A) \[\frac{f(x)+1}{f(x)+3}\]
B) \[\frac{3f(x)+1}{f(x)+3}\]
C) \[\frac{f(x)+3}{f(x)+1}\]
D) \[\frac{f(x+3)}{3f(x)+1}\]
Correct Answer: B
Solution :
\[\frac{f(x)}{1}=\frac{x-1}{x+1}\] Applying componendo - dividend \[\frac{f(x)+1}{f(x)-1}=\frac{x-1+x+1}{x-1-x-1}=\frac{x}{-1}\] \[x=\frac{f(x)+1}{1-f(x)}\] Now \[f(2x)=\frac{2x-1}{2x+1}\] \[=\frac{2\left( \frac{f(x)+1}{1-f(x)} \right)-1}{2\left( \frac{f(x)+1}{1-f(x)} \right)+1}=\frac{3f(x)+1}{f(x)+3}\]
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