Question

Let

$f(x) = \begin{cases} x^2 \cos \frac{1}{x}, & x < 0 \\ 0, & x = 0, \text{ then which of the following is (are) correct ?} \\ x^2 \sin \frac{1}{x}, & x > 0 \end{cases}$

• A. f(x) is continuous but not differentiable at x = 0
• B. f(x) is continuous and differentiable at x = 0
• C. f'(x) is continuous but not differentiable at x = 0
• D. f'(x) is discontinuous at x = 0

Answer 1

Answer: (B), (D)

B, D

Answer 2

1. Continuity of $f(x)$ at $x=0$: $\lim_{x \to 0^-} x^2 \cos \frac{1}{x} = 0$, $\lim_{x \to 0^+} x^2 \sin \frac{1}{x} = 0$, and $f(0) = 0$. Since all are equal, $f(x)$ is continuous at $x=0$.

2. Differentiability of $f(x)$ at $x=0$: $f'(0^-) = \lim_{h \to 0^-} h \cos \frac{1}{h} = 0$. $f'(0^+) = \lim_{h \to 0^+} h \sin \frac{1}{h} = 0$. Since LHD = RHD = 0, $f(x)$ is differentiable at $x=0$, and $f'(0)=0$.

3. Continuity of $f'(x)$ at $x=0$:

   For $x < 0$, $f'(x) = 2x \cos \frac{1}{x} + \sin \frac{1}{x}$. As $x \to 0^-$, $\lim_{x \to 0^-} \sin \frac{1}{x}$ does not exist. So $\lim_{x \to 0^-} f'(x)$ does not exist.

   For $x > 0$, $f'(x) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$. As $x \to 0^+$, $\lim_{x \to 0^+} \cos \frac{1}{x}$ does not exist. So $\lim_{x \to 0^+} f'(x)$ does not exist.

   Since the limits of $f'(x)$ as $x \to 0$ do not exist, $f'(x)$ is discontinuous at $x=0$.

Therefore, statements (B) and (D) are correct.