Question

Assertion (A): $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is continuous at $x = 0$.
Reason (R): When $x \to 0$, $\sin \frac{1}{x}$ is a finite value between -1 and 1.

• A. Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

For piecewise functions like this, use the Squeeze Theorem when checking continuity at points where the function is defined piecewise.

Answer 1

The Correct Option is A

The function $f(x)$ is given as: \[ f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} \] To check continuity at $x = 0$, we need to check if $\lim_{x \to 0} f(x) = f(0)$. We know that $\sin \frac{1}{x}$ oscillates between -1 and 1 as $x$ approaches 0. Hence: \[ -1 \leq \sin \frac{1}{x} \leq 1 \] Multiplying by $x$, we get: \[ -x \leq x \sin \frac{1}{x} \leq x \] As $x \to 0$, both $-x$ and $x$ approach 0. By the Squeeze Theorem, we conclude that: \[ \lim_{x \to 0} x \sin \frac{1}{x} = 0 \] Thus, $f(x)$ is continuous at $x = 0$. The assertion is correct, and the reason is also correct. Therefore, both the assertion and reason are true.