$f''(x) = g''(x) + 6x$ $\Rightarrow f'(x) = g'(x) + 3{x^2} + C$ $f'(1) = g'(1) + 3 + C$ $\Rightarrow g = 3 + 3 + C \Rightarrow C = 3$ $\Rightarrow f'(x) = g'(x) + 3{x^2} + 3$ $\Rightarrow f(x) = g(x) + {x^2} + 3x + C'$ $x = 2$ $f(2) = g(2) + 14 + C'$ $12 = 4 + 14 + C'$ $\Rightarrow C' = - 6$ $\Rightarrow f(x) = g(2) + {x^3} + 3x - 6$ $f( - 2) = g( - 2) - 8 - 6 - 6$ $g( - 2) - f( - 2) = 20$ $f'(x) - g'(x) = 3{x^2} + 3$ $x \in ( - 1,1)$ $3{x^2} + 3 \in (0,6)$ $\Rightarrow f'(x) - g'(x) \in (0,6)$ $f(x) - g(x) = {x^3} + 3x - 6$ At $x = - 1$ $|f( - 1) - g( - 1)| = 10$ $\therefore$ Option (4) is false.