Functions Question 589
Question: If f is an even function defined on the interval (-5, 5), then four real values of x satisfying the equation $ f(x)=f( \frac{x+1}{x+2} ) $ are
[IIT 1996]
Options:
A) $ \frac{-3-\sqrt{5}}{2},\ \frac{-3+\sqrt{5}}{2},\ \frac{3-\sqrt{5}}{2},\ \frac{3+\sqrt{5}}{2} $
B) $ \frac{-5+\sqrt{3}}{2},\ \frac{-3+\sqrt{5}}{2},\ \frac{3+\sqrt{5}}{2},\ \frac{3-\sqrt{5}}{2} $
C) $ \frac{3-\sqrt{5}}{2},\ \frac{3+\sqrt{5}}{2},\ \frac{-3-\sqrt{5}}{2},\ \frac{5+\sqrt{3}}{2} $
D) $ -3-\sqrt{5},\ -3+\sqrt{5},\ 3-\sqrt{5},\ 3+\sqrt{5} $
Show Answer
Answer:
Correct Answer: A
Solution:
Since f is an even function $ f,(-x)=f(x),,\forall x\in (-5,,5). $
We are given that $ f(x)=f,( \frac{x+1}{x+2} ) $
$ \Rightarrow f(-x)=f( \frac{-x+1}{-x+2} ),\Rightarrow f(x)=f,( \frac{-x+1}{-x+2} ) $ $ [\because ,f(-x)=f(x)] $
To find the values of x, we set $ x=\frac{-x+1}{-x+2}\Rightarrow x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2} $
Also $ f(x)=f,( \frac{x+1}{x+2} )=f(-x) $
To find the values of x, we set $ -x=\frac{x+1}{x+2}\Rightarrow x=\frac{-3\pm \sqrt{9-4}}{2}=\frac{-3\pm \sqrt{5}}{2} $
Thus the four required values of x are $ \frac{-3-\sqrt{5}}{2},,\frac{-3+\sqrt{5}}{2},,\frac{3-\sqrt{5}}{2},,\frac{3+\sqrt{5}}{2}. $
BETA
