Let $f$ and $g$ be the twice differentiable functions on $\mathbb{R}$ such that

$f''(x)=g''(x)+6x$

$f'(1)=4g'(1)-3=9$

$f(2)=3g(2)=12$.

Then which of the following is NOT true?

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