Answer
(c)Converse need not be true.
\[f(x)=\left\{\begin{array}{cc}1\;\;\;\;,x\geq 0\\
-1\;\;\;\;,x<0\end{array}\right.\]
Work Step by Step
(a) \[F(x)=|x|\]
\[|x|=\left\{\begin{array}{cc}x\;\;\;\;,x\geq 0\\
-x\;\;\;\;\;,x< 0\end{array}\right.\]
Clearly $|x|$ is continuous if $x\neq 0$ so we will check the continuity at $x=0$.
\[\lim_{x\rightarrow 0^{-}}|x|=\lim_{x\rightarrow 0}(-x)=0\]
\[\lim_{x\rightarrow 0^{+}}|x|=\lim_{x\rightarrow 0^{+}(x)=0}\]
\[\lim_{x\rightarrow 0^{-}}=\lim_{x\rightarrow 0^{+}}=|0|\]
$\Rightarrow |x|$ is continuous at $x=0$
Therefore $|x|$ is always continuous.
(b) Let $f(x)$ is continuous on the interval $[a,b]$
Let $c\in [a,b]$
$\Rightarrow f$ is continuous at $c$.
\[\lim_{x\rightarrow c^{-}}|f(x)|=|\lim_{x\rightarrow c}f(x)|=f(c)\]
$\Rightarrow |f|$ is continuous at $c$
Since $c$ is arbitrary therefore $|f|$ is continuous on $[a,b]$
(c) Converse need not be true.
Let \[f(x)=\left\{\begin{array}{cc}1\;\;\;\;\;\;\,x>0\\
-1\;\;\;\;\;\;\;\;\;, x\leq 0\end{array}\right.\]
\[|f(x)|=1\;\;\;\;\;,x\in (-\infty,\infty)\]
Clearly $|f(x)|$ is continuous at $x=0$ but $f$ is not continuous at $x=0$
