Let f: R-\0\ R be a function satisfying f ( xy )= f(x)f(y) for all x,y,f(y) 0. If f'(1)=2024, then
- xf'(x)+2024f(x)=0
- xf'(x)-2023f(x)=0
- xf'(x)-2024f(x)=0
- xf'(x)+f(x)=2024
Answer
(3) xf'(x)-2024f(x)=0
Solution
Differentiate w.r.t. x (keep y constant): 1yf' ( xy )= f'(x)f(y)
Put y=x: f'(1)x= f'(x)f(x) 2024x= f'(x)f(x)
∴ xf'(x)-2024f(x)=0
Answer: (3)