Q50 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

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

  • Correct answer
    A
    $\frac{1}{2}$
  • B
    $\frac{4}{3}$
  • C
    $\frac{3}{4}$
  • D
    $-\frac{1}{2}$

Solution

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, $x \in \mathbf{R}$ $\begin{aligned} & \because(4 y-2)^2-4(y-1)(10 y-5)>0 \\ & \Rightarrow 6 y^2-11 y+4 \leq 0 \\ & \Rightarrow(2 y-1)(3 y-4) \leq 0 \\ & \therefore y \in\left(\frac{1}{2}, \frac{4}{3}\right)\end{aligned}$ $\therefore$ Minimum value of $\frac{x^2+2 x+5}{x^2+4 x+10}$ is $\frac{1}{2}$

Hence, the correct answer is option: 'A'.

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