Question

Consider the equation $2+|x^2+4x+2|=m,m\in R$ Set of all real values of $m$ so that given equation has four distinct solutions, is

• A. $(0,1)$
• B. $(1,2)$
• C. $(1,3)$
• D. $(2,4)$

Answer 1

The correct option is D $(2,4)$

$2+|x^2+4x+2|=m$
let $f\left(x\right)=|x^2+4x+2|$ & $g\left(x\right)=m-2$
from the figure:
maximum value of $f\left(x\right)$ occur at $x=-\frac{b}{2a}=-2$.
Therefore, $max\left(f\left(x\right)\right)=\left|\frac{4ac-b^2}{4a}\right|=\frac{b^2-4ac}{4a}=2$
from the figure it is clear that, $f\left(x\right)$ & $g\left(x\right)$ will have four point of intersection when $0<g\left(x\right)<max\left(f\left(x\right)\right)$
$\Rightarrow 0<m-2<2$
Therefore, $m\in (2,4)$
Ans: D