Inequality Analysis for Differentiable Functions with Exponential Relation

To determine which statement(s) are true, we start by analyzing the given differential relation:

$f'(x) = e^{f(x) - g(x)} g'(x)$

This can be rewritten by separating the functions $f(x)$ and $g(x)$:

$e^{-f(x)} f'(x) = e^{-g(x)} g'(x)$

Integrating both sides with respect to $x$:

$\int e^{-f(x)} f'(x) \, dx = \int e^{-g(x)} g'(x) \, dx$

$-e^{-f(x)} = -e^{-g(x)} + C$

$e^{-g(x)} - e^{-f(x)} = C$

where $C$ is a constant of integration.

Now, we use the given boundary conditions $f(1) = 1$ and $g(2) = 1$:

1. At $x = 1$: $e^{-g(1)} - e^{-f(1)} = C \implies e^{-g(1)} - e^{-1} = C \quad \text{--- (1)}$

2. At $x = 2$: $e^{-g(2)} - e^{-f(2)} = C \implies e^{-1} - e^{-f(2)} = C \quad \text{--- (2)}$

Equating the expressions for $C$ from equations (1) and (2):

$e^{-g(1)} - e^{-1} = e^{-1} - e^{-f(2)}$

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

Since the exponential function is strictly positive for all real inputs, we have:

$e^{-f(2)} > 0 \quad \text{and} \quad e^{-g(1)} > 0$

Using $e^{-f(2)} > 0$: $e^{-g(1)} = \frac{2}{e} - e^{-f(2)} < \frac{2}{e}$

Taking the natural logarithm on both sides: $-g(1) < \ln\left(\frac{2}{e}\right) = \log_e 2 - 1$

Multiplying by $-1$ reverses the inequality: $g(1) > 1 - \log_e 2$

Similarly, using $e^{-g(1)} > 0$: $e^{-f(2)} = \frac{2}{e} - e^{-g(1)} < \frac{2}{e}$

Taking the natural logarithm on both sides: $-f(2) < \ln\left(\frac{2}{e}\right) = \log_e 2 - 1$

Multiplying by $-1$: $f(2) > 1 - \log_e 2$

Therefore, the correct statements are (B) and (C).