Question

If x is real, the expression $\frac{x+2}{2x^2+3x+6}$ takes all value in the interval

• A. $(-\frac{1}{13},\frac{1}{3})$
• B. $(\frac{1}{13},\frac{1}{3})$
• C. $(-\frac{1}{3},\frac{1}{13})$
• D. $[-\frac{1}{3},\frac{1}{13}]$

Answer 1

The correct option is A

$(-\frac{1}{13},\frac{1}{3})$

If the given expression be y, then
$\Rightarrow y=\frac{x+2}{2x^2+3x+6}$
$\Rightarrow 2x^{2}y+3xy+6y=x+2$
$\Rightarrow 2x^{2}y+(3y-1)x+(6y-2)=0$

If y ≠ 0 then Δ ≥ 0 for real x i.e. $B^2-4AC\ge 0$
$\Rightarrow (3y-1{)}^2-4\times 2y\times (6y-2)\ge 0$
$\Rightarrow 9y^2-6y+1-48y^2+16y\ge 0$
$\Rightarrow -39y^2+10y+1\ge 0or(13y+1)(3y-1)\le 0$

⇒ $-\frac{1}{13}\le y\le \frac{1}{3}$

if y = 0 then x = -2 which is real and this value of y is included in the above range.