Question

ƒ'(x) = e^{(ƒ(x) – g(x))} g'(x) $\forall x\in \mathrm{ℝ}$
$\Rightarrow$e^{–ƒ(x). ƒ'(x) – e–g(x)}g'(x) = 0
$\Rightarrow$$\int$(e^{-ƒ(x ) ƒ '(x) - eg(x)} .g '(x))dx = C
$\Rightarrow$ -e ^-ƒ(x) +e ^{-g(x )} )dx=c
$\Rightarrow$-e ^-ƒ(1) +e ^g(1) =-e ^-ƒ(2)+ e ^{-g(2)
$\Rightarrow -\frac{1}{e}+e^{-g\left(1\right)}=-e^{-f\left(2\right)}+\frac{1}{e}$
$$\begin{gathered} \Rightarrow e^{-f\left(2\right)}+e^{-g\left(1\right)}=\frac{2}{e} \\ \therefore e^{-f\left(2\right)}<\frac{2}{e} and <\frac{2}{e} \end{gathered}$$}
$\Rightarrow$ –ƒ(2) < ln2 – 1 and –g(1) < ln2 – 1
$\Rightarrow$ƒ(2) > 1– ln2 and g(1) > 1 – ln2