Q51 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

If $x$ is real, then the maximum and minimum values of $\frac{x^2+14 x+9}{x^2+2 x+3}$ are respectively

  • Correct answer
    A
    4,-5
  • B
    5,-4
  • C
    9,3
  • D
    24,6

Solution

$\begin{aligned} & \text { Let } y=\frac{x^2+14 x+9}{x^2+2 x+3} \\ & \Rightarrow \quad x^2 y+2 x y+3 y=x^2+14 x+9 \\ & \Rightarrow \quad x^2(y-1)+2 x(y-7)+3 y-9=0 \end{aligned}$ Here, $x \in \mathbf{R}$ $\begin{aligned} \therefore \quad 4(y-7)^2-4(y-1)(3 y-9) & \geq 0 \\ (y-7)^2-\left(3 y^2-12 y+9\right) & \geq 0 \\ y^2-14 y+49-3 y^2+12 y-9 & \geq 0 \\ \Rightarrow \quad 2 y^2+2 y-40 & \leq 0 \Rightarrow y^2+y-20 \leq 0 \\ (y+5)(y-4) & \leq 0 \quad \therefore \quad y \in[-5,4] \end{aligned}$ $\therefore$ Maximum and minimum value of $\frac{x^2+14 x+9}{x^2+2 x+3}$ are 4 and -5 respectively.

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

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