Question:
Let \(f,g:\mathbb{R}\to\mathbb{R}\) be two functions defined by \(f(x)=\begin{cases}x\sin\left(\frac{1}{x}\right), & x\ne0\\ 0, & x=0\end{cases}\) and \(g(x)=xf(x)\).
Statement I: \(f\) is a continuous function at \(x=0\).
Statement II: \(g\) is a differentiable function at \(x=0\).
Let \(f,g:\mathbb{R}\to\mathbb{R}\) be two functions defined by \(f(x)=\begin{cases}x\sin\left(\frac{1}{x}\right), & x\ne0\\ 0, & x=0\end{cases}\) and \(g(x)=xf(x)\).
Statement I: \(f\) is a continuous function at \(x=0\).
Statement II: \(g\) is a differentiable function at \(x=0\).
Statement I: \(f\) is a continuous function at \(x=0\).
Statement II: \(g\) is a differentiable function at \(x=0\).
Show Hint
Use \(|\sin(1/x)|\le1\) with the squeeze theorem for \(f\), and the definition of the derivative for \(g(x)=x^2\sin(1/x)\) at 0.
Updated On: Nov 11, 2026
- Both statement I and II are false.
- Both statement I and II are true.
- Statement I is true, statement II is false.
- Statement I is false, statement II is true.
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The Correct Option is B
Solution and Explanation
Statement I: \(|x\sin\frac{1}{x}|\le|x|\to0\) as \(x\to0\), so \(\lim_{x\to0}f(x)=0=f(0)\). \(f\) is continuous at 0: true.
Statement II: \(g(x)=x^2\sin\frac{1}{x}\) for \(x\ne0\), \(g(0)=0\).
\(g'(0)=\lim_{h\to0}\frac{g(h)-g(0)}{h}=\lim_{h\to0}h\sin\frac{1}{h}=0\) (squeeze). The limit exists, so \(g\) is differentiable at 0: true.
Both statements are true.
Answer: (2) Both statement I and II are true.
Statement II: \(g(x)=x^2\sin\frac{1}{x}\) for \(x\ne0\), \(g(0)=0\).
\(g'(0)=\lim_{h\to0}\frac{g(h)-g(0)}{h}=\lim_{h\to0}h\sin\frac{1}{h}=0\) (squeeze). The limit exists, so \(g\) is differentiable at 0: true.
Both statements are true.
Answer: (2) Both statement I and II are true.
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