If f(x) = \(\frac{1}{1+\frac{1}{x}}\) ; g(x) = \(\frac{1}{1+ \frac{

1.	If f(x) = \(\frac{1}{1+\frac{1}{x}}\) ; g(x) = \(\frac{1}{1+ \frac{1}{f(x)}}\) , then g(2) equals(a) \(\frac{1}{5}\) (b) \(\frac{1}{25}\) (c) \(\frac{2}{5}\)(d) \(\frac{1}{16}\)
Answer» Answer: (c) = \(\frac{2}{5}\) Given, f (x) = \(\frac{1}{1+\frac{1}{x}}\) and g(x) = \(\frac{1}{1+\frac{1}{f(x)}}\) ∴ f(x) = \(\frac{1}{\frac{x+1}{x}} = \frac{x}{x+1}\) ⇒ g(x) = \(\frac{1}{1+ \frac{1}{\frac{x}{x+1}}}\) = \(\frac{1}{1+ \frac{x+1}{x}}\) = \(\frac{x}{2x+1}\) ⇒ g(2) = \(\frac{2}{2\times 2+1}\) = \(\frac{2}{5}\)

If f(x) = \(\frac{1}{1+\frac{1}{x}}\) ; g(x) = \(\frac{1}{1+ \frac{1}{f(x)}}\) , then g(2) equals(a) \(\frac{1}{5}\) (b) \(\frac{1}{25}\) (c) \(\frac{2}{5}\)(d) \(\frac{1}{16}\)

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Answer: (c) = \(\frac{2}{5}\)

Given, f (x) = \(\frac{1}{1+\frac{1}{x}}\) and g(x) = \(\frac{1}{1+\frac{1}{f(x)}}\)

∴ f(x) = \(\frac{1}{\frac{x+1}{x}} = \frac{x}{x+1}\)

⇒ g(x) = \(\frac{1}{1+ \frac{1}{\frac{x}{x+1}}}\)

= \(\frac{1}{1+ \frac{x+1}{x}}\) = \(\frac{x}{2x+1}\)

⇒ g(2) = \(\frac{2}{2\times 2+1}\)

= \(\frac{2}{5}\)