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AP EAMCET · Maths · Quadratic Equation

The set of all real values ' \(a\) ' for which \(-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \lt 3\) holds for all real values of \(x\) is

  1. A \((-7,5)\)
  2. B \((5, \infty)\)
  3. C \((1,5)\)
  4. D \((-\infty, 1)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \((1,5)\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \qquad-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \lt 3 \\ & \text { Consider }-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \Rightarrow 3 x^2+(a+1) x+3\gt0 \\ & \Rightarrow \mathrm{D} \lt 0 \Rightarrow(a+1)^2-36 \lt 0 \Rightarrow a^2+2 a-35 \lt 0\end{aligned}\)…
From AP EAMCET
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