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CUET · MATHS · PYQ PAPER 2025

The interval(s), where the function \(f(x)=\left\{\begin{array}{l}\frac{1-e^x}{e^{2 x}-1}: x \neq 0 \\ -\frac{1}{2}: x=0\end{array}\right.\) is increasing, is/are:

  1. A \((1, \infty)\)
  2. B \((-\infty, 1)\)
  3. C \((-\infty, \infty)\)
  4. D \((-\infty, \infty)-\{0\}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \((-\infty, \infty)\)

Step-by-step Solution

Detailed explanation

\(f(x)=\frac{1-e^x}{e^{2x}-1} = \frac{-(e^x-1)}{(e^x-1)(e^x+1)} = \frac{-1}{e^x+1}\), for \(x \neq 0\) \(\lim_{x \to 0} f(x) = \frac{-1}{e^0+1} = -\frac{1}{2} = f(0)\). Thus, \(f(x)=\frac{-1}{e^x+1}\) for all \(x \in \mathbb{R}\).…
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