It seems that the title is misspelled. If it is "if the function y = x 3+ax+1monotonically decreases in the interval [-3, -2], the range of a is ()", then the answer is d.
If y = x 3+ax+1,then y' = 3x 2+a; If a≥0, then y' is not less than 0, and the function increases in the definition domain.
Therefore, if a < 0, let y'=0 and x = √-a/3, so when x∈(-∞, -√- a/3]∨[√- a/3,+∞) and y'≥0, y will increase monotonically;
When x∈[-√-a/3, √-a/3], y'≤0, and y monotonically decreases.
Therefore, if the function monotonically decreases in the interval [-3, -2], then [-3, -2] is contained in [-√-a/3, √-a/3], so -√-a/3≤-3 and √-a/3≥-2, the solution is a √.