AP EAMCET · Maths · Basic of Mathematics
The set of all real values of \(x\) for which \(\frac{x^2-1}{(x-4)(x-3)} \geq 1\) is
- A \([-1,1] \cup(3,4)\)
- B \(\left[\frac{13}{7}, 3\right) \cup(4, \infty)\)
- C \(\left(-\infty, \frac{13}{7}\right] \cup(3,4)\)
- D \(\mathbb{R}-[3,4]\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{13}{7}, 3\right) \cup(4, \infty)\)
Step-by-step Solution
Detailed explanation
\(\frac{x^2-1}{(x-4)(x-3)} \geq 1\) \(\frac{x^2-1-(x^2-7x+12)}{(x-4)(x-3)} \geq 0\) \(\frac{7x-13}{(x-4)(x-3)} \geq 0\) Critical points: \(x=\frac{13}{7}, x=3, x=4\) Sign analysis: \([\frac{13}{7}, 3) \cup (4, \infty)\)
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