Question

If $x\in [-4,-1]$, then $\frac{1}{x^2+4x+7}$ belongs to

• A. $[\frac{1}{7},\infty )$
• B. $(-\infty ,\frac{1}{7}]\cup [\frac{1}{3},\infty )$
• C. $[\frac{1}{7},1)$
• D. $[\frac{1}{7},\frac{1}{3}]$

Answer 1

The correct option is D $[\frac{1}{7},\frac{1}{3}]$

Here, $x\in [-4,-1]$
$\Rightarrow x^2+4x+7=(x+2{)}^2+3$
It will be minimum at $x=-2$ and maximum at $x=-4$
$\Rightarrow (x+2{)}^2+3\in [3,7]$
Since, $\left(3\right)\left(7\right)>0$
So, $\frac{1}{x^2+4x+7}\in [\frac{1}{7},\frac{1}{3}]$