If $(x)=x^3+a x^2+b x$ has a maximum at x = -2 and minimum at x = 1, then (a, b) is:

$\left(\frac{3}{2},-6\right)$

The correct answer is Option (4) → $\left(\frac{3}{2},-6\right)$

$f(x)=x^3+ax^2+bx$

$f'(x)=3x^2+2ax+b$

$\text{Extrema at } x=-2,1 \Rightarrow f'(x)=k(x+2)(x-1)$

$\text{Leading coefficient } 3 \Rightarrow k=3$

$f'(x)=3(x+2)(x-1)=3(x^2+x-2)=3x^2+3x-6$

$3x^2+2ax+b=3x^2+3x-6$

$2a=3 \Rightarrow a=\frac{3}{2}$

$b=-6$

$(a,b)=\left(\frac{3}{2},-6\right)$