CUET · MATHS · PYQ PAPER 2025
The function, \(f(x)=x-\frac{1}{x}\) is
- A increasing for all \(x \in(-\infty, 0) \cup(0, \infty)\)
- B decreasing for all \(x \in(-\infty, 0) \cup(0, \infty)\)
- C increasing for all \(x \in(-\infty, \infty)\)
- D neither increasing nor decreasing for \(x \in(-\infty, \infty)\)
Answer & Solution
Correct Answer
(A) increasing for all \(x \in(-\infty, 0) \cup(0, \infty)\)
Step-by-step Solution
Detailed explanation
\(f'(x) = \frac{d}{dx}(x - x^{-1}) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2}\) For \(x \in (-\infty, 0) \cup (0, \infty)\), \(x^2 > 0\), so \(\frac{1}{x^2} > 0\). \(f'(x) = 1 + \frac{1}{x^2} > 1\) for all \(x \in (-\infty, 0) \cup (0, \infty)\). Thus, \(f'(x) > 0\), so the function…
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