JEE Main 1999 Mathematics question on Limits Continuity and Differentiability

The only potential non-differentiability points of |x²-3x+2| = |(x-1)(x-2)| are x=1 and x=2 (where the expression inside is 0). At x=1: the factor (x²-1) = (x-1)(x+1) already has a zero, so (x²-1)|x²-3x+2| = (x-1)(x+1)|(x-1)(x-2)|. Near x=1, this equals (x-1)(x+1)|x-1|·|x-2|, which contains |x-1|·(x-1) = (x-1)|x-1|, a differentiable function (its derivative exists as 0 at x=1). So x=1 is fine. At x=2: (x²-1) = 3 ≠ 0, so the factor doesn't vanish to rescue the absolute value. We have f(x) = 3|x²-3x+2| + cos(x) near x=2, and the left/right derivatives of |x²-3x+2| at x=2 differ: (d/dx)(x²-3x+2) at x=2 is 2(2)-3=1, so the modulus creates a corner with derivative magnitude 1 but opposite signs. Thus f'(2⁻) = 3(-1)+0 = -3 and f'(2⁺) = 3(1)+0 = 3, which are unequal. Also, cos(|x|) = cos(x) for x near 0 and 2, contributing -sin(2) from both