A) AP
B) GP
C) HP
D) None of these
Correct Answer: B
Solution :
[b] since, f(x) is a polynomial function satisfying \[f(x).f\left( \frac{1}{x} \right)=f(x)+f\left( \frac{1}{x} \right),\] \[\therefore f(x)={{x}^{n}}+1\] or \[f(x)=-{{x}^{n}}+1\] If \[f(x)=-{{x}^{n}}+1,\] then \[f(4)=-{{4}^{n}}+1\ne 65\] So \[f(x)={{x}^{n}}+1\] since, \[f(4)=65\] \[\therefore {{4}^{n}}+1=65\] \[\Rightarrow n=3\therefore f(x)={{x}^{3}}+1\Rightarrow f'(x)=3{{x}^{2}}\] \[\therefore f'({{l}_{1}})=3{{l}^{2}}_{1},f({{l}_{2}})=3{{l}^{2}}_{2},f'({{l}_{3}})=3{{l}^{2}}_{3}\] Since \[{{l}_{1}},{{l}_{2}},{{l}_{3}}\] are in GP. \[\therefore f'({{l}_{1}}),f'({{l}_{2}}),f'({{l}_{3}})\] are also in GP.
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