Statement-1: If |f(x)| ≤|x| for all x ∈ R, then |f| is continuous at…

Let f be continuous at x = a. Then, for every $\in>0$, there exists $\delta>0$ such that

|f(x) - f(a)| < ∈ whenever $|x-a|<\delta$

⇒ ||f(x)| - |f(a)|| < |f(x) - f(a)| < ∈ whenever $|x-a|<\delta$

⇒ |f|(x) - |f|(a)| < ∈ whenever $|x-a|<\delta$

⇒ |f| is continuous at x = a

So, statement - 2 is true.

Now, $|f(x)| \leq|x|$

$\Rightarrow|f(0)| \leq 0$             [Replacing x by 0 ]

$\Rightarrow f(0)=0$

∴  $|f(x)| \leq|x|$

$\Rightarrow|f(x)-f(0)|<|x-0|$                 [∵ f(0) = 0]

$\Rightarrow|f(x)-f(0)|<\epsilon$ whenever $|x-0|<\delta(=\epsilon)$

$\Rightarrow f(x)$ is continuous at x = 0

$\Rightarrow|f(x)|$ is continuous at x = 0 [Using statement - 2]

Hence both the statements are true and statement - 2 is a correct explanation for statement - 1.