For the following two (02) items: Let the function f(x) = |x - 3| +...

The given function is $f(x) = |x - 3| + |x - 4|$ defined on $[0,5]$.

We can redefine the function by removing the absolute value signs in different intervals:

$f(x) =$\begin{cases} -(x - 3) - (x - 4), & 0 \le x < 3 \\ (x - 3) - (x - 4), & 3 \le x < 4 \\ (x - 3) + (x - 4), & 4 \le x \le 5 \end{cases}<br><br>Simplifying this, we get:<br><br>$f(x) = $\begin{cases} 7 - 2x, & 0 \le x < 3 \\ 1, & 3 \le x < 4 \\ 2x - 7, & 4 \le x \le 5 \end{cases}

Differentiating $f(x)$ with respect to $x$ in the open intervals:

$f'(x) =$\begin{cases} -2, & 0 < x < 3 \\ 0, & 3 < x < 4 \\ 2, & 4 < x < 5 \end{cases}$$<br><br>At$x = 3$:<br>Left Hand Derivative (LHD)$= -2$<br>Right Hand Derivative (RHD)$= 0$<br>Since LHD \neq RHD,$ f(x)$is not differentiable at$x = 3$.<br><br>At$x = 4$:<br>Left Hand Derivative (LHD)$= 0$<br>Right Hand Derivative (RHD)$= 2$<br>Since LHD \neq RHD,$ f(x)$is not differentiable at$x = 4$.

Therefore, neither statement I nor statement II is correct.

Answer: Neither I nor II