TS EAMCET · Maths · Quadratic Equation
If for any real \(x, \frac{11 x^2+12 x+6}{x^2+4 x+2}=y\) is such that \(y < a\) or \(y \geq b\), then \(a, b\) are
- A 3,5
- B \(-5,3\)
- C \(-4,5\)
- D \(-6,4\)
Answer & Solution
Correct Answer
(B) \(-5,3\)
Step-by-step Solution
Detailed explanation
We have, \( \begin{aligned} & y=\frac{11 x^2+12 x+6}{x^2+4 x+2} \\ & \Rightarrow \quad y x^2+4 x y+2 y=11 x^2+12 x+6 \\ & \Rightarrow \quad(y-11) x^2+4 x(y-3)+2(y-3)=0 \\ & \Rightarrow \quad x=\frac{-4(y-3) \pm \sqrt{16(y-3)^2-8}}{2(y-11)} \\ & \end{aligned} \) Since, \(x \in\)…
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