MHT CET · Maths · Application of Derivatives
The function \(\mathrm{f}(\mathrm{x})=\log (1+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}}\) is increasing on
- A \((-\infty, \infty)\)
- B \((-5, \infty)\)
- C \((-\infty, 0)\)
- D \((-1, \infty)\)
Answer & Solution
Correct Answer
(D) \((-1, \infty)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(\mathrm{x})=\log (1+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}} \Rightarrow \mathrm{x} \neq-2 \)
\( \therefore \mathrm{f}^{\prime}(\mathrm{x}) =\frac{1}{(1+\mathrm{x})}-\left[\frac{(2+\mathrm{x})(2)-(2 \mathrm{x})(1)}{(2+\mathrm{x})^2}\right] \)
\( =\frac{1}{1+\mathrm{x}}-\left[\frac{4}{(2+\mathrm{x})^2}\right]=\frac{(\mathrm{x}+2)^2-4(\mathrm{x}+1)}{(\mathrm{x}+2)^2(1+\mathrm{x})}\)
When \(f^{\prime}(x)>0\), we write
\(
\frac{x^2}{(x+2)^2(1+x)}>0
\)
Since \(x^2>0\) and \((x+2)^2>0\), we write \((1+x)>0 \Rightarrow x>-1\)
\( \therefore \mathrm{f}^{\prime}(\mathrm{x}) =\frac{1}{(1+\mathrm{x})}-\left[\frac{(2+\mathrm{x})(2)-(2 \mathrm{x})(1)}{(2+\mathrm{x})^2}\right] \)
\( =\frac{1}{1+\mathrm{x}}-\left[\frac{4}{(2+\mathrm{x})^2}\right]=\frac{(\mathrm{x}+2)^2-4(\mathrm{x}+1)}{(\mathrm{x}+2)^2(1+\mathrm{x})}\)
When \(f^{\prime}(x)>0\), we write
\(
\frac{x^2}{(x+2)^2(1+x)}>0
\)
Since \(x^2>0\) and \((x+2)^2>0\), we write \((1+x)>0 \Rightarrow x>-1\)
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