Question

Let $f$ be a polynomial function which satisfies the relation $f(x)+f\left(\frac{x}{y^2}\right)+f\left(\frac{x}{y}\right)=f(x) \cdot f\left(\frac{1}{y}\right)-\frac{1}{y^3}+\frac{x^3}{y^6}+2 \forall x \in R-\{0\}, f(1) \neq 1$ and $f(2)=9$ The value of $\displaystyle\sum_{ r =1}^{100} f ( r )$ equals

• A. 5050
• B. $(5050)^2$
• C. $100+(5050)^2$
• D. $100+(5050)^3$

Answer 1

Answer: (C)

$f(x)+f\left(\frac{x}{y^2}\right)+f\left(\frac{x}{y}\right)=f(x) \cdot f\left(\frac{1}{y}\right)+2$
Replacing y by $x$
$f ( x )+ f \left(\frac{1}{ x }\right)+ f (1)= f ( x ) \cdot f \left(\frac{1}{ x }\right)+2$.....(i)
Putting $x = y =1$ in the given relation
$3 f (1)=( f (1))^2+2 \Rightarrow( f (1))^2-3 f (1)+2=0 \Rightarrow f (1)=1 \text { or } 2 $
$\therefore f (1)=2$
Now, from (i)
$\Rightarrow f ( x )+ f \left(\frac{1}{ x }\right)= f ( x ) \cdot f \left(\frac{1}{ x }\right) $
$\Rightarrow f ( x )=1 \pm x ^{ n } $
$f (2)=1 \pm 2^{ n }=9 \Rightarrow \pm 2^{ n }=8$
Positive sign is to be taken and $n =3$
$\therefore f ( x )=1+ x ^3 $
$\displaystyle\sum_{ r =1}^{100} f ( r )=\displaystyle\sum_{ r =1}^{100}\left(1+ r ^3\right)=100+\left(1^3+2^3+\ldots \ldots+100^3\right)$
$=100+(1+2+\ldots \ldots+100)^2$
$=100+(5050)^2 $