Statement-1: If $|f(x)| \leq|x|$ for all $x \in R$, then |f| is continuous at x = 0. Statement-2: If f is continuous, then |f| is continuous. |
Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; statement-2 is not a correct explanation for Statement-1. Statement-1 is True, Statement-2 is False. Statement-1 is False, Statement-2 is True. |
Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for Statement-1. |
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. |