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