MHT CET · Maths · Application of Derivatives
The function \(f(x)=\log x-\frac{2 x}{x+2}\)
- A \(x \in(-\infty, 1)\)
- B \(x \in(-1, \infty)\)
- C \(x \in(-\infty, 0)\)
- D \(x\in(0, \infty)\)
Answer & Solution
Correct Answer
(D) \(x\in(0, \infty)\)
Step-by-step Solution
Detailed explanation
\(\text { Given } f(x) =\log x-\frac{2 x}{x+2} \)
\( \therefore f^{\prime}(x) =\frac{1}{x}-\left[\frac{(x+2)(2)-(2 x)(1)}{(x+2)^{2}}\right] \)
\( =\frac{1}{x}-\left[\frac{4}{(x+2)^{2}}\right]=\frac{x^{2}+4 x}{x(x+2)^{2}} \)
\( =\frac{x(x+4)}{x(x+2)^{2}}\)
For \(f^{\prime}(x)>0, x \neq 0\) and \(x>-4\)
Hence \(x \in(0, \infty)\) is permissible.
\( \therefore f^{\prime}(x) =\frac{1}{x}-\left[\frac{(x+2)(2)-(2 x)(1)}{(x+2)^{2}}\right] \)
\( =\frac{1}{x}-\left[\frac{4}{(x+2)^{2}}\right]=\frac{x^{2}+4 x}{x(x+2)^{2}} \)
\( =\frac{x(x+4)}{x(x+2)^{2}}\)
For \(f^{\prime}(x)>0, x \neq 0\) and \(x>-4\)
Hence \(x \in(0, \infty)\) is permissible.
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