MCQ
If $f(x) = |x - 3|,$ then $f$ is
- ADiscontinuous at $x = 2$
- BNot differentiable $x = 2$
- CDifferentiable at $x = 3$
- ✓Continuous but not differentiable at $x = 3$
✓
Answer
Correct option: D.
Continuous but not differentiable at $x = 3$
d
(d) $\mathop {\lim }\limits_{x \to {3^ - }} \,f(x) = \mathop {\lim }\limits_{h \to 0} f(3 - h) = \mathop {\lim }\limits_{h \to 0} \,\,|3 - h - 3|\,\, = 0$
(d) $\mathop {\lim }\limits_{x \to {3^ - }} \,f(x) = \mathop {\lim }\limits_{h \to 0} f(3 - h) = \mathop {\lim }\limits_{h \to 0} \,\,|3 - h - 3|\,\, = 0$
$\mathop {\lim }\limits_{x \to {3^ + }} \,f(x) = \mathop {\lim }\limits_{h \to 0} f(3 + h) = \mathop {\lim }\limits_{h \to 0} \,\,|3 + h - 3|\,\, = 0$
$\because \,\,\mathop {\lim }\limits_{x \to {3^ - }} \,f(x) = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = f(3)$
Hence $f$ is continuous at $x = 3$
Now $L\,f'(3) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(3 - h) - f(3)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{|3 - h - 3|\,\, - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \,\frac{h}{{ - h}} = - 1$
$R\,f'(3) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(3 + h) - f(3)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{|3 + h - 3|\,\, - 0}}{h} = 1$
$\because L\,{f}'(3)\,\ne \,R\,{f}'(3)$
Hence $f$ is not differentiable at $x = 3$.
Trick : Can be seen by graph it is continuous but tangent is not defined at $x = 3$.
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