Official solution
f(x) = (3x2
+ ax -- 2 -- a)ex
\[ \therefore \] f'(x) = ex(6x + a) + (3x2
+ ax -- 2 -- a)ex
= ex(3x2 + x(6 + a) -- 2)
f '(x) = 0 at x = 1
\[ \Rightarrow \] 3 + (6 + a) -- 2 = 0
\[ \Rightarrow \] a = -7
\[ \therefore \] f'(x) = ex(3x2 -- x -- 2)
= ex(x -- 1) (3x + 2)
x = 1 is a local minima and x = \[ - {2 \over 3}\] is a local
maxima of f.