Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

${\text{f}}^{″}\left(\text{x}\right)-2{\text{f}}^{\prime }\left(\text{x}\right)+\text{f}\left(\text{x}\right)\ge {\text{e}}^{\text{x}}$
${\text{f}}^{″}\left(\text{x}\right)·{\text{e}}^{-\text{x}}-{\text{f}}^{\prime }\left(\text{x}\right){\text{e}}^{-\text{x}}-{\text{f}}^{\prime }\left(\text{x}\right){\text{e}}^{-\text{x}}+\text{f}\left(\text{x}\right){\text{e}}^{-\text{x}}\ge 1$
$\frac{\text{d}}{\text{dx}}\left({\text{f}}^{\prime }\left(\text{x}\right){\text{e}}^{-\text{x}}\right)-\frac{\text{d}}{\text{dx}}\left(\text{f}\left(\text{x}\right)·{\text{e}}^{-\text{x}}\right)\ge 1$
$\frac{\text{d}}{\text{dx}}\left({\text{f}}^{\prime }\left(\text{x}\right){\text{e}}^{-\text{x}}-\text{f}\left(\text{x}\right){\text{e}}^{-\text{x}}\right)\ge 1$
$\Rightarrow \frac{{\text{d}}^2}{{\text{dx}}^2}\left({\text{e}}^{-\text{x}}\text{f}\left(\text{x}\right)\right)\ge 1$$\forall \text{x}\in \left[0\text{, }1\right]$
Let$\varphi \left(\text{x}\right)={\text{e}}^{-\text{x}}\text{f}\left(\text{x}\right)$
$\Rightarrow \varphi \left(\text{x}\right)$is concave upward
$f\left(0\right)=f\left(1\right)=0$
$\Rightarrow \varphi \left(0\right)=0=\varphi \left(1\right)$
$\Rightarrow \text{f}\left(\text{x}\right)<0$
$\Rightarrow \varphi \left(\text{x}\right)<0$