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Let $f : (0, \infty) \rightarrow (0, \infty)$ be a strictly decreasing function. Consider $h(x) = \dfrac{f(\frac{x}{1+x})}{1+f(\frac{x}{1+x})}$. Which one of the following is always true?

  1. $h$ is strictly decreasing
  2. $h$ is strictly increasing
  3. $h$ is strictly decreasing at first and then strictly increasing
  4. $h$ is strictly increasing at first and then strictly decreasing
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1 Answer

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It is given ,

$\large f(x)$ is strictly decreasing function so ,

$\large f'(x) <0$ for any x in $\large \left ( 0,\infty \right )$.

$\large h(x)=\frac{f(\frac{x}{x+1})}{1+f(\frac{x}{x+1})}$

$\large h(x)=1-\frac{1}{1+f(\frac{x}{x+1})}$

$\large h'(x)=-\frac{(-1)f'(\frac{x}{x+1})\frac{\mathrm{d} }{\mathrm{d} x}(\frac{x}{x+1})}{(1+f(\frac{x}{x+1}))^{2}}$

$\large h'(x)=\frac{f'(\frac{x}{x+1})\frac{1}{(1+x)^{2}}}{(1+f(\frac{x}{x+1}))^{2}}$

now here , $\large (1+f(\frac{x}{x+1}))^{2}>0$,$\large (\frac{1}{(x+1)^{2}})>0$

so now, the sign of $\large h'(x)$ is only depends on $\large f'(\frac{x}{x+1})$.

as it is given initially $\large f(x)$ is strictly decreasing function so $\large f'(x)$ is always less than zero .

so ,$\large f'(\frac{x}{x+1})<0$ .

so , $\large h'(x)<0$ for any x in $\large \left ( 0,\infty \right )$

so ,$\large h(x)$ is strictly decreasing .

so correct option is (A).

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