Question
Let $$f\left( x \right)$$ be a polynomial function satisfying $$f\left( x \right).f\left( {\frac{1}{x}} \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right).$$ If $$f\left( 4 \right) = 65$$ and $${l_1},\,{l_2},\,{l_3}$$ are in GP, then $$f'\left( {{l_1}} \right),\,f'\left( {{l_2}} \right),\,f'\left( {{l_3}} \right)$$ are in :
A.
AP
B.
GP
C.
HP
D.
none of these
Answer :
GP
Solution :
Since, $$f\left( x \right)$$ is a polynomial function satisfying
$$\eqalign{
& f\left( x \right).f\left( {\frac{1}{x}} \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right), \cr
& \therefore \,f\left( x \right) = {x^n} + 1{\text{ or }}f\left( x \right) = - {x^n} + 1 \cr
& {\text{If }}f\left( x \right) = - {x^n} + 1,{\text{ then }}f\left( 4 \right) = - {4^n} + 1 \ne 65 \cr
& {\text{So, }}f\left( x \right) = {x^n} + 1{\text{ Since, }}f\left( 4 \right) = 65 \cr
& \therefore \,{4^n} + 1 = 65\,\, \Rightarrow \,n = 3 \cr
& \therefore \,f\left( x \right) = {x^3} + 1\,\, \Rightarrow f'\left( x \right) = 3{x^2} \cr
& \therefore \,f'\left( {{l_1}} \right) = 3l_1^2,\,f'\left( {{l_2}} \right) = 3l_2^2,\,f'\left( {{l_3}} \right) = 3l_3^2 \cr
& {\text{Since, }}{l_1},\,{l_2},\,{l_3}{\text{ are in GP}}{\text{.}} \cr
& \therefore \,f'\left( {{l_1}} \right),\,f'\left( {{l_2}} \right),\,f'\left( {{l_3}} \right){\text{ are also in GP}}{\text{.}} \cr} $$