Let a function f(x) defined on [3, 6] be given by $f(x)= \begin{cases}\log _e[x], & 3 \leq x<5 \\ \left|\log _e x\right|, & 5 \leq x<6\end{cases}$, then f(x) is

none of these

We have,

$f(x)= \begin{cases}\log _e 3, & 3 \leq x<4 \\ \log _e 4, & 4 \leq x<5 \\ \log _e x, & 5 \leq x<6\end{cases}$

Clearly, f(x) is continuous and differentiable on $[3,4) \cup(4,5) \cup(5,6)$.

At x = 4, we have

$\lim\limits_{x \rightarrow 4^{-}} f(x)=\log _e 3$  and  $\lim\limits_{x \rightarrow 4^{+}} f(x)=\log _e 4$

∴  $\lim\limits_{x \rightarrow 4^{-}} f(x) \neq \lim\limits_{x \rightarrow 4^{+}} f(x)$

Thus, f(x) is neither continuous nor differentiable at x = 4. At x = 5, we have

$\lim\limits_{x \rightarrow 5^{-}} f(x) =\log _e 4$  and  $\lim\limits_{x \rightarrow 5^{+}} f(x)=\log _e 5$

∴  $\lim\limits_{x \rightarrow 5^{-}} f(x) \neq \lim\limits_{x \rightarrow 5^{+}} f(x)$

So, f(x) is neither continuous nor differentiable at x = 5.