Let $f: R \rightarrow R$ be a differentiable function with $f(0)=1$ and satisfying the equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y), \forall x, y \in R$, then the value of $\log (f(4))$ is

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Aicharya · Accepted Answer

$f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y) \forall x, y \in R$ putting $x=y=0$ we get $\begin{aligned} & f(0)=2 f(0) \cdot f^{\prime}(0) \ & \Rightarrow f^{\prime}(0)=\frac{1}{2}[\because 	ext { given } f(0)=1]\end{aligned}$ Now putting $x=x$ and $y=0$ $\begin{aligned} & f(x)=f(x) \cdot f^{\prime}(0)+f^{\prime}(x) \cdot f(0) \ & \Rightarrow f(x)=\frac{1}{2} f(x)+f^{\prime}(x)\left[\because f(0)=1 	ext { and } f^{\prime}(0)=\frac{1}{2}ight] \ & \Rightarrow \frac{1}{2} f(x)=f^{\prime}(x) \ & \Rightarrow \int \frac{f^{\prime}(x)}{f(x)} \mathrm{d} x=\int \frac{1}{2} \mathrm{~d} x \ & \Rightarrow \log (f(x))=\frac{1}{2} x+c \ & \because f(0)=1 \ & \Rightarrow c=0\end{aligned}$ i.e., $\log (f(x))=\frac{1}{2} x$ putting $x=4$ $\log (f(4))=\frac{1}{2} 	imes 4=2$

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

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