Q. If $f$ is a real valued function such that $f(x + y) = f(x) + f(y)$ and $f(1) = 5$, then the value of $f(100)$ is
A) 200
B) 300
C) 350
D) 500
Explanation
(d)
Given,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x + y) = f(x) + f(y)$
For $x = 1,\ \ \ y = 1$ we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(2) = f(1) + f(1)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.f(1) = 10$
Also
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(3) = f(2) + f(1) = 15$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(n) = 5n$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ f(100) = 500$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.