AP EAMCET · Maths · Application of Derivatives
If \(x>0\), then \(\frac{x}{1+x}-\log (1+x)\)
- A is less than zero
- B is greater than zero
- C is equal to zero
- D takes all the real values
Answer & Solution
Correct Answer
(A) is less than zero
Step-by-step Solution
Detailed explanation
Let a function \(f(x)=\frac{x}{1+x}-\log (1+x)\), for \(x>0\) \[ \begin{aligned} \because \quad f^{\prime}(x) & =\frac{(x+1)-x}{(1+x)^2}-\frac{1}{(1+x)} \\ & =\frac{1}{(1+x)^2}-\frac{1}{1+x}=\frac{1-(1+x)}{(1+x)^2} \\ & =\frac{-x}{(1+x)^2} 0 \end{aligned} \]…
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