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

Question

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

MARKS App · Accepted Answer

$[a, b]$ 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, $\quad a=-\frac{1}{13}, b=\frac{1}{3}$ $\therefore \quad a 0$

Search any question & find its solution

Looking for more such questions to practice?