Question

If $x\in R$, then $\frac{1}{x^2+4x+9}$ lies in the interval

• A. $[5,\infty )$
• B. $(0,\frac{1}{5}]$
• C. $(0,\infty )$
• D. $(0,\frac{1}{2}]$

Answer 1

The correct option is B $(0,\frac{1}{5}]$

Here, $x\in R$
$\Rightarrow x^2+4x+9=(x+2{)}^2+5$
$=(x+2{)}^2+5$
Now
$-\infty <x<\infty \Rightarrow -\infty <x+2<\infty \Rightarrow 0\le (x+2{)}^2<\infty \Rightarrow 5\le (x+2{)}^2+5<\infty$
$\Rightarrow 0<\frac{1}{(x+2{)}^2+5}\le \frac{1}{5}$
​​​​​​​$\therefore \frac{1}{x^2+4x+9}\in (0,\frac{1}{5}]$