Let f be a function such that 3f(x) + 2fleft( | vedclass

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

Question

(C) Given $m = \sum_{i=1}^9 i^2 = \frac{9(9+1)(2 	imes 9 + 1)}{6} = \frac{9 	imes 10 	imes 19}{6} = 285$. Since $m = 285$, the expression $\frac{m}

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) Given $m = \sum_{i=1}^9 i^2 = \frac{9(9+1)(2 	imes 9 + 1)}{6} = \frac{9 	imes 10 	imes 19}{6} = 285$. Since $m = 285$, the expression $\frac{m}{19x}$ becomes $\frac{285}{19x} = \frac{15}{x}$. The given equation is $3f(x) + 2f\left(\frac{15}{x}ight) = 5x$. Replacing $x$ with $\frac{15}{x}$, we get $3f\left(\frac{15}{x}ight) + 2f(x) = 5\left(\frac{15}{x}ight) = \frac{75}{x}$. We have a system of two equations: $1) 3f(x) + 2f\left(\frac{15}{x}ight) = 5x$ $2) 2f(x) + 3f\left(\frac{15}{x}ight) = \frac{75}{x}$ Multiply $(1)$ by $3$ and $(2)$ by $2$: $9f(x) + 6f\left(\frac{15}{x}ight) = 15x$ $4f(x) + 6f\left(\frac{15}{x}ight) = \frac{150}{x}$ Subtracting the second from the first: $5f(x) = 15x - \frac{150}{x} \implies f(x) = 3x - \frac{30}{x}$. Now, $f(5) = 3(5) - \frac{30}{5} = 15 - 6 = 9$. $f(2) = 3(2) - \frac{30}{2} = 6 - 15 = -9$. Therefore, $f(5) - f(2) = 9 - (-9) = 18$.

Let $f$ be a function such that $3f(x) + 2f\left(\frac{m}{19x}\right) = 5x$, $x \neq 0$, where $m = \sum_{i=1}^9 (i)^2$. Then $f(5) - f(2)$ is equal to

Similar Questions

Let $f\left( \frac{x + 8y}{9} \right) = \frac{f(x) + 8f(y)}{9}$ for all real $x$ and $y$. If $f'(0)$ exists and is equal to $2$,and $f(0) = -5$,then $f(7)$ is equal to:

If $f(x + y) = f(x)f(y)$ and $\sum_{x=1}^{\infty} f(x) = 2$,where $x, y \in N$ and $N$ is the set of all natural numbers,then the value of $\frac{f(4)}{f(2)}$ is

If $f(x)$ is a polynomial function satisfying $f(x) \cdot f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$ and $f(4) = 65$,then the value of $f(6)$ is:

Exactly how many functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ exist such that $f(x+y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$ for all $x, y \in \mathbb{Q}$?

If $f(x_1) - f(x_2) = f\left( \frac{x_1 - x_2}{1 - x_1 x_2} \right)$ for $x_1, x_2 \in [-1, 1]$,then $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module