Question:
Let \(f\) be a real valued function, defined on \(\mathbb{R} - \{-1, 1\}\) 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?
Let \(f\) be a real valued function, defined on \(\mathbb{R} - \{-1, 1\}\) 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?
\(f(x) = 3\log_e\left|\frac{x-1}{x+1}\right| - \frac{2}{x-1}\).
Then in which of the following intervals, function \(f(x)\) is increasing?
Show Hint
Differentiate f and find where \(f'(x)\) is non-negative.
Updated On: Nov 4, 2026
- \((-\infty, -1) \cup \left(\left[\frac{1}{2}, \infty\right) - \{1\}\right)\)
- \((-\infty, \infty) - \{-1, 1\}\)
- \(\left(-\infty, \frac{1}{2}\right] - \{-1\}\)
- \(\left(-1, \frac{1}{2}\right]\)
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The Correct Option is A
Solution and Explanation
\(f(x) = 3\log_e|x-1| - 3\log_e|x+1| - \frac{2}{x-1}\).
\(f'(x) = \frac{3}{x-1} - \frac{3}{x+1} + \frac{2}{(x-1)^2}\).
\(\frac{3}{x-1} - \frac{3}{x+1} = \frac{6}{(x-1)(x+1)}\), so \(f'(x) = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} = \frac{8x - 4}{(x-1)^2(x+1)} = \frac{4(2x-1)}{(x-1)^2(x+1)}\).
Since \((x-1)^2 > 0\) for \(x \ne 1\), \(f'(x)\ge 0\) iff \(\frac{2x-1}{x+1}\ge 0\), i.e. \(x < -1\) or \(x \ge \frac{1}{2}\).
Excluding \(x = 1\) (not in the domain): \(x \in (-\infty, -1)\cup\left(\left[\frac{1}{2},\infty\right) - \{1\}\right)\).
Answer: (1) \((-\infty, -1) \cup \left(\left[\frac{1}{2}, \infty\right) - \{1\}\right)\)
\(f'(x) = \frac{3}{x-1} - \frac{3}{x+1} + \frac{2}{(x-1)^2}\).
\(\frac{3}{x-1} - \frac{3}{x+1} = \frac{6}{(x-1)(x+1)}\), so \(f'(x) = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} = \frac{8x - 4}{(x-1)^2(x+1)} = \frac{4(2x-1)}{(x-1)^2(x+1)}\).
Since \((x-1)^2 > 0\) for \(x \ne 1\), \(f'(x)\ge 0\) iff \(\frac{2x-1}{x+1}\ge 0\), i.e. \(x < -1\) or \(x \ge \frac{1}{2}\).
Excluding \(x = 1\) (not in the domain): \(x \in (-\infty, -1)\cup\left(\left[\frac{1}{2},\infty\right) - \{1\}\right)\).
Answer: (1) \((-\infty, -1) \cup \left(\left[\frac{1}{2}, \infty\right) - \{1\}\right)\)
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