MHT CET · Maths · Application of Derivatives
The function \(f(x)=\log (1+x)-\frac{2 x}{2+x}\) is increasing on
- A \((0, \infty)\)
- B \((-\infty, 0)\)
- C \((-\infty, \infty)\)
- D None of these
Answer & Solution
Correct Answer
(A) \((0, \infty)\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\log (1+x)-\frac{2 x}{2+x}\)
\(\therefore f^{\prime}(x) =\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x}{(2+x)^{2}} \)
\( =\frac{x^{2}}{(1+x)(x+2)^{2}}\)
Clearly, \(f^{\prime}(x)>0\) for all \(x>0\). Hence, \(f(x)\) is increasing on \((0, \infty)\).
\(\therefore f^{\prime}(x) =\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x}{(2+x)^{2}} \)
\( =\frac{x^{2}}{(1+x)(x+2)^{2}}\)
Clearly, \(f^{\prime}(x)>0\) for all \(x>0\). Hence, \(f(x)\) is increasing on \((0, \infty)\).
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