Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a differentiable function with \(f(0)=1\) and satisfying the equation \[ f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y) \text { for all } x, y \in \mathbb{R} . \] Then, the value of \(\log _{e}(f(4))\) is ____.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Differential Equations

Let

f : R \to 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

lo g_{e} (f (4))

is ____.

Answer: 2

Taking the derivative with respect to

y

and setting

y = 0

f^{'} (x) = f (x) f^{''} (0) + f^{'} (x) f (0)

Setting

x = 0

in the original equation:

f (y) = f (0) f^{'} (y) + f^{'} (0) f (y)

f (y) = f^{'} (y) + f^{'} (0) f (y)

Since

f (0) = 1

, we get

f^{'} (0) = 1/2

.

From the limit definition:

f^{'} (x) = h \to 0 lim \frac{f ( x ) f ^{'} ( h ) + f ^{'} ( x ) f ( h ) - f ( x )}{h}

For the limit to exist:

f^{'} (x) = \frac{f ( x )}{2}

Solving:

ln (f (x)) = \frac{x}{2} + C

Using

f (0) = 1

C = 0

Thus

f (x) = e^{x /2}

ln (f (4)) = ln (e^{2}) = 2

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer