Question

Let $f$ be a polynomial function which satisfies the relation $f(x)+f\left(\frac{x}{y^2}\right)+f\left(\frac{x}{y}\right)=f(x)\cdot f\left(\frac{1}{y}\right)-\frac{1}{y^3}+\frac{x^3}{y^6}+2\forall x\in R-\{0\},f(1)\ne 1$ and $f(2)=9$ The value of $\sum _{r=1}^{100}f(r)$ 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
$3f(1)=(f(1){)}^2+2\Rightarrow (f(1){)}^2-3f(1)+2=0\Rightarrow f(1)=1\text{ or }2$
$\therefore f(1)=2$
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 $n=3$
$\therefore f(x)=1+x^3$
$\sum _{r=1}^{100}f(r)=\sum _{r=1}^{100}\left(1+r^3\right)=100+\left(1^3+2^3+\dots \dots +100^3\right)$
$=100+(1+2+\dots \dots +100{)}^2$
$=100+(5050{)}^2$