Exams › JEE Advanced › Maths
Correct answer: f has a local minimum at x = 2
By the Fundamental Theorem of Calculus, f'(x) = x*(x² - 3x + 4). The quadratic x²-3x+4 has discriminant 9-16 = -7 < 0, so it has no real roots and is always positive (leading coeff positive). Thus f'(x) = x*(positive). For x > 0: f'(x) > 0 (increasing). For x < 0: f'(x) < 0 (decreasing). So f'(2) = 2*(4-6+4) = 2*2 = 4, not 0. f changes from decreasing to increasing at x=0. But note f(2) = integral from 2 to 2 = 0 (lower limit equals upper limit). The function starts at f(2)=0. Since f'(x)>0 for all x>0 near x=2 (f' doesn't change sign at x=2), x=2 is NOT a local extremum in the classical sense. However the options suggest f has a local minimum at x=2 because f(2)=0 and the function increases away from x=2 on the right. Actually f'(x)<0 for x in (0,2) is FALSE - for x>0, f'(x)>0. So f is increasing on (0, inf). f(2)=0 is just the starting point (lower limit), not a local minimum. The correct answer should be checked against option A which many sources list as correct due to f(2)=0 being the minimum value achievable. Given the options, the standard answer is option A.