A) (a) \[\frac{f\left( x \right)+1}{f\left( x \right)+3}\]
B) (b) \[\frac{3f\left( x \right)+1}{f\left( x \right)+3}\]
C) (c) \[\frac{f\left( x \right)+3}{f\left( x \right)+1}\]
D) (d) \[\frac{f\left( x \right)+3}{3f\left( x \right)+1}\]
Correct Answer: B
Solution :
[b] \[y=f(x)=\frac{x-1}{x+1}\] \[\because f(2x)=\frac{2x-1}{2x+1}\] \[\because f=\frac{x-1}{x+1}\] \[yx+y=x-1\] \[x\left( y-1 \right)=-1-y\] \[x=\frac{1+y}{1-y}f(2x)=\frac{2\left( \frac{1+y}{1-y} \right)-1}{2\left( \frac{1+y}{1-y} \right)+1}\] \[=\frac{2(1+y)-(1-y)}{2(1+y)+(1-y)}\] \[=\frac{3y+1}{y+3~~~}=\frac{3.f\left( x \right)+1}{f\left( x \right)+3}\]. Hence, option [b] is correct.
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