Mathematics - Definite Integration Question with Solution | TestHub

$f^{\prime }\left(x\right)=e^{\left(f\left(x\right)-g\left(x\right)\right)}{g}^{\prime }\left(x\right)\forall x\in \mathbb{R}$
$\Rightarrow e^{-f\left(x\right)}.f^{\prime }\left(x\right)-e^{-g\left(x\right)}{g}^{\prime }\left(x\right)=0$
$\Rightarrow \int \left(e^{-f\left(x\right)}{f}^{\prime }\left(x\right)-e^{-g\left(x\right)}.g^{\prime }\left(x\right)\right)dx=C$
$\Rightarrow -e^{-f\left(x\right)+e^{-g\left(x\right)}}=C$
$\Rightarrow -e^{-f\left(1\right)}+e^{-g\left(1\right)}=-e^{-f\left(2\right)}+e^{-g\left(2\right)}$
$\Rightarrow -\frac{1}{e}+e^{g\left(1\right)}=-e^{-f\left(2\right)}+\frac{1}{e}$
$\Rightarrow e^{-f\left(2\right)}+e^{-g\left(1\right)}=\frac{2}{e}$
$\therefore e^{-f\left(2\right)}<\frac{2}{e}\mathrm{a}\mathrm{n}\mathrm{d} e^{-g\left(1\right)}<\frac{2}{e}$
$\Rightarrow -f\left(2\right)<\ln 2-1 \mathrm{a}\mathrm{n}\mathrm{d}-g\left(1\right)<\ln 2-1$
$\Rightarrow f\left(2\right)>1-\ln 2 \mathrm{a}\mathrm{n}\mathrm{d} g\left(1\right)>1-\ln 2$