Q105 / 160 · TS EAMCET 2013 · Mathematics

If $f(x)=\frac{1}{1+\frac{1}{x}}$ and $g(x)=\frac{1}{1+\frac{1}{f(x)}}$, then $g^{\prime}(2)$ is equal to

  • A
    $\frac{1}{5}$
  • Correct answer
    B
    $\frac{1}{25}$
  • C
    $5$
  • D
    $\frac{1}{16}$

Solution

$\begin{aligned} & \quad f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x} \\ & \text { and } \quad g(x)=\frac{1}{1+\frac{1}{f(x)}}=\frac{1}{1+\frac{1+x}{x}}=\frac{x}{2 x+1} \\ & \therefore \quad g^{\prime}(x)=\frac{(2 x+1) \cdot 1-x(2)}{(2 x+1)^2}=\frac{1}{(2 x+1)^2} \\ & \text { Now, } g^{\prime}(2)=\frac{1}{(4+1)^2}=\frac{1}{25}\end{aligned}$

Hence, the correct answer is option: 'B'.

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