A) \[2{{e}^{2}}\]
B) 4e
C) 2e
D) \[4{{e}^{2}}\]
Correct Answer: B
Solution :
Given,\[\frac{f'(x)}{f(x)}=1\forall x\in R\] Integrating both sides, we get \[\log f(x)=x+C\Rightarrow f(x)={{e}^{x+C}}={{e}^{x}}.{{e}^{C}}\] \[\because \]\[f(1)=2\Rightarrow f(1)=e.{{e}^{C}}=2\] \[\Rightarrow \]\[{{e}^{C}}=\frac{2}{e}\Rightarrow C=\log \frac{2}{e}\] \[\therefore \]\[f(x)={{e}^{x}}.\frac{2}{e}=2{{e}^{x-1}}\]\[\Rightarrow \]\[f'(x)=2{{e}^{x-1}}\] Now,\[h(x)=(f(x))\] \[\therefore \]\[h'(x)=f'(f(x)).f'(x)\] \[\therefore \]\[h'(1)=f'(f(1)).f'(1)=f'(2)f'(1)\] \[=2e.2=4e\]
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