^3}\right)}^2+e^{\frac{-1}{x^2}}\left[2(-3x^{-4})\right] \\ \Rightarrow f^{11}\left(x\right)=e^{-\sqrt{x}}(\frac{1}{4x}+\frac{1}{4x\sqrt{x}})+e^{\frac{-1}{x^2}}(\frac{4}{x^6}-\frac{6}{x^4}) \\ \Rightarrow f^{11}\left(x\right)=\frac{e^{-\sqrt{x}}}{4x}(1+\frac{1}{\sqrt{x}})+e^{\frac{-1}{x^2}}\left(\frac{-2}{x^4}\right)(3-\frac{2}{x^2}) \\ \therefore \alpha =\frac{1}{4},\beta =-2 \end{gathered}$$