Statement I: If  $f\left(x\right)=x^2+bx+c$ and $f\left(2+t\right)=f\left(2-t\right)$ for all real numbers t, then  $f\left(1\right)<f\left(2\right)<f\left(4\right)$
Statement II: If $f\left(x\right)=x^2+bx+c$  and $f\left(2+t\right)=f\left(2-t\right)$ for all real numbers t, then  $f\left(2\right)<f\left(1\right)<f\left(4\right)$
 

 $f\left(x\right)=x^2+bx+c$, $f\left(2+t\right)=f\left(2-t\right)$
$\Rightarrow$ f is symmetric about a line x=2
 $\therefore \frac{-b}{2}=2\Rightarrow b=-4$
 $\therefore f\left(x\right)=x^2-4x+c$
Now, 3 graphs are possible.
 
In (1) and (2) ‘c’ is positive and in (3) ‘c’ is negative.
 $f\left(0\right)=c$
Let c is positive.
Now, $f\left(1\right)=c-3$
 $f\left(2\right)=c-4$
 $f\left(4\right)=c$
Say  $c=3$
then  $f\left(1\right)=0;f\left(2\right)=-1;f\left(3\right)=3\Rightarrow f\left(2\right)<f\left(1\right)<f\left(3\right)$
Again c is negative. Let  $c=-3$
 $f\left(1\right)=-6;f\left(2\right)=-7;f\left(4\right)=-3$
 $\therefore f\left(2\right)<f\left(1\right)<f\left(4\right)\Rightarrow \left(B\right)$
Also,if c=0 the statement ‘2’ is true.