If f(x) = l , , , , , , , e^x ; , , , ,x 0 |1 - x|; , ,x > 0 ., t… - Vidyadip

MCQ

If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,{e^x};\,\,\,\,x \le 0\\|1 - x|;\,\,x > 0\end{array} \right.$, then

A
$f(x)$ is continuous at $x = 1$
B
$f(x)$ is continuous at $x = 0$
✓
$(a)$ and $(b)$ both
D
None of these

✓

Answer

Correct option: C.

$(a)$ and $(b)$ both

c
(c) $f(x) = \left\{ \begin{array}{l}{e^x}\,\,;\,\,\,x \le 0\\1 - x;\,\,0 < x \le 1\\x - 1\,;\,\,x > 1\end{array} \right.$

$Rf'(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - h - 1}}{h} = - 1$

$Lf'(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - h}} - 1}}{{ - h}} = 1$

So, it is not differentiable at $x = 0$.

Similarly, it is not differentiable at $x = 1$.

But it is continous at $x = 0$, $1$.

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