Q29 / 54 · JEE Advanced 2018 (Paper 2) · Mathematics

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 $\log_{e} (f (4))$ is

Correct answer: 2

Solution

P (x, y) : f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y) \forall x, y \in R

P (0,0) : f (0) = f (0) f^{'} (0) + f^{'} (0) f (0)

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

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

P (x, 0) : 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 f (x) = e^{\frac{1}{2} x}

\Rightarrow \ln (f (4)) = 2

Hence, the correct answer is 2.

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