Lf' (-1) = -3 and Rf' (-1) = 3 ⇒ f' (-1) does not exist At x = - 1 , LHL = RHL = 0 and f (-1) = 0 ∴ f is contiuous at x = - 1 Lf'(0) = - 1 and Rf' (0) = 1 ⇒ f'(0) does not exist Rf'(1) = [(x + 1) 2 + (2x - 1) (1) at x = 1] = 5 (B) (C) (D) is also correct