Functions Question 168
Question: Let the function f be defined by $ f(x)=\frac{2x+1}{1-3x} $ , then $ {f^{-1}}(x) $ is
[Kerala (Engg.) 2002]
Options:
A) $ \frac{x-1}{3x+2} $
B) $ \frac{3x+2}{x-1} $
C) $ \frac{x+1}{3x-2} $
D) $ \frac{2x+1}{1-3x} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y=f(x) $
Þ $ y=\frac{2x+1}{1-3x} $
Þ $ y-3xy=2x+1 $
Þ $ x=\frac{y-1}{3y+2} $
Þ $ {f^{-1}}(y)=\frac{y-1}{3y+2}\Rightarrow {f^{-1}}(x)=\frac{x-1}{3x+2} $ .
BETA
