Limits, Continuity & Differentiability
Functional Equation — Power Function
nta_pyq_2024_jan
Grade 12
Question:
Let $f:\mathbb{R}-\{0\}\to\mathbb{R}$ be a function satisfying $f\left(\dfrac{x}{y}\right)=\dfrac{f(x)}{f(y)}$ for all $x,y$, $f(y)\ne 0$. If $f'(1)=2024$, then
$xf'(x)-2024f(x)=0$
$xff'(x)+2024f(x)=0$
$xf'(x)+f(x)=2024$
$xf'(x)-2023f(x)=0$
Step-by-Step Solution
Key Concept: The functional equation $f(x/y)=f(x)/f(y)$ implies $f$ is a power function $f(x)=x^k$. Differentiate the functional equation partially w.r.t. $x$, then set $y\to x$ to find a differential equation relating $f$ and $f'$.
Differentiate $f(x/y)=f(x)/f(y)$ w.r.t. $x$: $\frac{1}{y}f'(x/y)=\frac{f'(x)}{f(y)}$.
Set $y=x$: $\frac{1}{x}f'(1)=\frac{f'(x)}{f(x)}\Rightarrow\frac{2024}{x}=\frac{f'(x)}{f(x)}\Rightarrow xf'(x)=2024f(x)\Rightarrow xf'(x)-2024f(x)=0$.
Correct Answer: 1