Y = √(4x-x²+12)
√(3-x)
{4x-x²+12≥0
{3-x>0
4x-x²+12≥0
x² -4x -12≤0
x² -4x -12=0
D=16+48=64
x₁=4-8 = -2
2
x₂ =4+8 =6
2
+ - +
-------------- -2 -------------- 6 --------------------
\\\\\\\\\\\\
x∈[-2; 6]
3-x>0
-x>-3
x<3<br> \\\\\\\\\\\\\\\\\\\\\\\
------------- -2 -------- 3 ------------------ 6 -------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[-2; 3)
Ответ: 2)