If f:R to R satisfies f(x + y) = f(x) + f(y) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the functional equation $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$. Since

Vedclass · Accepted Answer

$\frac{7n(n + 1)}{2}$

Vedclass · Answer

(D) Given the functional equation $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$. Since $f(1) = 7$,we have $c(1) = 7$,so $c = 7$. Thus,$f(x) = 7x$. Now,we need to calculate the sum $\sum_{r = 1}^n f(r) = \sum_{r = 1}^n 7r$. This simplifies to $7 \sum_{r = 1}^n r$. Using the formula for the sum of the first $n$ natural numbers,$\sum_{r = 1}^n r = \frac{n(n + 1)}{2}$. Therefore,$\sum_{r = 1}^n f(r) = 7 	imes \frac{n(n + 1)}{2} = \frac{7n(n + 1)}{2}$.

If $f:R \to R$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 7$,then $\sum_{r = 1}^n f(r)$ is

Similar Questions

If $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 1$,then find the value of $\lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{f(\tan x) - f(\sin x)}$.

If $f(x + ay, x - ay) = axy$,then $f(x, y)$ is equal to

If $g:[-2, 2] \to R$ where $g(x) = x^3 + \tan x + \left[ \frac{x^2 + 1}{P} \right]$ is an odd function,then the value of the parameter $P$ is:

If $f(x)$ is a differentiable function such that $f(xy) = f(x) + f(y)$ for all $x, y > 0$,then $f(e) + f(1/e) = ?$

If $f(x)$ satisfies $f(7 - x) = f(7 + x)$ for all $x \in R$ such that $f(x)$ has exactly $5$ real roots which are all distinct,and the sum of the real roots is $S$,then $S/7$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module