- ANot continuous at $x = 2$
- BDifferentiable at $x = 2$
- ✓Continuous but not differentiable at $x = 2$
- DNone of these
$\mathop {\lim }\limits_{h \to {0^ + }} 5 - (2 + h) = 3$, $f(2) = 3$
Hence, $f$ is continuous at $x = 2$
Now $Rf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{5 - (2 + h) - 3}}{h} = - 1$
$Lf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{1 + (2 - h) - 3}}{{ - h}} = 1$
$\because Rf'(x) \ne Lf'(x)$; $f$ is not differentiable at $x = 2$.
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