AP EAMCET · Maths · Quadratic Equation
If \(\mathrm{x}^2-4 \mathrm{ax}+5+\mathrm{a}>0\) for all \(\mathrm{x} \in \mathrm{R}\) whenever \(\mathrm{a} \in(\alpha, \beta)\), then \(4 \beta+\alpha=\)
- A \(0\)
- B \(4\)
- C \(5\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(\Delta \((-4a)^2 - 4(1)(5+a) \(16a^2 - 20 - 4a \(4a^2 - a - 5 \((4a-5)(a+1) \(-1 \(\alpha = -1, \beta = \frac{5}{4}\) \(4\beta + \alpha = 4\left(\frac{5}{4}\right) + (-1)\) \(4\beta + \alpha = 5 - 1 = 4\)
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