Maximum absolute value

Completing the square, we see $f(x)=\left|(x-1)^2-1-t \right|$ , so that $f$ is symmetric about $x=1$ . This means the extreme values of $f$ in the interval can only occur at $x=0$ , $x=1$ or $x=3$ . This gives three conditions on $t$ :

$f(0)\leq 2 \Rightarrow |-t| \leq 2$ , ie $-2 \leq t \leq 2$

$f(1)\leq 2 \Rightarrow |-1-t| \leq 2$ , ie $-3 \leq t \leq 1$

$f(3)\leq 2 \Rightarrow |3-t| \leq 2$ , ie $1 \leq t \leq 5$

Combining the last two of these conditions, we find that the only possible value of $t$ is $t=1$ ; this also satisfies the first condition, and $f$ attains its maximum value at $f(1)=f(3)=2$ , so the answer is $\boxed1$ .