AP EAMCET · Maths · Basic of Mathematics
The set of all real values of \(x\) satisfying the inequation \(\frac{8 x^2-14 x-9}{3 x^2-7 x-6}>2\) is
- A \((-\infty, 1) \cup(3, \infty)\)
- B \(\left(-\infty,-\frac{2}{3}\right) \cup(2, \infty)\)
- C \(\left(-\frac{2}{3}, 2\right)\)
- D \(\left(-\infty,-\frac{2}{3}\right) \cup(3, \infty)\)
Answer & Solution
Correct Answer
(D) \(\left(-\infty,-\frac{2}{3}\right) \cup(3, \infty)\)
Step-by-step Solution
Detailed explanation
\(\frac{8 x^2-14 x-9}{3 x^2-7 x-6}-2>0\) \(\frac{8 x^2-14 x-9 - (6 x^2-14 x-12)}{3 x^2-7 x-6} > 0\) \(\frac{2 x^2 + 3}{3 x^2-7 x-6} > 0\) \(2x^2+3 > 0\) for all real \(x\). \(3x^2-7x-6 > 0\)…
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