The function f(x) = x^2 , , 1 x , ,x ,0, , ,f(0) , = 0 at x = 0 - Vidyadip

MCQ

The function $f(x) = {x^2}\,\,\sin \frac{1}{x},\,x \ne \,0,\,\,f(0)\, = 0$ at $x = 0$

A
Is continuous but not differentiable
B
Is discontinuous
C
Is having continuous derivative
✓
Is continuous and differentiable

✓

Answer

Correct option: D.

Is continuous and differentiable

(d)
$\mathop {\lim }\limits_{x \to 0} f(x) = {x^2}\sin \left( {\frac{1}{x}} \right)$, but $ - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1$ and $x \to 0$
$\therefore $ $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 0 = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0)$
Therefore $f(x)$ is continuous at $x = 0$.
Also, the function $f(x) = {x^2}\sin \frac{1}{x}$ is differentiable because
$Rf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin \frac{1}{h} - 0}}{h} = 0$,
$Lf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin (1/ - h)}}{{ - h}} = 0$.

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