If \([a, b]\) is the range of the function \(\frac{x+2}{2 x^2+3 x+6}\) for \(x \in \mathbf{R}\), then

If \([a, b]\) is the range of the function \(\frac{x+2}{2 x^2+3 x+6}\) for \(x \in \mathbf{R}\), then
  1. \(a < 0, b < 0\)
  2. \(a < 0, b>0\)
  3. \(a>0, b>0\)
  4. \(a>0, b < 0\)

Solution

\([a, b]\) is range of \(\frac{x+2}{2 x^2+3 x+6}\) and \(x \in \mathbf{R}\) Let \(y=\frac{x+2}{2 x^2+3 x+6}\) \(\begin{aligned} & \Rightarrow \quad 2 y x^2+3 x y+6 y=x+2 \\ & \Rightarrow \quad 2 y x^2+(3 y-1) x+6 y-2=0 \\ & x \in \mathbf{R} \text { So, } D \geq 0 \\ & \Rightarrow \quad(3 y-1)^2-4(6 y-2)(2 y) \geq 0 \\ & \Rightarrow \quad-39 y^2+10 y+1 \geq 0 \\ & \Rightarrow \quad 39 y^2-10 y-1 \leq 0 \\ & \Rightarrow \quad(3 y-1)(13 y+1) \leq 0 \Rightarrow y \in\left[-\frac{1}{13}, \frac{1}{3}\right] \\ \end{aligned}\) So, \(\quad a=-\frac{1}{13}, b=\frac{1}{3}\) \(\therefore \quad a < 0, b > 0\)

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

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