Discussion Forum

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

Answer:

Explanation:

We have,

    $f'(x)=e^{(f(x)-g(x))}g'^{(x)}  \forall x \in R$

$\Rightarrow$        $f'(x)= \frac{e^{f(x)}}{e^{g(x)}}g'(x)$

$\Rightarrow$     $\frac{f'(x)}{e^{f(x)}}= \frac{g'(x)}{e^{g(x)}}$

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

  On integrating both sides, we get

   $e^{-f(x)}=e^{-g(x) }+c$

  At x=1,

  $e^{-f(1)}=e^{-g(1) }+c$

   $e^{-1}=e^{-g(1) }+c$    [ $\because$ f(1)= 1]  ......(i)

At x=2,

 $e^{-f(2)}=e^{-g(2) }+c$

$\Rightarrow$    $e^{-f(2)}=e^{-1}+C$  [   $\because$ g(2)= 1]    ..........(ii)

 From Eqs. (i)   and (ii)

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

$\Rightarrow$       $e^{-f(2)}>2e^{-1}$

 We know that , e^-x  is decreasing

   $\because$   $-f(2)<\log_{e}{2}-1$

                   $f(2)>1-\log_{e}{2}$

  $\Rightarrow$  $e^{-g(1)}+e^{-f(2)}=2e^{-1}$    [from Eq,(iii)]

$\Rightarrow$ $e^{-g(1)}<2e^{-1}$

    $-g(1)<\log_{e}{2}-1$

$\Rightarrow$     $g(1)>-\log_{e}{2}$