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. None of these
• D. $(-\frac{1}{3},\frac{1}{‘3})$

Answer 1

The correct option is A $[-\frac{1}{13},\frac{1}{3}]$

Given: $\frac{x+2}{2x^2+3x+6}$

If the given expression be equal to $y$, then $2x^{2}y+(3y-1)x+(6y-2)=0$

If $y\ne 0$ then $△\ge 0$ for real $x$ i.e., $b^2-4ac\ge 0$
or $-39y^2+10y+1\ge 0$ or $(13y+1)(3y-1)\le 0$

$\Rightarrow -\frac{1}{13}\le y\le \frac{1}{3}$

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