A) \[f\left( x \right)\]
B) 0
C) \[f\left( x \right)f'\left( nx \right)\]
D) none of these
Correct Answer: C
Solution :
We have, \[{{\left[ f\left( x \right) \right]}^{n}}=f\left( nx \right)\]for all \[x\] \[\Rightarrow n{{\left[ f\left( x \right) \right]}^{n-1}}f'\left( x \right)=n\,f'\left( nx \right)\]\[\Rightarrow n{{\left[ f\left( x \right) \right]}^{n}}f'\left( x \right)=n\,f'\left( x \right)f'\left( nx \right)\] \[\left[ Multiplying both sides by f \left( x \right) \right]\]\[\Rightarrow f\left( nx \right)f'\left( x \right)=f\left( x \right)f'\left( nx \right)\]\[\left[ \because {{\left[ f\left( x \right) \right]}^{n}}=f\left( nx \right) \right]\]\[\Rightarrow f\left( nx \right)f'\left( x \right)=f\left( x \right)f'\left( nx \right)\]
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