Functions Question 12
Question 12 - 30 January - Shift 1
Suppose $f: R \to(0, \infty)$ be a differentiable function such that $5 f(x+y)=f(x) \cdot f(y), \forall x, y \in R$. If $f(3)=320$, then $\sum _{n=0}^{5} f(n)$ is equal to :
(1) 6875
(2) 6575
(3) 6825
(4) 6528
Show Answer
Answer: (3)
Solution:
Formula: Operations on functions, Sum of terms in G.P.
Option (3)
$ \begin{aligned} & 5 f(x+y)=f(x) \cdot f(y) \\ & 5 f(0)=f(0)^{2} \Rightarrow f(0)=5 \\ & 5 f(x+1)=f(x) \cdot f(1) \\ \Rightarrow \quad & \frac{f(x+1)}{f(x)}=\frac{f(1)}{5} \\ \Rightarrow \quad & \frac{f(1)}{f(0)} \cdot \frac{f(2)}{f(1)} \cdot \frac{f(3)}{f(2)}=(\frac{f(1)}{5})^{3} \text{ } \\ \Rightarrow \quad & \frac{320}{5}=\frac{(f(1))^{3}}{5^{3}} \Rightarrow f(1)=20 \\ \therefore \quad & 5 f(x+1)=20 \cdot f(x) \Rightarrow f(x+1)=4 f(x) \\ & \sum _{n=0}^{5} f(n)=5+5.4+5.4^{2}+5.4^{3}+5.4^{4}+5.4^{5} \\ & =\frac{5[4^{6}-1]}{3}=6825 \end{aligned} $
