Exams › JEE Advanced › Maths
Correct answer: 1
f'(x) = c*e^(-x)*(1-x) - x + 1. For x <= 0, we need f'(x) <= 0 for all x in (-inf, 0]. At x = 0: f'(0) = c*1*(1) - 0 + 1 = c + 1 <= 0 => c <= -1. For x < 0: -x > 0, so -x + 1 > 1 > 0. Also (1-x) > 0 for x <= 0. So c*e^(-x)*(1-x) <= x - 1 for all x <= 0 means c <= (x-1)/(e^(-x)*(1-x)) = -(1-x)/(e^(-x)*(1-x)) = -e^x. For x <= 0, -e^x ranges: at x=0 gives -1; as x->-inf, -e^x -> 0⁻. So inf{-e^x for x <= 0} = -1. Thus c <= -1. The minimum value of c satisfying c <= -1 is c = -1. Then c² = 1. But checking at x = 0: f'(0) = c + 1 = -1 + 1 = 0 (not strictly negative, just 0, which is okay for non-increasing). So c can be -1 and c² = 1. But this should be the LEAST value of c², meaning smallest c². For c <= -1, c² >= 1. Minimum c² = 1. However, we must verify: for c = -1, is f'(x) <= 0 for all x <= 0? f'(x) = -e^(-x)*(1-x) - x + 1 = -(1-x)*e^(-x) + (1-x) = (1-x)(1 - e^(-x)). For x < 0: (1-x) > 0; (1 - e^(-x)) where x < 0 means -x > 0, e^(-x) > 1, so 1 - e^(-x) < 0. Thus (1-x)(1-e^(-x)) < 0 for x < 0. At x = 0: (1)(1-1) = 0. So f'(x) <= 0 for all x <= 0. c = -1 works. Minimum c² = 1.