Definite Integration Question 96
Question: The function $ L(x)=\int_1^{x}{\frac{dt}{t}} $ satisfies the equation
[IIT 1996; DCE 2001]
Options:
A) $ L(x+y)=L(x)+L(y) $
B) $ L( \frac{x}{y} )=L(x)+L(y) $
C) $ L(xy)=L(x)+L(y) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given function $ L(x)=\int_1^{x}{\frac{1}{t}dt=[\log t]_1^{x}}=\log x-\log 1 $
therefore $ L(x)=\log x $ ,
Hence $ L(xy)=L(x)+L(y) $ .
BETA
