Question

A function $y=f\left(x\right)$ satisfies the equation $f(x+y)=f\left(x\right)\cdot f\left(y\right)$ where $x,y\in R$. It is known that $f\left(1\right)=25$. If $S=f\left(2\right)+f\left(1\right)+f\left(0\right)+f(-1)+...\infty$, then the value of $\left[\right(f\left(1\right)-1)S{]}^{1/2}$ is

• A. $25$
• B. $125$
• C. $625$
• D. $5^6$

Answer 1

The correct option is B $125$

$f(x+y)=f\left(x\right)\cdot f\left(y\right)$
Put $x=y=1$
$\Rightarrow f\left(2\right)=[f\left(1\right){]}^2={25}^2$
Put $x=1,y=2$
$\Rightarrow f\left(3\right)=[f\left(1\right){]}^3={25}^3$

$\therefore f\left(x\right)={25}^x\Rightarrow f\left(2\right)={25}^2=625\Rightarrow f(-1)={25}^{-1}=1/25\Rightarrow f(-2)={\left(a^{}\right)}^{-2}={\left(\frac{1}{25}\right)}^2$
Sum of the given series is
$S=625+25+1+\frac{1}{25}+...$
clearly it is a G.P. with $1^{st}$ term $625$ and common ratio $\frac{1}{25}$
$\Rightarrow S=\frac{625}{1-\frac{1}{25}}=\frac{625\times 25}{24}\Rightarrow S=\frac{5^6}{24}$
Now,
$\left(f\right(1)-1)S=(25-1)\frac{5^6}{24}\Rightarrow \left(f\right(1)-1)S=5^6\therefore \left[\right(f\left(1\right)-1)S{]}^{1/2}=125$

Alternate Solution:
$f(x+y)=f\left(x\right)\cdot f\left(y\right)$
Take $f\left(x\right)=a^{kx}$

Given $f\left(1\right)=25$
$\Rightarrow a^k=25\therefore f\left(x\right)={25}^x$