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
- A \(\frac{1}{2}\)
- B \(\frac{4}{3}\)
- C \(\frac{3}{4}\)
- D \(-\frac{1}{2}\)
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, \(x \in \mathbf{R}\)…
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