A) \[\sqrt{2}\]
B) \[-\sqrt{2}\]
C) 1
D) ?1
Correct Answer: D
Solution :
\[f(f(x))=\frac{\alpha \,f(x)}{f(x)+1}=\frac{\alpha \left( \frac{\alpha x}{x+1} \right)}{\left( \frac{\alpha x}{x+1}+1 \right)}=\frac{{{\alpha }^{2}}.x}{\alpha x+x+1}\] \\[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 \[\alpha +1=0,\,\,1-{{\alpha }^{2}}=0\], \ \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{h\,.\,\frac{{{e}^{-1/h}}-{{e}^{1/h}}}{{{e}^{-1/h}}+{{e}^{1/h}}}-0}{-h}=-1\].
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