QJEEAdv18Ma4aad: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma4aad Solution: Let f : R R be a differentiable function with f(0) = 1 and…

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)

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

Put x = x and y = 0, we get

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

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

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

On integrating, we get

$\log f(x) = {1 \over 2}x + C$

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

If f(0) = 1, then A = 1

Hence, $f(x) = {e^{{1 \over 2}x}}$

$\Rightarrow {\log _e}f(x) = {1 \over 2}x$

$\Rightarrow {\log _e}f(4) = {1 \over 2} \times 4 = 2$