Question

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

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

Answer 1

Answer: (B), (C)

$g\left(1\right)>1-lo{g}_{e}2$
$f\left(2\right)>1-lo{g}_{e}2$

$f^{\prime }\left(x\right)=e^{\left(f\right(x)-g(x\left)\right)}{g}^{\prime }\left(x\right)\forall xϵR$

$\Rightarrow e^{-f\left(x\right)}\cdot f^{\prime }\left(x\right)-e^{-g\left(x\right)}{g}^{\prime }\left(x\right)=0$

$\Rightarrow \int \left(e^{-f\left(x\right)}{f}^{\prime }\right(x)-e^{-g\left(x\right)}\cdot g^{\prime }(x\left)\right)dx=C$

$\Rightarrow -e^{-f\left(x\right)}+e^{-g\left(x\right)}=C$

$\Rightarrow -e^{-f\left(1\right)}+e^{-g\left(1\right)}=-e^{-f\left(2\right)}+e^{-g\left(2\right)}$

$\Rightarrow -\frac{1}{e}+e^{-g\left(1\right)}=-e^{-f\left(2\right)}+\frac{1}{e}$

$\Rightarrow e^{-f\left(2\right)}+e^{-g\left(1\right)}=\frac{2}{e}$

$\therefore$ $e^{-f\left(2\right)}<\frac{2}{e}$ and $e^{-g\left(1\right)}<\frac{2}{e}$

$\Rightarrow -f\left(2\right)<lo{g}_{e}2-1$ and $-g\left(1\right)<lo{g}_{e}2-1$

$\Rightarrow f\left(2\right)>1-lo{g}_{e}2$ and $g\left(1\right)>1-lo{g}_{e}2$.