Question

If a polynomial function 'f' satisfaction the realation.
${\log }_2\left(f\right(x\left)\right)={\log }_2(2+\frac{2}{3}+\frac{2}{9}+.......\infty )\cdot {\log }_3(1+\frac{f\left(x\right)}{f\frac{1}{x}})$
and $f\left(10\right)=1001$,
the the values of $f\left(20\right)$ is

• A. $2002$
• B. $7999$
• C. $8001$
• D. $16001$

Answer 1

Answer: (C)

$8001$

${\log }_2\left(f\right(x\left)\right)$
$={\log }_2\left(\frac{2}{1-\frac{1}{3}}\right)\cdot {\log }_3(1+\frac{f\left(x\right)}{f\left(\frac{1}{x}\right)})$
$\Rightarrow {\log }_2\left(f\right(x\left)\right)={\log }_{2}3\cdot \log (1+\frac{f\left(x\right)}{f\left(\frac{1}{x}\right)})$
$f\left(x\right)=1+\frac{f\left(x\right)}{f\left(\frac{1}{x}\right)}$
$f\left(x\right)\cdot f\left(\frac{1}{x}\right)=f\left(\frac{1}{x}\right)+f\left(x\right)\Rightarrow f\left(x\right)=1\pm x^n$
$f\left(x\right)=1\pm {10}^n=1001$
$\Rightarrow \pm {10}^n=1000\Rightarrow positivesignmustbetakenandn=3$
$\therefore f\left(x\right)=1+x^3$
$f\left(x\right)=1+(20{)}^3=8001$