A function satisfies f(x-y)=(f(x))/(f(y)) , where f(y) ≠ 0 . If f(1)=0.5 , then what is f(2)+f(3)+f(4)+f(5)+f(6) equal to?

Correct answer: D. (31)/(64) aligned & Since, f(2-1)=f(1)= f(2) f(1) & f(2)=(f(1))^2=(5)/(10) × (5)/(10)=(1)/(4) & f(3-2)=f(1)= f(3) f(2) (1)/(2)= f(3) (1)/(4) f(3)=(1)/(8) & Similarly we get f(4)=(1)/(16), f(5)=(1)/(32), f(6)=(1)/(64) & Now, f(2)+f(3)+f(4)+f(5)+f(6) & =(1)/(4)+(1)/(8)+(1)/(16)+(1)/(32)+(1)/(64)=(31)/(64) aligned

NDA 2023

Exam: NDA
Year: 2023
Subject: Mathematics
Chapter: Functions

NDA 2023 · Mathematics · Functions

A function satisfies f(x-y)=(f(x))/(f(y)) , where f(y) ≠ 0 . If...

A function satisfies

f(x-y)=\frac{f(x)}{f(y)}

, where

f(y) \neq 0

. If

f(1)=0.5

, then what is

f(2)+f(3)+f(4)+f(5)+f(6)

equal to?

Options

A $\frac{15}{32}$
B $\frac{17}{32}$
C $\frac{29}{64}$
D $\frac{31}{64}$

Correct answer

D. $\frac{31}{64}$

Step-by-step solution

\begin{aligned} & \text{ Since, } \mathrm{f}(2-1)=\mathrm{f}(1)=\frac{\mathrm{f}(2)}{\mathrm{f}(1)} \\ & \Rightarrow \mathrm{f}(2)=(\mathrm{f}(1))^2=\frac{5}{10} \times \frac{5}{10}=\frac{1}{4} \\ & \mathrm{f}(3-2)=\mathrm{f}(1)=\frac{\mathrm{f}(3)}{\mathrm{f}(2)} \Rightarrow \frac{1}{2}=\frac{\mathrm{f}(3)}{\frac{1}{4}} \Rightarrow \mathrm{f}(3)=\frac{1}{8} \\ & \text{Similarly we get } \mathrm{f}(4)=\frac{1}{16}, \mathrm{f}(5)=\frac{1}{32}, \mathrm{f}(6)=\frac{1}{64} \\ & \text{Now, } \mathrm{f}(2)+\mathrm{f}(3)+\mathrm{f}(4)+\mathrm{f}(5)+\mathrm{f}(6) \\ & \quad=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{31}{64}\end{aligned}

Practice Functions on Quantrex Academy →

More from Functions