Let $f(x)=x^2+x g'(1)+g''(2)$ and, $g(x)=x^2+x f'(2)+f''$ (3). Then,

all the above

We have,

$f(x)=x^2+x g'(1)+g''(2)$  and  $g(x)=x^2+x f'(2)+f''(3)$

$\Rightarrow f'(x)=2 x+g'(1)$ and $g'(x)=2 x+f'(2)$          ........(i)

Putting x = 1 in (i), we get

$f'(1)=2+g'(1)$ and $g'(1)=2+f'(2)$

$\Rightarrow f'(1)=4+f'(2)$

Putting  x = 2 in (i), we get

$f'(2) =4+g'(1)$ and $g'(2)=4+f'(2)$

$\Rightarrow g'(2)=4+4+g'(1)=8+g'(1)$

Differentiating (i) w.r.t. x, we get

$f''(x)=2$ and $g''(x)=2$ for all x

$\Rightarrow f''(3)=2$ and $g''(2)=2$

$\Rightarrow g''(2)+f''(3)=2+2=4$