Question

Let $f:\mathbb{R}\to \mathbb{R}$ and $g:\mathbb{R}\to \mathbb{R}$ be two non-constant differentiable functions.If
$f\prime (x)=(e^{(f(x)-g(x)))g\prime (x)}$ for all $x\in \mathbb{R}$,
and $f(1)=g(2)=1$, then which of the following statement(s) is (are) TRUE?

• A. $f(2)<1-lo{g}_{e}2$
• B. $f(2)>1-lo{g}_{e}2$
• C. $g(1)>1-lo{g}_{e}2$
• D. $g(1)<1-lo{g}_{e}2$

Answer 1

Answer: (B), (C)

$f^{\prime }\left(x\right)=e^{\left(f\left(x\right)-g\left(x\right)\right)}{g}^{\prime }\left(x\right)\forall x\in \mathbb{R}$
$\Rightarrow e^{-f\left(x\right)}.f^{\prime }\left(x\right)-e^{-g\left(x\right)}{g}^{\prime }\left(x\right)=0$
$\Rightarrow \int \left(e^{-f\left(x\right)}{f}^{\prime }\left(x\right)-e^{-g\left(x\right)}.g^{\prime }\left(x\right)\right)dx=C$
$\Rightarrow -e^{-f\left(x\right)}+e^{-g\left(x\right)}=C$
$\Rightarrow -e^{-f\left(1\right)}+e^{-g\left(1\right)}=-e^{-f\left(2\right)}+e^{-g\left(2\right)}$
$\Rightarrow -\frac{1}{e}+e^{-g\left(1\right)}=\frac{2}{e}$
$\therefore e^{-f\left(2\right)}<\frac{2}{e}and{e}^{-g\left(1\right)}<\frac{2}{e}$
$\Rightarrow -f\left(2\right)<In2-1$ and $-g\left(1\right)<In2-1$
$\Rightarrow f\left(2\right)>1-In2$ and $g\left(1\right)>1-In2$