Question:
Let \(f(x)= \frac{x^2+1}{x^2-1} \) if \(x \neq 1,-1,\) and \(1\) \(x=1,-1\). Let \(g(x)= \frac{x+1}{x-1} \) if \(x \neq 1\) and \(3\) if \(x=1\).What is the minimum possible values of \( \frac{f(x)}{g(x)} \)?
Let \(f(x)= \frac{x^2+1}{x^2-1} \) if \(x \neq 1,-1,\) and \(1\) \(x=1,-1\). Let \(g(x)= \frac{x+1}{x-1} \) if \(x \neq 1\) and \(3\) if \(x=1\).What is the minimum possible values of \( \frac{f(x)}{g(x)} \)?
Updated On: Aug 9, 2024
- \( \frac{1}{2} \)
- -1
- \( \frac{1}{4} \)
- \( \frac{1}{3} \)
- 1
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The Correct Option is D
Solution and Explanation
The correct answer is option (D):\( \frac{1}{3} \)
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