JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f(x)=\left\{\begin{array}{cl}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{array} ;\right.\) Then at \(x=0\)
- A \(f\) is continuous but not differentiable
- B \(f\) is continuous but \(f ^{\prime}\) is not continuous
- C \(f\) and \(f\) ' both are continuous
- D \(f ^{\prime}\) is continuous but not differentiable
Answer & Solution
Correct Answer
(B) \(f\) is continuous but \(f ^{\prime}\) is not continuous
Step-by-step Solution
Detailed explanation
\(\text { Continuity of } f(x): f\left(0^{+}\right)=h^2 \cdot \sin \frac{1}{h}=0\) \(f\left(0^{-}\right)=(-h)^2 \cdot \sin \left(\frac{-1}{h}\right)=0\) \(f(0)=0\) \(f(x)\) is continuous…
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