A) \[\sqrt{2}\]
B) \[-\sqrt{2}\]
C) 1
D) 2
E) ?1
Correct Answer: E
Solution :
\[f(x)=\frac{\alpha x}{x+1}\]; \[f(f(x))=f\left( \frac{\alpha x}{x+1} \right)=\frac{\alpha \left( \frac{\alpha x}{x+1} \right)}{\frac{\alpha x}{x+1}+1}\] But \[f(f(x))=x\], \\[\frac{{{\alpha }^{2}}x}{\alpha x+x+1}=x\] Put \[\alpha =-1\], \[\frac{{{(-1)}^{2}}x}{(-1)x+x+1}=\frac{x}{-x+x+1}=x\]; \\[\alpha =-1\].
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