Discussion Forum

1)

Let  $f:R\rightarrow R$  be a differentiable function with f(0) =1 and satisfying the equation f(x+y) =f(x)  f ' (y)+ f ' (x) f(y)  for 

all x , $y \in R$, Then, the value of log_e (f(4)) is..............

Answer:

Option A

Explanation:

Given, $f(x+y) = f  (x) f ' (y) +f '(x) f (y), \forall x,y \in R$  and f (0)=1

Put x=y=0, we get

f(0) = f(0) f '(0) +f '(0) f (0)

$\Rightarrow$      1= 2 f '(0) $\Rightarrow$  f ' (0)= $\frac{1}{2}$

 Put x=x and y=0 , we get

          f (x) = f (x) f '(0) + f '(x) f(0)

 $\Rightarrow$   $f (x)= \frac{1}{2} f(x)+f '(x)$

 $\Rightarrow$   $f' (x)= \frac{1}{2} f(x) \Rightarrow \frac{f'(x)}{f(x)}=\frac{1}{2}$ 

On integrating , we get

    $\log_{}{f(x)}=\frac{1}{2}x+C$

 $\Rightarrow$  $f (x)= Ae^{\frac{1}{2}x}$ where e^C =A

  If  f(0)=1   . then A=1

  Hence,   $f(x)= e^{\frac{1}{2}x}$

 $\Rightarrow$    $\log_{e}{f(x)}=\frac{1}{2}x$

$\Rightarrow$    $\log_{4}{f(4)}=\frac{1}{2}\times 4=2$