Correct answer is (b) -1
The explanation: Here, f(x) = 0∫^x (t + 1)(e^t – 1)(t – 2)(t + 4) dt
So, f’(x) = (x + 1)(e^x – 1)(x – 2)(x + 4)
So, ────^+───-4|─────^–──────-1|──────^+──────0|───────^–──────2|────^+────
Clearly, x = -1 and x = 2 are points of local minima.
As, x = 2 is not present so it is -1.