AP EAMCET · Maths · Basic of Mathematics
If \(\frac{1}{2} \leq \frac{x^2+x+a}{x^2-x+a} \leq 2 \forall x \in R\), then \(a=\)
- A \(\frac{3}{4}\)
- B \(\frac{-3}{4}\)
- C \(\frac{9}{4}\)
- D \(\frac{-9}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{9}{4}\)
Step-by-step Solution
Detailed explanation
Let \( y = \frac{x^2+x+a}{x^2-x+a} \). For \( x^2-x+a \) to be always positive (ensuring the expression is defined and the sign of the denominator is consistent), its discriminant must be \( \( (-1)^2 - 4(1)(a) \frac{1}{4} \). Consider the inequality…
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