QJEEM23Mae763: JEE Main 2023 Mathematics - Class 12 JEEM23Mae763 Solution: Let and be the twice differentiable functions on such that .…

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

$\Rightarrow f'(x) = g'(x) + 3{x^2} + C$

$f'(1) = g'(1) + 3 + C$

$\Rightarrow g = 3 + 3 + C \Rightarrow C = 3$

$\Rightarrow f'(x) = g'(x) + 3{x^2} + 3$

$\Rightarrow f(x) = g(x) + {x^2} + 3x + C'$

$x = 2$

$f(2) = g(2) + 14 + C'$

$12 = 4 + 14 + C'$

$\Rightarrow C' = - 6$

$\Rightarrow f(x) = g(2) + {x^3} + 3x - 6$

$f( - 2) = g( - 2) - 8 - 6 - 6$

$g( - 2) - f( - 2) = 20$

$f'(x) - g'(x) = 3{x^2} + 3$

$x \in ( - 1,1)$

$3{x^2} + 3 \in (0,6)$

$\Rightarrow f'(x) - g'(x) \in (0,6)$

$f(x) - g(x) = {x^3} + 3x - 6$

At $x = - 1$

$|f( - 1) - g( - 1)| = 10$

$\therefore$ Option (4) is false.