Continuity And Differentiability Question 8
Question 8 - 2024 (30 Jan Shift 2)
Let $\quad \mathrm{f}: \mathbb{R}-{0} \rightarrow \mathbb{R}$ be a function satisfying $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ for all $x, y, f(y) \neq 0$. If $f^{\prime}(1)=2024$, then
(1) $x f^{\prime}(x)-2024 f(x)=0$
(2) $x f f^{\prime}(x)+2024 f(x)=0$
(3) $x f^{\prime}(x)+f(x)=2024$
(4) $x f^{\prime}(x)-2023 f(x)=0$
Show Answer
Answer (1)
Solution
$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$
$f^{\prime}(1)=2024$
$f(1)=1$
Partially differentiating w. r. t. $x$
$f^{\prime}\left(\frac{x}{y}\right) \cdot \frac{1}{y}=\frac{1}{f(y)} f^{\prime}(x)$
$y \rightarrow x$
$f^{\prime}(1) \cdot \frac{1}{x}=\frac{f^{\prime}(x)}{f(x)}$
$2024 f(x)=x f^{\prime}(x) \Rightarrow x f^{\prime}(x)-2024 f(x)=0$
