Let f: R → 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 ∈ R . Then, the value of _ e (f (4)) is

Correct answer: 0 P(x, y): f(x + y) = f(x) f ^ ' (y) + f ^ ' (x)f(y) x, y ∈ R P(0,0): f(0) = f(0) f ^ ' (0) + f ^ ' (0)f(0) ⇒ 1 = 2 f ^ ' (0) ⇒ f ^ ' (0) = (1)/(2) P(x, 0): f(x) = f(x) . f ^ ' (0) + f ^ ' (x) . f(0) ⇒ f(x) = (1)/(2)f(x) + f ^ ' (x) ⇒ f ^ ' (x) = (1)/(2)f(x) ⇒ f(x) = e ^ (1)/(2)x ⇒ (f(4)) = 2

JEE Advanced 2018Paper 2

Exam: JEE Advanced
Year: 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Functions

JEE Advanced 2018 · Paper 2 · Mathematics · Functions

Let f: R → R be a differentiable function with f(0) = 1 and...

Let

f: R \to R

be a differentiable function with

f\left(0\right) = 1

and satisfying the equation

f\left(x + y\right) = f\left(x\right){f}^{'}\left(y\right) + {f}^{'}\left(x\right)f\left(y\right)

for all

x, y \in R

. Then, the value of

\log_{e}

\left(f

\left(4\right)\right)

Correct answer

Step-by-step solution

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

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

⇒ 1 = 2{f}^{'}\left(0\right)

⇒ {f}^{'}\left(0\right) = \dfrac{1}{2}

P\left(x, 0\right): f\left(x\right) = f\left(x\right) . {f}^{'}\left(0\right) + {f}^{'}\left(x\right) . f\left(0\right)

⇒ f\left(x\right) = \dfrac{1}{2}f\left(x\right) + {f}^{'}\left(x\right)

⇒ {f}^{'}\left(x\right) = \dfrac{1}{2}f\left(x\right)

⇒ f\left(x\right) = {e}^{\dfrac{1}{2}x}

⇒ \ln \left(f\left(4\right)\right) = 2

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