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
- \(a < 0, b < 0\)
- \(a < 0, b>0\)
- \(a>0, b>0\)
- \(a>0, b < 0\)
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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