Q175 / 270 · NDA 2023 (Phase 2) · Mathematics

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?

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

Solution

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

Hence, the correct answer is option: 'D'.

Relevant Questions