Functions Question 490
Question: If $ f(x)=\frac{\alpha ,x}{x+1},\ x\ne -1 $ . Then, for what value of $ \alpha $ is $ f(f(x))=x $
[IIT Screening 2001; UPSEAT 2001]
Options:
A) $ \sqrt{2} $
B) $ -\sqrt{2} $
1
?1
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(f(x))=\frac{\alpha ,f(x)}{f(x)+1}=\frac{\alpha ( \frac{\alpha x}{x+1} )}{( \frac{\alpha x}{x+1}+1 )}=\frac{{{\alpha }^{2}}.x}{\alpha x+x+1} $ \ $ x=\frac{{{\alpha }^{2}}.x}{(\alpha +1)x+1} $ or $ x((\alpha +1)x+1-{{\alpha }^{2}})=0 $ or $ (\alpha +1)x^{2}+(1-{{\alpha }^{2}})x=0 $ . This should hold for all x.
Þ $ \alpha +1=0,\ 1-{{\alpha }^{2}}=0 $ , \ $ =\underset{h\to 0}{\mathop{\lim }}\frac{h,.,\frac{{e^{-1/h}}-{e^{1/h}}}{{e^{-1/h}}+{e^{1/h}}}-0}{-h}=-1 $ .
BETA
