ExamBro
ExamBro
AP EAMCET · Maths · Quadratic Equation

For \(x \in \mathbb{R}\), the minimum value of \(\frac{x^2+2 x+5}{x^2+4 x+10}\) is

  1. A \(\frac{1}{2}\)
  2. B \(\frac{4}{3}\)
  3. C \(\frac{3}{4}\)
  4. D \(-\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

Let \(\frac{x^2+2 x+5}{x^2+4 x+10}=y, y \in \mathrm{R}\) \(\begin{aligned} & \Rightarrow x^2+2 x+5=y\left(x^2+4 x+10\right) \\ & \Rightarrow(y-1) x^2+(4 y-2) x+(10 y-5)=0\end{aligned}\) Since, \(x \in \mathbf{R}\)…