Functions Question 463
Question: Let $ f(x)=x^{2}+3x-3,x>0. $ If n points $ x_1,x_2,x_3,…x_{n} $ are so chosen on the x-axis such that (i) $ \frac{1}{n}\sum\limits_{i=1}^{n}{{f^{-1}}(x_{i})}=f( \frac{1}{n}\sum\limits_{i=1}^{n}{x_{i}} ) $ (ii) $ \sum\limits_{i=1}^{n}{{f^{-1}}}(x_{i})=\sum\limits_{i=1}^{n}{x_{i}}, $ where $ {f^{-1}} $ denotes the inverse of f. The value of $ \frac{x_1+x_2+…+x_{n}}{n}= $
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{{f^{-1}}(x_1)+{f^{-1}}(x_2)+…+{f^{-1}}(x_{n})}{n} $ $ =f( \frac{x_1+x_2+…+x_{n}}{n} ) $ and $ \frac{{f^{-1}}(x_1)+{f^{-1}}(x_2)+…+{f^{-1}}(x_{n})}{n} $ $ =f( \frac{x_1+x_2+…+x_{n}}{n} ) $
$ \Rightarrow f(\bar{x})=\bar{x}, $ Where $ \bar{x}=\frac{x_1+x_2+…+x_{n}}{n} $
$ \Rightarrow {{\bar{x}}^{2}}+3\bar{x}-3=\bar{x}\Rightarrow {{\bar{x}}^{2}}+2\bar{x}-3=0 $
$ \Rightarrow \bar{x}=-3,1\Rightarrow \bar{x}=1as\bar{x}>0 $
BETA
