Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

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.

Options:

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.

Correct Answer:

Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for Statement-1.

Explanation:

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.