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Statement I: is continuous at but is not continuous at . Statement II: The derivative of a…
Statement I: $f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$ but $f'(x)$ is not continuous at $x = 0$.
Statement II: The derivative of a continuous function need not be a continuous function.
- (a)Both (I) and (II) are correct and (II) is the correct explanation of (I).
- (b)Both (I) and (II) are correct and (II) is not the correct explanation of (I).
- (c)(I) is correct but (II) is incorrect.
- (d)(II) is correct but (I) is incorrect.
Answer
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From ISC 2025 Practice Mathematics, question 32.