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Q.
Let a function $f(x)$ be defined by $f(x)=\frac{x-|x-1|}{x}$, then which of the following is not true?
Continuity and Differentiability
Solution:
We have
$f(x) =\frac{x-|x-1|}{x}=
\begin{cases}
\frac{x+x+1}{x}, & \text{$x<1, x \neq 0$ } \\[2ex]
\frac{x-(x-1)}{x}, & \text{if $x \geq 1$ }
\end{cases}$
$=
\begin{cases}
\frac{2x-1}{x}, & \text{$x<1, x \neq 0$ } \\[2ex]
\frac{1}{x}, & \text{$x \geq 1$ }
\end{cases}$
Clearly, $f(x)$ is discontinuous at $x=0$ as it is not defined at $x=0 .$ Since $f(x)$ is not defined at $x=0, f(x)$ cannot be differentiable at $x=0$. Clearly, $f(x)$ is continuous at $x=1$, but it is not differentiable at $x=1$, because $Lf'(1)$ and $Rf' (1)=1$