TS EAMCET · Maths · Application of Derivatives
If \(f(x)=x+\log \left(\frac{x-1}{x+1}\right)\) is a well-defined real valued function then \(f\) is
- A monotonically decreasing function
- B monotonically increasing function
- C increasing in \((1, \infty)\) and decreasing in \((-\infty,-1)\)
- D decreasing in \((1, \infty)\) and increasing in \((-\infty,-1)\)
Answer & Solution
Correct Answer
(B) monotonically increasing function
Step-by-step Solution
Detailed explanation
\(f'(x) = \frac{d}{dx}\left(x+\log \left(\frac{x-1}{x+1}\right)\right)\) \(f'(x) = 1 + \frac{1}{x-1} - \frac{1}{x+1}\) \(f'(x) = 1 + \frac{(x+1)-(x-1)}{(x-1)(x+1)}\) \(f'(x) = 1 + \frac{2}{x^2-1}\) Domain: \(\frac{x-1}{x+1} > 0 \implies x \in (-\infty, -1) \cup (1, \infty)\) For…
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