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(B) Given $\frac{f'(x)}{f(x)} = 1$ for all $x \in R$.Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = Ae^x$

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$4e$

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(B) Given $\frac{f'(x)}{f(x)} = 1$ for all $x \in R$.Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = Ae^x$.Using the condition $f(1) = 2$,we have $Ae^1 = 2$,so $A = 2e^{-1}$.Thus,$f(x) = 2e^{-1} \cdot e^x = 2e^{x-1}$.Consequently,$f'(x) = 2e^{x-1}$.Given $h(x) = f(f(x))$,by the chain rule,$h'(x) = f'(f(x)) \cdot f'(x)$.At $x = 1$,$h'(1) = f'(f(1)) \cdot f'(1)$.Since $f(1) = 2$,we have $h'(1) = f'(2) \cdot f'(1)$.Substituting the values,$f'(2) = 2e^{2-1} = 2e$ and $f'(1) = 2e^{1-1} = 2$.Therefore,$h'(1) = (2e) \cdot (2) = 4e$.

Let $f$ be a differentiable function such that $f(1) = 2$ and $f'(x) = f(x)$ for all $x \in R$. If $h(x) = f(f(x))$,then $h'(1)$ is equal to

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