JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(f\) be a real valued function, defined on \(R -\{-1,1\}\) and given by \(f(x)=3 \log _{e}\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}\) Then in which of the following intervals, function \(f ( x )\) is increasing?
- A \((-\infty,-1) \cup\left(\left[\frac{1}{2}, \infty\right)-\{1\}\right)\)
- B \((-\infty, \infty)-\{-1,1\}\)
- C \(\left(-1, \frac{1}{2}\right]\)
- D \(\left(-\infty, \frac{1}{2}\right]-\{-1\}\)
Answer & Solution
Correct Answer
(A) \((-\infty,-1) \cup\left(\left[\frac{1}{2}, \infty\right)-\{1\}\right)\)
Step-by-step Solution
Detailed explanation
\(f(x)=3 \ell n(x-1)-3 \ell n(x+1)-\frac{2}{x-1}\) \(f^{\prime}(x)=\frac{3}{x-1}-\frac{3}{x+1}+\frac{2}{(x-1)^{2}}\) \(f^{\prime}(x)=\frac{4(2 x-1)}{(x-1)^{2}(x+1)}\) \(f^{\prime}(x) \geq 0\) \(\Rightarrow \quad x \in(-\infty,-1) \cup\left[\frac{1}{2}, 1\right) \cup(1, \infty)\)
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