$f(x)=x^3+p x^2+q x+10$ has a maximum at x = -3 and a minimum at x = 1. The values of p and q are:

p = 3, q = -9

The correct answer is Option (4) → $p = 3, q = -9$

$f(x)=x^3+p x^2+q x+10$

It has maximum at $x=-3$ and a minimum at $x=1$

To find critical points,

$f'(x)=0$

$⇒3x^2+2px+q=0$

and, $f'(-3)=0$

$⇒3(-3)^2+2p(-3)+q=0$

$⇒6p-q=27$   ...(1)

$f'(1)=0$

$⇒3(1)+2p+q=0$

$⇒2p+q=-3$   ...(2)

Solving both (1) and (2), we get

$p=3$ and $q=-9$