Differentiation Question 472
Question: Let f(x) be a polynomial function satisfying $ f(x).f( \frac{1}{x} )=f(x)+f( \frac{1}{x} ). $ if $ f(4)=65 $ and $ l_1,l_2,l_3 $ are in $ GP, $ then $ f’(l_1),f’(l_2),f’(l_3) $ are in
Options:
A) AP
B) GP
C) HP
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] since, f(x) is a polynomial function satisfying $ f(x).f( \frac{1}{x} )=f(x)+f( \frac{1}{x} ), $
$ \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)=3x^{2} $
$ \therefore f’(l_1)=3l^2_1,f(l_2)=3l^2_2,f’(l_3)=3l^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.
BETA
