MCQ
The function $f(x)\, = \,|x| + |x - 1|$ is
- ✓Continuous at $x = 1,$ but not differentiable at $x = 1$
- BBoth continuous and differentiable at $x = 1$
- CNot continuous at $x = 1$
- DNot differentiable at $x = 1$
$ = \left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}&{}\\{x - x + 1,}&{0 \le x < 1}& = \\{x + x - 1,}&{x \ge 1}&{}\end{array}} \right.\left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}\\1&{0 \le x < 1}\\{2x - 1,}&{x \ge 1}\end{array}} \right.$
Clearly, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$
and $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1$.
So, $f(x)$ is continuous at $x = 0,\,\,1.$
Now $f'(x) = \left\{ {\begin{array}{*{20}{l}}{ - 2,\,\,\,\,x < 0}\\{\,\,0,\,\,\,\,\,0 \le x < 1}\\{\,\,2,\,\,\,\,\,x \ge 1}\end{array}} \right.$
Here $x = 0$, $f'({0^ + }) = 0$ while $f'({0^ - }) = - 2$
and at $x = 1$, $f'({1^ + }) = 2$ while $f'({1^ - }) = 0$
Thus, $f(x)$ is not differentiable at $x = 0$ and $1.$
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