$\text { The number of real solution(s) of the equation } x^2+3 x+2=\min \{|x-3|,|x+2|\} \text { is : }$

• A

  1
• B

  0
• Correct answer
  C

  2
• D

  3

Solution

$x^2+3 x+2=\min \{|x-3|,|x+2|\}$
$\begin{aligned} & y=x^2+3 x+2 \\ & y=x^2+2\left(\frac{3}{2}\right) x+\frac{9}{4}-\frac{9}{4}+2 \\ & y=\left(x+\frac{3}{2}\right)^2-\frac{1}{4} \\ & y+\frac{1}{4}=\left(x+\frac{3}{2}\right)^2 \\ & \Rightarrow \text { Parabola vertex }\left(\frac{-3}{2}, \frac{-1}{4}\right) \end{aligned}$ $\Rightarrow$ By graph 2 solution possible

Hence, the correct answer is option: 'C'.