1
Hence, (B) is atleast correct.
Multiple Correct Answer Type
Let f(x)=\left\{\begin{matrix} x^{2}sin\frac{1}{x}\; \; x\neq0 \\ 0\; \; \; \; \; \; x=0 \end{matrix}\right.
then:
A) f and f' are continuous at x=0
B) f is derivable at x=0
C) f is derivable at x=0 and f' is not continuous at x=0
D) f is derivable at x=0 and f' is continuous at x=0
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4 Answers
1Please see:
http://www.math.umn.edu/~rogness/mathlets/xSquaredSin1overX/
66B), C)
First of all note that f(x) is continuous at x=0.
Further, we have
f'(0)=\lim_{\Delta x\to0}\dfrac{f(0+\Delta x)-f(0)}{\Delta x}=\lim_{\Delta x\to0} \Delta x\sin\left(\dfrac{1}{\Delta x}\right)=0
Hence, f is differentiable at x=0.
Next, for x≠0,
f'(x)=2x\sin\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)
which has no limit as x→0.
Hence f'(x) is not continuous at x=0.