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 $x\in R$, and $f\left(1\right)=g\left(2\right)=1$, then which of the following statement(s) is (are) TRUE?

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

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(2\right)>1-{\log }_{e}2$
C $g\left(1\right)>1-{\log }_{e}2$

Given:
$f^{\prime }\left(x\right)=\left(e^{\left(f\right(x)-g(x\left)\right)}\right)g^{\prime }\left(x\right)$
$\Rightarrow e^{-f\left(x\right)}\cdot f^{\prime }\left(x\right)=e^{-g\left(x\right)}\cdot g^{\prime }\left(x\right)$
Integrating w.r.t. $x$, we get
$\Rightarrow e^{-f\left(x\right)}=e^{-g\left(x\right)}+C$

Putting $x=1$, we get
$\frac{1}{e}=\frac{1}{e^{g\left(1\right)}}+C\Rightarrow e^{-1}-e^{-g\left(1\right)}=C\cdots \left(i\right)$
Putting $x=2$, we get
$\frac{1}{e^{f\left(2\right)}}=\frac{1}{e}+C\Rightarrow e^{-f\left(2\right)}-e^{-1}=C\cdots \left(ii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$, we get
$\Rightarrow -e^{-g\left(1\right)}-e^{-f\left(2\right)}+2e^{-1}=0\Rightarrow e^{-g\left(1\right)}+e^{-f\left(2\right)}=2e^{-1}$

As exponential function is always positive, so
$e^{-f\left(2\right)}<2e^{-1};e^{-g\left(1\right)}<2e^{-1}$
Taking $\log$, we get
$-f\left(2\right)<{\log }_{e}2-1;-g\left(1\right)<{\log }_{e}2-1$

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