Question

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

• A. $𝑓(2) < 1 - log_e \, 2$
• B. $𝑓(2) > 1 - log_e \, 2$
• C. $g(1) > 1 - log_e \, 2$
• D. $g(1) < 1 - log_e \, 2$

Answer 1

Answer: (B), (C)

$f'\left(x\right) = e^{\left(f\left(x\right) - g\left(x\right)\right)} g'\left(x\right) \forall x \in \mathbb R $
$ \Rightarrow e^{-f\left(x\right)}. f'\left(x\right) - e^{-g\left(x\right)}g'\left(x\right)=0 $
$\Rightarrow \int\left(e^{-f\left(x\right)} f'\left(x\right) -e^{-g\left(x\right)} .g'\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 $